CODEWITHSUNDEEP
python
tadeufontes
tadeufontes

Reputation: 477

How to iterate over each position of a character of a string, for a list of strings?

So I have a list of strings, all of the same length, like this:

list_strings=["a-a--","-ab-b","a---a","b-b-a","aab-a"]

What I want to do it iterate over each position of the strings inside the list, so to calculate the number of times the character "-" appears in each position. In this case for example, position 0 has 1 "-"s, position 1 has 3 "-"s, position 2 has 1 "-"s, position 3 has 5 "-"s. But I want to do this for a file that has more than 100,000 strings

So far I have:

for i in range(0,len(list_strings)):
    for j in range(0,len(list_strings[i])):
        if list_strings[i][j]=="-":
            #count how many "-"s appear in this position and maybe save it in a list?

Thanks in advance for any answer

Upvotes: 2

Views: 836

Answers (3)

tygzy
tygzy

Reputation: 716

I won't do much explaining, so this is the code to start:

list_strings=["a-a--","-ab-b","a---a","b-b-a","aab-a"]

for string in list_strings:
    occurrence = 0
    check_letter = '-'
    for letter in string:
         if letter == check_letter:
             occurrence += 1
    print('string: ' + string)
    print('occurrences: ' + occurrence)
    print('\n')

Upvotes: 0

Andrej Kesely
Andrej Kesely

Reputation: 195438

list_strings=["a-a--","-ab-b","a---a","b-b-a","aab-a"]

# if every string in `list_strings` is same length:
out = [v.count('-') for v in zip(*list_strings)]
print(out)

Prints:

[1, 3, 1, 5, 1]

If some strings are different length:

from itertools import zip_longest
out = [v.count('-') for v in zip_longest(*list_strings)]

Upvotes: 3

marksman123
marksman123

Reputation: 399

you are good. just add counter=0 variable who will add himself every time your if clause is true, and you will have the number of '-' in your list.

counter =0
for i in range(0,len(list_strings)):
    for j in range(0,len(list_strings[i])):
        if list_strings[i][j]=="-":
           counter = counter +1

Upvotes: 1

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