Question

我有一个字符串：

seq = '01234567890123456789'

我想首先分成4个区间。我找到了之前的答案

n = 4    
[seq[i:i+n] for i in range(0,len(seq,n)]

给了我

['0123', '4567', '8901', '2345', '6789']

现在我想将我的chunked up字符串与子字符串列表中的每个条目进行比较：

mylist = ['0123', '1111' '2345']

并返回一个数组，该数组计算每个子字符串在原始字符串中出现的次数。我看到很多在较大的字符串中找到一个子字符串的例子，但我对如何用一个子字符串列表这样做感到困惑

Answer 1

generated = ['0123', '4567', '8901', '2345', '6789']
mylist =  ['0123', '1111' '2345']
result = [generated.count(element) for element in mylist ]

list.count(e)会返回e中list的出现次数，因此我们可以为mylist中的每个项目执行此次。

在这种情况下，结果为[1,0,1]，这意味着'0123'出现generated一次，'1111'未显示，'2345'也出现过一次。

无法从您的描述中找出您想要比较事件的方式。如果我以错误的方式说出来并且我很乐意编辑。

Answer 2

chunked = ['0123', '4567', '8901', '2345', '6789']
mylist = ['0123', '1111', '2345']
my_count = {}
for m in mylist:
   for c in chunked:
      if m == c:
          try:
              my_count[m] += 1
          except KeyError:
              my_count[m] = 1

>>> chunked
['0123', '4567', '8901', '2345', '6789']
>>> mylist
['0123', '1111', '2345']
>>> my_count
{'2345': 1, '0123': 1}
>>>

Answer 3

seq = '01234567890123456789'
chunks = ['0123', '4567', '8901', '2345', '6789']
dict = {}
for elt in chunks:
    dict[elt] = seq.count(elt)

print(dict)

Answer 4

在列表中添加包含块及其计数的元组，在本例中称为计数。

In [37]: seq = '01234567890123456789'

In [38]: chunks = ['0123', '4567', '8901', '2345', '6789']

In [39]: mylist =  ['0123', '1111' '2345']

In [40]: counts = [(i, mylist.count(i)) for i in chunks]

In [41]: counts
Out[41]: [('0123', 1), ('4567', 0), ('8901', 0), ('2345', 0), ('6789', 0)]

计算列表中子字符串出现在字符串

4 个答案: