Python：尝试在索引处比较两个字符串时，列表索引超出范围？

Question

I have this function to check if a string contains three or more lowercase letters.

def lowerCaseValid(word):
lowCharList = ['abcdefghijklmnopqrstuvwxyz']
i = 0
flag = 0
while i <= len(word):
    j = 0
    while j <= len(lowCharList):
        if lowCharList[j] == word[i]:
            flag += 1
            j = 0
        else:
            j += 1
    i += 1
if flag >= 3:
    return True

In simple terms, I pass in a string (word) and create a list of acceptable characters (lowCharList).

Then, I set up a nested while loop that checks word[i] at every index of lowCharList, until it finds a match.

Then it resets lowCharList counter and adds 1 to flag, and moves on to word[i+1].

If it doesn't find a match by the time it reaches z, then it moves onto word[i+1] anyways.

For some reason, why I try my sample input in my main function.

def main():
  word = 'corRe!33'
  print(lowerCaseValid(word))

I get this error:

in lowerCaseValid
if lowCharList[j] == word[i]:
IndexError: list index out of range

Why is it throwing this error? Thank you.

Answer 1

using python's in operator is easier...

    def lowerCaseValid(word):
       cnt = 0
       lowCharList = ['abcdefghijklmnopqrstuvwxyz']
       chars = lowCharList.pop ()
       for ch in word:
           if ch in chars:
           cnt += 1

       return  cnt >= 3

or with using sets just 2 lines of code

def lowerCaseValid(word):

    lowCharList = ['abcdefghijklmnopqrstuvwxyz']

    return len(set(lowCharList.pop()) & set(word)) >=3

or one liner with map and lambda

def lowerCaseValid(word):

    return len(list(map(lambda x: x.islower, list(word)))) >=3

Answer 2

Another alternative approach using a list comprehension and string.ascii_lowercase instead of redefining the lowercase letters:

from string import ascii_lowercase

def lowerCaseValid(word):
    return sum[x in ascii_lowercase for x in word] >= 3

How this works is that the list comprehension goes through each letter in word. The x in ascii_lowercase will return a boolean value of either True (1) or False (0), then sum up the True s

Answer 3

Change

lowCharList = ['abcdefghijklmnopqrstuvwxyz']

to

lowCharList = list('abcdefghijklmnopqrstuvwxyz')

I believe this should help, since you have a list containing only 1 item, whereas this way, you create a list with 24 items (all different letters).

Answer 4

As heemayl pointed out in the comments, lowCharList is 1 element long! Now you have 2 options: make lowCharList an actual list ( lowCharList = list ("abcd...") ), or keep lowCharList a string, which will work just fine (remove the brackets in the definition.

Might I suggest another method: checking if str.islower count adds up to >=3. So:

lower_count = 0
for letter in word: 
    if letter.islower(): lower_count += 1
if lower_count >= 3: return True

Also, as suggested in the comments:

return len ([letter for letter in word if letter.islower()]) >= 3 

would work (better than my answer, which is just the expanded form of it)