需要计算if语句中给定字符串的出现次数

Question

so in my function one of my if statements is that the string does not have more than 1 occurrence of a string that is in a list.

for example:

list = ['a','b']

i need to check if a string has more than one of those 2 characters in it or else it will move on to the next if statement.

s = 'aloha'

s will pass the if statement because b is not in s.

s = 'abcd'

s should fail this if statement because both a and b are in s.

more examples if it's not clear enough.

s = 'aaab'

this will fail the if statement and move on.

s = 'aaloh'

this will also fail

my if statement was:

if s.count('a') == 1 or s.count('b') == 1:

This if statement doesn't work

my question is, is there a way to check this without doing a for loop before hand?

Answer 1

You'll need a for loop, but you can still do it one line.

if sum(char in s for char in list) > 1:

If you wanted to look for specific numbers of a given character with some threshold, put your list of characters into a dict like d = {'a': 1, 'b': 4} .

if sum(s.count(char) >= count for char, count in d.items) > THRESHOLD:

Update

The OP commented that he wants a

a function that does something with the string if and only if that string doesn't have both 'a' and 'b' in the string. It can only have one of the two and not both and also only one occurrence of it.

Because OP's original post wanted to generalize 'a' and 'b' to a sequence of characters list , I interpret his meaning to be he wants an expression that returns True for a string s if and only if s has exactly one occurrence of exactly one element of a list of characters list .

sum(s.count(char) for char in list) <= 1

In OP's simple example, list = ['a', 'b'] .