如何在python中增加元组值并在循环中搜索字符串

我有这个密码

 arfffile = []

inputed = raw_input("Enter Evaluation for name including file extension...")

reader = open(inputed, 'r')

verses = []

for line in reader:
    verses.append(line)

for line in verses:
    if line.split('@') == "@":
        verses.pop(line)


numclusters = int(raw_input("Enter the number of clusters"))

clusters = {}

for i in range(1,numclusters+1):
    clusters["cluster"+str(i)] = 0



print clusters
 # If verse belongs to a cluster, increment the cluster count by one in the dictionary value.
for verse in verses:
    for k in clusters:
        if k in verse:
            clusters[k] += 1
        else:
            print "not in"

print clusters

yeslist = []

for verse in verses:
    for k in clusters:
        if k not in yeslist:
            yeslist.append((k,0))
        elif k in yeslist:
            print "already in" + k


for verse in verses:
    for k in clusters:
        if k in verse and "Yes" in verse:
            yeslist.append(yeslist.index(k), +1)


    # iterate through dictionary and iterate through the lines
    # need to read in file line by line, 



    # if "yes" and cluster x increment cluster 
    # need to work out percentage of possitive verses in each cluster. 

arff文件的一个示例是

@relation tester999.arff_clustered

@attribute Instance_number numeric
@attribute allah numeric
@attribute day numeric
@attribute lord numeric
@attribute people numeric
@attribute earth numeric
@attribute men numeric
@attribute truth numeric
@attribute verily numeric
@attribute chapter numeric
@attribute verse numeric
@attribute CLASS {Yes,No}
@attribute Cluster {cluster1,cluster2,cluster3}

@data
0,1,0,0,0,0,0,0,0,1,1,No,cluster3
1,1,0,0,0,0,0,0,0,1,2,No,cluster3
2,0,0,0,0,0,0,0,0,1,3,No,cluster3
3,0,1,0,0,0,1,0,0,1,4,No,cluster3
4,0,0,0,0,0,0,0,0,1,5,No,cluster3
5,0,0,0,0,0,0,0,0,1,6,No,cluster3
6,0,0,0,0,0,0,0,0,1,7,No,cluster3
7,0,0,0,0,0,0,0,0,2,1,No,cluster3
8,1,0,0,0,0,0,0,0,2,2,No,cluster3
9,0,0,0,0,0,0,0,0,2,3,No,cluster3
10,0,0,0,0,0,0,0,0,2,4,No,cluster3
11,0,0,1,0,0,0,0,0,2,5,No,cluster2

当它处于静止状态时，程序读取数据行，例如

0,1,0,0,0,0,0,0,0,1,1,No,cluster3

我还创建了一个字典来检测数据文件中有多少集群。在这个例子中有3个。集群1集群2和集群3。然后，代码将每个集群作为键值追加到字典“clusters”中的字符串中
然后我迭代所有的诗句，数一数每一行，看看它属于哪个集群

我的下一步是尝试计算每个集群中出现“Yes”行的次数。因此，假设数据中的每一行有10行字符串中带有“yes”，那么代码应该能够计算出出现这种情况的次数

到目前为止，我所做的代码就在这里

for verse in verses:
        for k in clusters:
            if k in verse and "Yes" in verse:
                yeslist.append(yeslist.index(k), +1)

我正在创建一个名为“yeslist”的元组列表，其值如下[（cluster1,0），（cluster2,3）]

因此，对于每一行（表示为字符串），检查其中是否有“是”，如果有检查它属于哪个集群，则将该元组值增加1

我想不出怎么做的逻辑。。。有人能帮忙吗

谢谢

import collections

inputed = raw_input("Enter Evaluation for name including file extension...")

reader = open(inputed, 'r')

verses = [ line.strip() for line in reader.readlines() if line[0] != '@' ]

reader.close()

cluster_count = collections.defaultdict(int)
yes_count = collections.defaultdict(int)

verse_infos = [ (split_verse[-1],split_verse[-2]) for split_verse \
                 in verses.split(",") ]

for verse in verse_infos:
    cluster_count[verse[0]]+=1
    if verse[1] == 'yes':
        yes_count[verse[0]]+=1

您最终会得到两本词典：

cluster_count : keys = cluster#, values = count
yes_count     : keys = cluster#, values = #yes

如果确实需要元组列表：

yes_tuples = ( x for x in sorted(yes_count.iteritems()) )

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问题的简短变体是？我很确定元组是不可变的。