计算列表中的子字符串出现在字符串中的次数

Question

I have a string:

seq = '01234567890123456789'

that I want to first break up into intervals of 4. I found a previous answer

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

which gives me

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

And now I want to compare my chunked up string to every entry in a list of substrings:

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

and return an array that counts how many times each substring appeared in the original string. I see a lot of examples of finding one substring in a larger string but I'm confused as to how to do this with a list of substrings

Answer 1

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

list.count(e) returns number of occurrences of e in list , so we can execute it for each item in mylist .

In this case result is [1,0,1] which means '0123' appeared in generated once, '1111' did not appear and '2345' appeared once as well.

Couldn't figure out from your description which way you want to compare occurrences. If I got it the wrong way around just say and I'll be happy to edit.

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

Add a tuple containing the chunk and its count in a list, in this case called counts.

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)]