Categories and Subject Descriptors H.2.8 [Database Management] Database Applications- Data

Categories and Subject Descriptors H.2.8 [Database Management] Database Applications- Data
Categories and Subject Descriptors H.2.8 [Database Management] Database Applications- Data

CARPENTER:Finding Closed Patterns in Long Biological

Datasets

Feng Pan,Gao Cong, Anthony K.H.Tung??Natl.University of Singapore {panfeng,conggao,atung}

@https://www.360docs.net/doc/d63698514.html,.sg

Jiong Y ang

University of Illinois,Urbana

Champaign

jioyang@https://www.360docs.net/doc/d63698514.html,

Mohammed J.Zaki?

Rensselaer Polytechnic

Institute

zaki@https://www.360docs.net/doc/d63698514.html,

Categories and Subject Descriptors

H.2.8[Database Management]:Database Applications-Data Mining

Keywords

frequent pattern,closed pattern,row enumeration

ABSTRACT

The growth of bioinformatics has resulted in datasets with new characteristics.These datasets typically contain a large number of columns and a small number of rows.For ex-ample,many gene expression datasets may contain10,000-100,000columns but only100-1000rows.

Such datasets pose a great challenge for existing(closed) frequent pattern discovery algorithms,since they have an exponential dependence on the average row length.In this paper,we describe a new algorithm called CARPENTER that is specially designed to handle datasets having a large number of attributes and relatively small number of rows. Several experiments on real bioinformatics datasets show that CARPENTER is orders of magnitude better than pre-vious closed pattern mining algorithms like CLOSET and CHARM.

1.INTRODUCTION

The growth of bioinformatics has resulted in datasets with new characteristics.These datasets typically contain a large number of columns and a small number of rows.For ex-ample,many gene expression datasets may contain10,000-100,000columns or items but usually have only100-1000 rows or transactions.Such datasets pose a great challenge Contact Author

?This work was supported in part by NUS ARF grant R252-000-121-112and R252-000-142-112.

?This work was supported in part by NSF CAREER Award IIS-0092978,DOE Career Award DE-FG02-02ER25538,and NSF grant EIA-0103708.

Permission to make digital or hard copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for pro£t or commercial advantage and that copies bear this notice and the full citation on the£rst page.To copy otherwise,to republish,to post on servers or to redistribute to lists,requires prior speci£c permission and/or a fee.

SIGKDD’03,August24-27,2003,Washington,DC,USA

Copyright2003ACM1-58113-737-0/03/0008...$5.00.for existing frequent pattern discovery algorithms.While there are a large number of algorithms that had been de-veloped for frequent pattern mining[1,5,11],their run-ning time increases exponentially with increasing average row length,thus such high-dimensional data renders most current algorithms impractical.This also holds for extant methods for mining closed patterns[6,2,7,10].

Previous(closed)frequent pattern mining methods work well for datasets with small average row length,since if i is the maximum row size,there could be2i potential frequent itemsets;usually i<100.However,for the gene expresion datasets in bioinformatics domain i can be in the range of tens of thousands.As a result,search over the itemset space is impractical.Since these datasets have a small number of rows(say m)usually in the range of a hundred or a thousand (i.e.,m i),it appears reasonable to design an algorithm that searches the row set space instead of the usual itemset space.

In this paper,we describe a new algorithm called CAR-PENTER1,that is specially designed to handle datasets having a large number of items and relatively small number of rows.CARPENTER is a novel algorithm which discov-ers frequent closed patterns by performing depth-?rst row-wise enumeration instead of the usual itemset enumeration, combined with e?cient search pruning techniques,to yield a highly optimized algorithm.Our experiments show that this unconventional approach produces good results when min-ing long biological datasets and outperforms current meth-ods like CHARM[10]and CLOSET[7]by more than an order of magnitude.

2.PRELIMINARIES

Let F={f1,f2,..,f m}be a set of items,also called fea-tures.Our dataset D consists of a set of rows R={r1,..,r n}, where each row r i is a set of features,i.e.,r i?F.Figure 1(a)shows an example dataset in which the features are rep-resented using alphabets a through t.There are altogether 5rows,r1,...,r5.The?rst row r1contains the feature set {a,b,c,l,o,s}.For convenience,in the sequel,we drop set notation and denote a set of features{a,c,f}as acf,and we denote a row set{r2,r3,r5}as235.

Given a set of features F ?F,we de?ne the feature support set,denoted R(F )?R,as the maximal set of rows that contain F .Likewise,given a set of rows R ?R, we de?ne the row support set,denoted F(R )?F,as the maximal set of features common to all the rows in R . 1CARPENTER stands for Closed Pattern Discovery by Transposing Tables that are Extremely Long;the“ar”in the name is gratuitous.

i r i

1a,b,c,l,o,s

2a,d,e,h,p,l,r 3a,c,e,h,o,q,t 4a,e,f,h,p,r

5b,d,f,g,l,q,s,t f j R(f j)

a1,2,3,4

b1,5

c1,3

d2,5

e2,3,4

f4,5

g5

h2,3,4

l1,2,5

o1,3

p2,4

q3,5

r2,4

s1,5

t3,5

Transposed Table,T T

Figure1:Running Example

f j R(f j)

a4

e4

h4

Figure2:T T|{2,3}

As an exmaple consider Figure1(a).Let F =aeh,then R(F )=234since these are all the rows that contain F . Also let R =23,then F(R )=aeh since it is the maximal set of features common to both r2and r3.

Given a set of features F ,the number of rows in the dataset that contain F is called the support of F .By de?nition,the support of F is given as|R(F )|.A set of features F ?F is called a closed pattern if there exists no F such that F ?F and|R(F )|=|R(F )|,i.e.,there is no superset of F with the same support.Put another way, the row set that contains superset F must not be exactly the same as the row set of F .A feature set F is called a fre-quent closed pattern,if it is i)closed,ii)|R(F )|≥minsup, where minsup is a user speci?ed lower support threshold. For example,given minsup=2,the feature set aeh is a frequent closed pattern in Figure1(a)since it occurs three times.ae,on the other hand,is not a frequent closed pat-tern,since it is not closed(|R(aeh)|=|R(ae)|),although its support is more than minsup.

Given a dataset D which contains records that are subset of a set of features F,our problem is to discover all fre-quent closed patterns with respect to a user support thresh-old minsup.In addition we assume that the dataset satis?es the condition|R| |F|.

3.THE CARPENTER ALGORITHM

To illustrate CARPENTER,we will use the tables in Figure1as a running example.Table1(b)is a transposed version of Table1(a),denoted T T.In T T,each tuple lists a feature,along with the row ids where that feature occurs in the original table.As an example15is the set of rows that contain feature b,which produces the

second tuple in T T. In the sequel we always refer to entries of the transposed table as tuples,and entries of the original table as rows. Unlike existing algorithms which perform their search by enumeration of feature sets[6,7],CARPENTER performs search by enumeration of row sets.Figure3illustrates the

15

{l}

12

1

{}

{}

{}

{}

{}

345

12345

34

35

45

3

2345

5

4

2

{dl}

25

24

23

{}

145

14

{}

Figure3:The Row Enumeration Tree.

complete row set enumeration tree without application of any pruning strategies.Each node in the tree represents a row set R ;also shown is F(R ).For example,the node 12represents the row set{r1,r2},along with it supporting feature set F(12)=al.

To?nd frequent closed patterns,CARPENTER performs a depth?rst search(DFS)of the row set enumeration tree. By imposing a total order,such as lexicographic order on the row sets,we are able to perform a systematic search for closed patterns.For example,DFS on the row enumeration tree in Figure3will be{1,12,123,1234,12345,1235, (45)

5}(in absence of any optimization and pruning strategies). Lemma 3.1.Let F be a closed pattern and R(F)be the set of rows that contain F,then R(F)is unique.In other words, there does not exist a closed pattern F ,F =F,that satis-?es R(F)=R(F ).

Proof:We prove by contradiction.Assume there exists a closed pattern F that satis?es R(F)=R(F )but F =F. Let pattern CF=F ∪F.Then R(CF)=R(F)=R(F ). Since F ?CF,this contradicts the fact that F is closed.

2 By Lemma3.1,each closed pattern corresponds to a unique set of rows.By enumerating all combinations of rows as shown in Figure3,we ensure that all closed patterns in the datasets are enumerated.It is obvious that a complete traversal of the row enumeration tree is not e?cient and pruning techniques must be introduced to prune o?unnec-essary searches.

Let X be a subset of rows.Given the transposed table T T, a X-conditional transposed table,denoted as T T|X,is a subset of tuples from T T such that:1)For each tuple x in T T,there exist a corresponding tuple x in T T|X.2)x contains all rows in x with row ids larger than any row in X.As an example,let the transposed table in Figure1(b) be T T and let X=23.The X-conditional transposed table, T T|X is as shown in Figure2.

Our formal algorithm is shown in Figure4.For clarity,we assume that the database is already transposed with infre-quent features removed.This pre-processing step is trivial

Algorithm CARPENTER

Input:Transposed table T T,features set F and support level minsup

Output:Complete set of frequent closed patterns,F CP Method:

1.Initialization.F CP=?.Let R be the(numerically sorted)

set of rows in the original table;

2.Mine Frequent Closed Pattern.MinePattern(T T|?,R,

F CP);

Subroutine:MinePattern(T T |X,R ,F CP).

Parameters:

?T T |X:A X-conditional transposed table;

?R :A subset of rows which have not been considered in the enumeration;

?F CP:The set of frequent closed patterns that have been found;

Method:

1.Scan T T |X and count the frequency of occurrences for each

row,r i∈R .Y=?.

2.Pruning1:Let U?R be the set of rows in R which

occur in at least one tuple of T T |X.If|U|+|X|≤minsup, then return;else R =U;

3.Pruning2:Let Y be the set of rows which are found

in every tuple of the X-conditional transposed table.Let R =R ?Y and remove all rows of Y from T T |X;

4.Pruning3:If F(X)∈F CP,then return;

5.If|X|+|Y|≥minsup,add F(X)into F CP;

6.For each r i∈R ,

R =R ?{r i}

MinePattern(T T |X|r

i

,R ,F CP);

Figure4:The CARPENTER Algorithm

and take negligible time since our datasets usually?t in main memory.CARPENTER does recursive generation of conditional transposed tables,performing a depth-?rst traversal of the row enumeration tree.Each computed con-ditional table represents a node in the enumeration tree of Figure3.For example,the23-conditional table represents the node23.After initializing F CP,the set of frequent closed pattern,to be empty and letting R to be the set of rows in the original table,CARPENTER calls the sub-routine MineP attern to recursively generate X-conditional tables.

The subroutine MineP attern takes in three parameters T T |X,R and F CP.T T X is a X-conditional table.R contains the set of rows that will be used to enumerate the next level of conditional transposed table while F CP con-tains the frequent closed patterns that have been discovered so far.

Steps1to4in the subroutine perform the counting and pruning.They are extremely important for e?ciency of CARPENTER.Before we explain these4steps,we will ?rst show that the MineP attern subroutine will only out-put a pattern if and only if it is a frequent closed patterns(in the absence of these4steps).This is done at Step5which checks whether F(X)is a frequent closed pattern before in-serting F(X)into F CP,and at Step6which continues the next level of enumeration in the search tree.We prove the correctness of the two steps by two lemmas as follows: Lemma 3.2.Let X be a subset of rows from the original table,then F(X)must be a closed pattern(not necessarily frequent).

Proof:We will prove by contradiction.Assuming F(X)is not a closed pattern,then there exists a feature f i such that R(F(X))=R(F(X)∪f i).Since X contains all features of F(X),then X?R(F(X)).This means that f i belongs to in every row in X,which contradicts the de?nition that F(X)is the maximal set of features common to rows of X.

2 Lemma3.2ensures that Step5only inserts closed patterns that are frequent into F CP.The main observation used in the proof is that F(X)cannot be a maximal feature set that is common in all rows of X unless it is a closed pattern.A check on|X|+|Y|is needed to ensure that the minimum support threshold is satis?ed(Note that Y is an empty set if the Steps1to4of MineP attern are not executed).To-gether with Lemma3.1,we know that the complete and correct set of frequent closed patterns will be in F CP.

Lemma 3.3.T T |X|r

i

=T T |X∪r

i2 Lemma3.3is useful for explaining Step6.It simply states that a X∪r i conditional transposed table can be computed from a X conditional transposed table,T T |X,by selecting those tuples that contain r i in T T |X.This is utilized in Step6where a recursive call on MineP attern is called with T T |X|r

i

as the conditional transposed table.This is in fact generating the X+r i conditional transposed table that is needed to represent the next level of row set enumeration. Note that Step6implicitly represents a form of pruning too since it is possible to have R =?.It can be observed from the enumeration tree that there exist some combina-tions of rows,X,such that F(X)=?(an example is node “134”).This implies that there is no feature which exists in all the rows in X.When this occurs,R will be empty and no further enumeration will be performed.

We next look at the pruning techniques that are used in CARPENTER to enhance its e?ciency.Our emphasis here is to show that our pruning steps do not prune o?any fre-quent closed patterns,while preventing unnecessary traver-sal of the enumeration tree.This guarantees correctness. The?rst pruning step is executed in Step2of MineP attern. The pruning is essentially aimed at removing search branches which can never yield closed patterns that satisfy the minsup threshold.The following lemma is applied in the pruning. Lemma 3.4.Let T T |X be a X conditional transposed table. Let U be a set of rows which occur in at least one tuple of T T |X.If|U|+|X|

Lemma 3.5.Let T T |X be a X conditional transposed table and Y be a set of rows which occur in every tuple of T T |X. Given any subset R ?R,we have F(X∪R )=F(X∪Y∪R ).

Proof:By de?nition,F(X∪R )contains a set of features which occur in every row of X∪R .Since the rows in Y occur

in every tuple of T T |X ,this means that these rows also occur in every tuples of T T |(X ∪R )(Note:T T |(X ∪R )?T T |X ).Thus,the set of tuples in T T |(X ∪R )is exactly the set of tuples in T T |(X ∪R ∪Y ).From this,we can conclude that F (X ∪R )=F (X ∪Y ∪R ).2As an example to illustrate Lemma 3.5,let us consider the 23conditional transposed table in Figure 2.Since row 4occurs in every tuple of T T |23,we can conclude that F (23)=F (234)=aeh .Thus,we need not create T T |234in our search and row 4need not be considered for further enumeration down that branch of the enumeration tree.Our ?nal and most complex pruning strategy is shown in Step 4of MineP attern .This step will prune o?any further search down the branch of node X if it is found that F (X )was already discovered previously in the enumeration tree.The inituitive reasoning which we will prove later is as follows:the set of closed patterns that will be enumerated from the descendants of node X must have been enumerated previously.Unlike feature-based mining algorithms,such as CHARM and CLOSET,we need not perform detection of superset-subset relationship among the patterns since Lemma 3.2al-ready shows that only closed patterns will be enumerated in our search tree.For example,in Figure 3,it is not possible for the pattern {a,c }to be enumerated although both {a }and {a,c,o }are closed patterns with support of 80%and 40%respectively.This is unlike CHARM and CLOSET,both of which will enumerate {a,c }and check that it has the same support as a superset {a,c,o }before discarding it as a non-closed pattern.Another important thing to note here is that the correct-ness of the third pruning strategy (Step 3)is dependent on the second pruning criteria .This is essential because of the following lemma.

Lemma 3.6.Let X be the set of rows in the current search node and X be the set of rows that result in F (X )(which is the same as F(X’))being inserted into F CP in earlier enu-meration.If pruning strategy 2is applied consistently in the algorithm,then the node representing X in the enumeration tree will not be the descendent of the node representing X in the enumeration tree.

Proof:Assume otherwise,then X ?X .Let Z =X ?X .Since F (X )=F (X ),all rows in Z must be contained in all tuples of the X conditional transposed table.Based on pruning strategy 2,the rows in Z would be added to X and will be removed from subsequent transposed table down that search branch.Thus the node representing X will not be visited,which contradicts the fact that node X is currently being processed in the enumeration tree.2Consider the node 23in Figure 3.As mentioned earlier,its descendant node 234will not be visited since row 4occurs in every tuple of 23-conditional transposed table.Without pruning strategy 2,this will not be the case.

We next try to prove that all branches from a node X in the enumeration tree can be pruned o?if F (X )is already in F CP .We have the following lemma.

Lemma 3.7.Let T T |X be the conditional transposed table in the current search node.Let X be the set of rows which result in F (X )(which equals to F (X ))being inserted into F CP in earlier enumeration.Let x f i and x f i be the two

tuples that represent feature f i in T T |X and T T |X respec-

tively.We will have x f i ?x

f i for all f i ∈F (X ).

Proof:We know that F (X )=F (X )which implies that the set of features represented by tuples in both the conditional transposed tables are the same.Let the maximal set of rows that contains the feature set F (X )be R max ={r 1,...,r n }which is sorted in numerical order.Let X ={r 1,...,r m }be the ?rst row set that causes F (X )to be inserted into F CP and X ={r 1,...,r m }is also sorted in numerical order.Based on Lemma 3.6,X cannot be a descendent of X in the enumeration tree.Thus,X must be of the form (X ?A )∪B where A ?X and B ?R max ?X ,A =?,B =?.By lexicographic row set search,we can conclude from here that there exists a row r i such that i >m and r i ∈X .By de?nition of a conditional transposed table,we know that all rows which occur before r m will be removed in T T |X .Likewise,all rows occurring before r i will be removed in T T X .Since i >m ,a tuple x f i representing feature f i in T T |X will have less rows being removed than the corre-sponding tuple x f i representing feature f i in T T |X .Hence the proof.2In a less formal term,Lemma 3.7shows that if X is the ?rst combination of rows that cause F (X )to be inserted into F CP ,then the conditional transposed table T T |X will be more “general”than any other conditional transposed table T T |X in which F (X )=F (X ).“General”in this case,refers to the fact that each tuple in T T |X is in fact a superset of the corresponding tuple in T T |X .We will now formalize our third pruning strategy as a theorem.Theorem 3.1.Given a node representing a set of rows X in the enumeration tree,if F (X )is already in F CP ,then all enumeration down that node can be pruned o?.Proof Let X be the combination of rows that ?rst cause F (X )to be inserted into F CP .From Lemma 3.7,we know that any tuple x f i in the X conditional table will be a super-set of a corresponding tuple x f i in the X conditional table and X conditional table has the same number of tuples with X conditional table.Since we know that the next level of search at node X in the enumeration tree is based on the set of rows in the X conditional transposed table,it is easy to conclude that the possible enumeration at the node X is a subset of the possible enumeration at node X .Since X had been visited,it is thus not necessary to perform any enumeration from the node X onwards.2Consider the node 23in Figure 3which is the ?rst node that results in the insertion of aeh based on the enumeration order.Thus,R min =23.Next look at node 34.We have F (34)=F (23)=aeh .It can be seen that node 34is not a descendant of node 23and that 34satis?es the formula (R min ?A )∪B ,A =2,B =4.In this case,pruning strategy 3can be applied and no further enumeration will be done from node 34.CARPENTER is implemented by adopting the in-memory,pointer-based approach in BUC [3]2.With the in-memory pointer,CARPENTER does not construct conditional trans-posed table physically,thus saving space.Due to space lim-itation,we will not give details on the implementation of CARPENTER .2We note also that there are other alternatives for imple-mentation including building a FP-tree [4]on the trans-posed table and adopting the vertical data representation in the row-wise manner [10].However,our central theme of

row enumeration is independant of these techniques and we leave it to interested readers to explore them during their

own implementation.

4.PERFORMANCE STUDIES

In this section,we compare the performance of CARPEN-TER against other algorithms.All experiments were per-formed on a PC with a Pentium III1.4Ghz CPU,1GB RAM and a80GB hard-disk.Algorithms were coded in Standard C.

Algorithms:We compare CARPENTER against two other closed pattern discovery algorithms,CHARM[10]and CLOSET [7]Experiments in[7,10]have shown that depth-?rst mining algorithms like CHARM and CLOSET are substantially bet-ter than levelwise mining algorithms like Close[6]and Pascal [2].To make a fair comparison,CHARM and CLOSET are also run in the main memory after one disk scan is done to load the datasets.The run time for CARPENTER includes the time for transposing the datasets.

Datasets:We choose3real-life gene/protein expression datasets to analyze the performance of CARPENTER.The Lung Cancer(LC)dataset3is a gene expression dataset. The rows in the dataset represent sample tissues and these tissues can come from either malignant pleural mesothe-lioma(MPM)or adenocarcinoma(ADCA)of the lung.There are181tissue samples and each sample is described by the activity level of12533genes or features.The Acute Lym-phoblastic Leukemia(ALL)dataset4is also a gene expression dataset containing tissues from cancerous/non-cancerous cells.There are215tissue samples described by activity level of12533genes.The Ovarian Cancer(OC) dataset5is for identifying proteomic patterns in serum that distinguish ovarian cancer from non-cancer cases.There are 253samples each described by activity level of15154pro-teins.These expression datasets have real valued entries, which have to be discretized to obtain binary features;we

do an equal-depth partition for each attribute using20buck-ets.6

Parameters:Two parameters are varied in our experiment, minimum support(minsup)and length ratio(l).We use a default value of minsup=4%.The parameter length ra-tio,l,has a value between0and1.It is used to generate new datasets with di?erent average row size from the orig-inal datasets.A dataset with a length ratio of l retains

on average l?100%of the columns in the original dataset. Columns to be retained are randomly selected for each row. The default value used is l=0.6,unless otherwise stated. 4.1Varying Minimum Support

Figure5shows how CARPENTER compares against CHARM and CLOSET as minsup is varied with l=0.6. Note that the y-axis is in logarithmic scale.There is a large variation in the running time for both CHARM and CLOSET even though the variation in absolute minsup is small.This is because the average length of each row af-ter removing the infrequent features can increase(decrease) substantially due to a small decrease(increase)in minsup value.This increases(decreases)the search space of both CHARM and CLOSET substantially,resulting in a large

di?erence in running time.

Among the three algorithms,we?nd that CLOSET is the slowest and has the steepest increases in run time as minsup

3available from https://www.360docs.net/doc/d63698514.html,

4https://www.360docs.net/doc/d63698514.html,/data/ALL1/

5https://www.360docs.net/doc/d63698514.html,/

6Fewer buckets results in a extremely high running time(up

to a few days)for CHARM and CLOSET.is decreased.CHARM on the other hand is generally2to

3orders of magnitude slower than CARPENTER and only outperforms CARPENTER at higher support level,where the di?erence in time is under10seconds.

4.2Varying Length Ratio

Figure6shows the performance comparison of the meth-ods as we vary l,with minsup=4.The growth in run time of all the algorithms is exponential with respect to the length ratio(note the log scaled y-axes).CARPENTER is however substantially faster than CHARM and CLOSET. While CHARM outperforms CLOSET,CARPENTER can be up to a100times faster than CHARM and1000times faster than CLOSET.

As we can see,in all the experiments we conducted,CAR-PENTER outperforms CHARM and CLOSET in most cases. These results clearly demonstrate that CARPENTER is very e?cient in?nding frequent closed patterns on datasets with small number of rows and large number of features.

5.RELATED WORK

Frequent pattern mining[1,5,11,8]as a vital topic has received a signi?cant amount of attention during the past decade.The number of frequent patterns in a large data set can be very large and many of these frequent patterns may be redundant.To reduce the frequent patterns to a compact size,mining frequent closed patterns has been pro-posed.The followings are some new advances for mining closed frequent patterns.

Close[6]and Pascal[2]are two algorithms which discover closed patterns by performing breadth-?rst,column enumer-ation.Close[6]is an Apriori-like algorithm to determine a closed itemset.Due to the level-wise approach of Close and Pascal,the number of feature sets enumerated will be ex-tremely large when they are run on long biological datasets.

In[7],the CLOSET algorithm was proposed for mining closed frequent patterns.Unlike Close and Pascal,CLOSET performs depth?rst,column enumeration.CLOSET uses a novel frequent pattern tree(FP-structure)for a compressed representation of the datasets.It then performs recursive computation of conditional tables to simulate the search on the column enumeration tree.CLOSET is unable to handle long biological datasets because of two reasons.First,the FP-tree is unable to give good compression for long rows. Second,there are too many combinations when performing column enumerations.CLOSET+[9]is a recent improve-ment on CLOSET.Our study show that CARPENTER still outperforms CLOSET+on average by around500-600 times.7

Another algorithm for mining frequent closed pattern is CHARM[10].Like CLOSET,CHARM performs depth-

?rst,column enumeration.However,unlike CLOSET,CHARM stores the dataset in a vertical format where a list of row ids

is stored for each feature.These row id lists are then merged during the column enumeration to generate new rows id lists that represent nodes in the enumeration tree.In addition,

a technique called di?set is used to reduce the size of the row id lists and the computational complexity for merg-ing them.Although performance studies in[10]shows that CHARM is substantially faster than all other algorithm on most datasets,CHARM is still unable to handle long biolog-

7We will like to thank Jianyong Wang and Jiawei Han for making the executable code of CLOSET+available to us. They have indicated that the version of CLOSET+that they pass to us at press time is not optimized for our datasets and an optimized version will be released in the future.

(完整版)2018全国二卷高考生物真题(含答案)

2018全国生物卷(Ⅱ卷) 1.下列关于人体中蛋白质功能的叙述,错误的是 A.浆细胞产生的抗体可结合相应的病毒抗原 B.肌细胞中的某些蛋白质参与肌肉收缩的过程 C.蛋白质结合Mg2+形成的血红蛋白参与O2运输 D.细胞核中某些蛋白质是染色体的重要组成成分 2.下列有关物质跨膜运输的叙述,正确的是 A.巨噬细胞摄入病原体的过程属于协助扩散 B.固醇类激素进入靶细胞的过程属于主动运输 C.神经细胞受到刺激时产生的Na+内流属于被动运输 D.护肤品中的甘油进入皮肤细胞的过程属于主动运输 3.下列有关人体内激素的叙述,正确的是 A.运动时,肾上腺素水平升高,可使心率加快,说明激素是高能化合物 B.饥饿时,胰高血糖素水平升高,促进糖原分解,说明激素具有酶的催化活性 C.进食后,胰岛素水平升高,其即可加速糖原合成,也可作为细胞的结构组分 D.青春期,性激素水平升高,随体液到达靶细胞,与受体结合可促进机体发育 4.有些作物的种子入库前需要经过风干处理。与风干前相比,下列说法错误的是

A.风干种子中有机物的消耗减慢 B.风干种子上微生物不易生长繁殖 C.风干种子中细胞呼吸作用的强度高 D.风干种子中结合水与自由水的比值大 5.下列关于病毒的叙述,错误的是 A.从烟草花叶病毒中可以提取到RNA B.T2噬菌体可感染肺炎双球菌导致其裂解 C.HIV可引起人的获得性免疫缺陷综合征 D.阻断病毒的传播可降低其所致疾病的发病率 6.在致癌因子的作用下,正常动物细胞可转变为癌细胞。有关癌细胞特点的叙述错误的是 A.细胞中可能发生单一基因突变,细胞间黏着性增加 B.细胞中可能发生多个基因突变,细胞的形态发生变化 C.细胞中的染色体可能受到损伤,细胞的增殖失去控制 D.细胞中遗传物质可能受到损伤,细胞表面糖蛋白减少 29.(8分)为研究垂体对机体生长发育的作用,某同学用垂体切除法进行实验。在实验过程中,用幼龄大鼠为材料,以体重变化作为生长发育的检测指标。回答下列问题: (1)请完善下面的实验步骤 ①将若干只大鼠随机分为A、B两组后进行处理,A组(对照组)的处理是_____________;B组的处理是_____________。 ②将上述两组大鼠置于相同的适宜条件下饲养。 ③___________。

初中五种基本句型的用法及其运用讲义(无答案)

学科教师辅导教案 学员编号:年级:新初三课时数:学员姓名:辅导科目:英语学科教师:课程主题:初中五种基本句型授课时间:2018.7.24 学习目标 1.掌握五种基本句型的用法以及它们在用法上的区别 教学内容

一、知识精讲 五种基本句型是句子最基本的组成部分。掌握了这五种基本句型,在阅读中当我们遇到较复杂的句子时,运用这些基本句型,对句子的分析就会变得容易多了。在写作中,首先要能运用好这些基本句型,才能得到高分。 (一)五种基本句型的句子成分: 1. 句子成分的定义:构成句子的各个部分叫做句子成分。句子成分包括主要成分和次要成分;主要成分有主语和谓语;次要成分有表语、宾语、定语、状语、补足语等。 2. 主语(subject):主语是一个句子所叙述的主体,一般位于句首。主语可由名词、代词、数词、不定式、动名词、名词化的形容词和主语从句等表示。 【例句】 We often speak English in class. 我们在课上经常说英语。 Smoking does harm to the health. 吸烟对健康有害。 The rich should help the poor. 富有的人应该帮助贫困的人。 3.谓语(verb):谓语可用来说明主语所做的动作或具有的特征和状态。动词常在句中作谓语,一般放在主语之后。分及物动词和不及物动词两种。 【例句】 He practices running every morning. 他每天早晨练习跑步。 I have caught a bad cold. 我得了重感冒。 We like helping the people in trouble 我们喜欢帮助那些处于困境中的人。 4.宾语(object):宾语在句中表示动作的对象或承受者,一般位于及物动词或介词后面。 【例句】 They went to see a film yesterday. 他们昨天看了一场电影。 She often helps her mother with their housework. 她经常帮助她的妈妈做家务。 I enjoy listening to popular music. 我喜欢听流行音乐。 5. 表语(predicative):表语用以说明主语的身份、特征和状态,它一般位于系动词(如be, become, get, look, grow, turn, seem等)之后。表语一般由名词、代词、形容词、分词、数词、不定式、动名词、介词短语、副词构成。 【例句】 The weather has turned cold. 天气变得冷了。 His job is to teach English.

雅思口语part2话题万能模板:事件类

雅思口语part2话题万能模板:事件类今天三立在线教育雅思网为大家带来的是雅思口语part2话题万能模板:事件类的相关资讯,备考的烤鸭们,赶紧来看看吧! 首先我们来说说“一件从数学中学到的事情”,拿到这个话题,先不要急着回答,可以用笔稍微构思一下思路,该怎么回答。首先,可以说一下这件事是什么、发生在什么时候,也就是作文中的事件和事件,这里需要注意的就是具体描述一下什么事但是又不能啰嗦累赘,而时间则可以大致概括点明即可。我们就从资料中的例子来说:高中的时候从乘法表中学到的一件事。 然后,要具体叙述一下这件事,即是最重要的部分,发生的地方、牵扯到的人物、以及这件事情影响到你的原因,为什么影响到你…… 资料例子来说:在高中的一次数学竞赛中失败,以至于灰心丧气,甚至对自己产生怀疑,而此时数学老师拿乘法的规则来比喻人生中的失败与成功的关系来鼓励我,以至于我从失败中走出来,从新赢得了下一次的数学竞赛。这件事情使我明白了,生活就如同数学乘法,有无数种可能,多种多样,我们要坦然面对。 而这里的例子看似像我们中文作文中的思路,但是关键在于它表述方式以及用词造句,不能太中式化。这里的重点,就是描述事件经过,即“在高中的一次数学竞赛中失败,以至于灰心丧气,甚至对自己产生怀疑,而此时数学老师拿乘法的规则来比喻人生中的失败与成功的关系来鼓励我,以至于我从失败中走出来,从新赢得了下一次的数学竞赛。”而老师的话则是重中之重,因为他让我明白了道理“生活就如同数学乘法,有无数种可能,多种多样,我们要坦然面对。”也是我深深牢记的原因,以及这件事对我影响深刻的原因。

而最后,只需要简单概括,从这件事中学到的东西即可。其实这个话题思路并不是很难,关键在于烤鸭们的表述以及词汇的积累。 其实,大家都知道事件类话题是雅思口语考试中最简单的、最容易描述,却也是最常见的话题,所以大家要多注意积累素材、思考思路。其实“A way of communication”和“A good parent”这两个话题也可以用这个例子来阐述回答,而最关键的则是把老师的那段话稍作转述即可。例如,“A way of communication”我们可以先描述手机及其性能作用,以及现今的流行趋势,然后从一件事引到交流方式,还是那件事,还是那写话,只不过要突出老师用短信的方式告知我,最后,简单概述手机的用处及好处即可。是不是很简单呢?同样的,“A good parent”只需要根据情况把那些话转述为父母亲对你说的话,已经对你的影响,所以一个模板真的是百用哦! 那么我们赶紧来看看还有哪些话题可以这样转换呢? 多米诺骨牌效应:=Something you learned from math=Something you learned from your family=A piece of advice=A change you had=An important stage in your life=A text message you got=A letter you got=An E-mail you got=A way of communication=An important decision you made=A person who helped you=A kind thing someone did to you=A family member you love talking to=A good parent=An elderly family member=A neighbor=A good friend=A teacher=A science course you took=A subject you love/d at school=Your personality=First day of college=Your experience of getting money as a gift=Your experience of getting congratulations=A person you helped=A thing your friend did made you admire.

感官动词和使役动词

感官动词和使役动词 默认分类2010-05-28 23:14:26 阅读46 评论0 字号:大中小订阅 使役动词,比如let make have就是3个比较重要的 have sb to do 没有这个用法的 只有have sb doing.听凭某人做某事 have sb do 让某人做某事 have sth done 让某事被完成(就是让别人做) 另外: 使役动词 1.使役动词是表示使、令、让、帮、叫等意义的不完全及物动词,主要有make(使,令), let(让), help(帮助), have(叫)等。 2.使役动词后接受词,再接原形不定词作受词补语。 He made me laugh. 他使我发笑。 I let him go. 我让他走开。 I helped him repair the car. 我帮他修理汽车。 Please have him come here. 请叫他到这里来。 3.使役动词还可以接过去分词作受词补语。 I have my hair cut every month. 我每个月理发。 4.使役动词的被动语态的受词补语用不定词,不用原形不定词。 (主)He made me laugh. 他使我笑了。 (被)I was made to laugh by him. 我被他逗笑了。 使役动词有以下用法: a. have somebody do sth让某人去做某事 ??i had him arrange for a car. b. have somebody doing sth.让某人持续做某事。 ??he had us laughing all through lunch. 注意:用于否定名时,表示“允许” i won't have you running around in the house. 我不允许你在家里到处乱跑。 ******** 小议“使役动词”的用法 1. have sb do 让某人干某事 e.g:What would you have me do? have sb/sth doing 让某人或某事处于某种状态,听任 e.g: I won't have women working in our company. The two cheats had the light burning all night long. have sth done 让别人干某事,遭受到 e.g:you 'd better have your teeth pulled out. He had his pocket picked. notes: "done"这个动作不是主语发出来的。 2.make sb do sth 让某人干某事 e.g:They made me repeat the story. What makes the grass grow?

2017年江苏生物高考真题(含答案)

绝密★启用前 2017年普通高等学校招生全国统一考试(卷) 生物 1.下列关于糖类化合物的叙述,正确的是 A.葡萄糖、果糖、半乳糖都是还原糖,但元素组成不同 B.淀粉、糖原、纤维素都是由葡萄糖聚合而成的多糖 C.蔗糖、麦芽糖、乳糖都可与斐林试剂反应生成砖红色沉淀 D.蔗糖是淀粉的水解产物之一,麦芽糖是纤维素的水解产物之一 2.下列关于探索DNA 是遗传物质的实验,叙述正确的是 A.格里菲思实验证明DNA 可以改变生物体的遗传性状 B.艾弗里实验证明从S 型肺炎双球菌中提取的DNA 可以使小鼠死亡 C.赫尔希和蔡斯实验中离心后细菌主要存在于沉淀中 D.赫尔希和蔡斯实验中细菌裂解后得到的噬菌体都带有32P 标记 3.下列关于肽和蛋白质的叙述,正确的是 A.琢鄄鹅膏蕈碱是一种环状八肽,分子中含有8 个肽键 B.蛋白质是由2 条或2 条以上多肽链构成的 C.蛋白质变性是由于肽键的断裂造成的 D.变性蛋白质不能与双缩脲试剂发生反应

4.下列关于用显微镜观察细胞的实验,叙述正确的是 A.转换物镜时应该手握物镜小心缓慢转动 B.以洋葱鳞片叶表皮为材料不能观察到质壁分离 C.丹芋染色后的花生子叶细胞中可观察到橘黄色颗粒 D.在新鲜黑藻小叶装片中可进行叶绿体形态观察和计数 5.某小组开展酵母菌培养实验,下图是摇瓶培养中酵母种群变化曲线。下列相关叙述正确的是 A.培养初期,酵母因种竞争强而生长缓慢 B.转速150 r/min 时,预测种群增长曲线呈“S冶型 C.该实验中酵母计数应采用稀释涂布平板法 D.培养后期,酵母的呼吸场所由胞外转为胞 6.下列关于人类遗传病的叙述,正确的是 A.遗传病是指基因结构改变而引发的疾病 B.具有先天性和家族性特点的疾病都是遗传病 C.杂合子筛查对预防各类遗传病具有重要意义 D.遗传病再发风险率估算需要确定遗传病类型 7.下列关于生物进化的叙述,错误踿踿的是 A.某物种仅存一个种群,该种群中每个个体均含有这个物种的全部基因 B.虽然亚洲与澳洲之间存在地理隔离,但两洲人之间并没有生殖隔离 C.无论是自然选择还是人工选择作用,都能使种群基因频率发生定向改变 D.古老地层中都是简单生物的化石,而新近地层中含有复杂生物的化石 8.下图为突触结构示意图,下列相关叙述正确的是

subject_to_浅谈

首先将subject 和to 单独理解,subject的主要和常用意思是: 1. As a noun, a person owing loyalty to a certain state or royal ruler: a subject of the United Kingdom. 作名词,指对某个国家或统治者表示忠诚的人,用常用汉语表述为“臣民、国民”;如:英国国民。 2. As a noun, something being considered, as in a c onversation: Don’t change the subject; answer the question. 作名词,指如在一个谈话中正在讨论、考虑的东西,汉语表述为“话题、主题、论题”;如:不要改变话题,回答这个问题。从翻译的角度来讲,这句话可以用更好的中文表述:不要东拉西扯,回答这个问题;不要罔顾左右而言它,回答这个问题。 3. As a noun, a branch of knowledge studied, as in a system of education: she’s taking 3 subjects in her examinations. 作名词,所学知识的一个分支,如在教育系统中的学科,汉语表述为“科目、学科”:她在参加三门学科的考试。 4. As a noun, a cause: his strange cloth was a subject for amusement. 作名词,(事情的)起因,汉语表述为“起因、原因、缘由”:他奇怪的服饰是大家取笑的原因(直译);他奇怪的服饰成了大家的笑料。 5. As a noun, the main area of interest treated in a work, esp. written: a book on the subject of love. 作名词,著作中的主题思想,尤其是在书面作品中,汉语表述“主题、主体、题材”:一本以爱情为主题的书。 6. As an adjectiv e, governed by some one else; not independent: a subject race. 作形容词,(被)统治的,不独立的;汉语表述为“隶属的、统治的、管辖的、服从的”:一个隶属民族。 7. As a verb, to cause to be controlled or ruled: These people have been subjected by another tribe. 作动词,被统治、控制;汉语表述为“统治、控制”:这些人被另外一个部落统治着。 我们都知道“to”在英语中都作介词使用,主要意思是:向、去、到、达到、比、对、在等等。 当subject 和to连用,构成短语subject to,其意思发生了很大变化,尤其是和它的名词形式。因而Subject to 是英语中一个很常见、很重要但很难把握的短语。特别是在商务合同、法律和税收等中出现的频率很高,不少人由于没有真正了解它的用法,而不能正确理解、翻译和使用它。下面谈谈subject to的用法: 一、作形容词用 1.基本含义是“应服从……的……”、“受制于……”。英语解释:owing obedience to, 示例: All the foreigner friends are subject to the law of our country. 所有外国朋友都应服从我们国家法律。 2.基本含义是“有……倾向”、“易受……”。英语解释为:prone to,vulnerable to。 Wenchuan and Qingchuan counties in Sichuan province are subject to earthquakes as they are located on Longmenshan Earthquake Zone. 四川省的汶川和青川县易受地震的影响,因为它们位于龙门山地震带上。 A weak man is subject to cold. 体质不好的人易感冒。

雅思口语part2中不建议背诵范文

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