如何返回具有特定要求的字符串在列表中出现的次数？

Question

Given a list of strings, return the count of the number of strings where the string length is 3 or more and the first and last chars of the string are the same.

To solve this problem I created the following function,

def match_ends(words):
  FirstL=words[0:1]
  LastL=words[-1:]
  x=len(words)>3 and FirstL == LastL
  count=0
 for x in words:
    count+=1
    return count

then tested it here,

def test(got, expected):
  if got == expected:
    prefix = ' OK '
  else:
    prefix = '  X '
  print ('%s got: %s expected: %s' % (prefix, repr(got), repr(expected)))


# Calls the above functions with interesting inputs.
def main():
  print ('match_ends')
  test(match_ends(['abaa', 'xyzax', 'aa', 'x', 'bbb']), 3)
  test(match_ends(['', 'x', 'xy', 'xyx', 'xx']), 1)
  test(match_ends(['aaa', 'be', 'abc', 'hello']), 1)


  print

Result:

X  got: 1 expected: 3
OK  got: 1 expected: 1
OK  got: 1 expected: 1

Answer 1

You have a few problems here:

1. When you're looping over each of the words, you return count within the loop, instead of at the end when the loop finishes. This is why you were always getting 1 .

2. You always count += 1 even if x is False .

3. x is taking the first and last item of the list, rather than the first and last letter in each word in the list.

4. Finally you shadow the bool x in your for loop.

---

Tips

Why not split this into two functions?

 def match_end(word): first, last = word[0], word[-1:] return True if first == last and len(word) >= 3 else False def match_ends(words): count = 0 for word in words: count += 1 if match_end(word) else 0 return count

The first function, match_end(word) returns a bool of either True or False .

• First, It sets the variables first and last to the first and last letter of the string via slicing.

• Next, it return s True if the first letter is the same as the last, and if the length of the word is less than three. Otherwise, it return s False . This is done with Python's Ternary Operator .

The second function, match_ends(words) , takes in a list of strings(like your original one) iterates over each word in the list .

• For each word in the list, it tests to see if match_end(word) returns True .

  • If so, it increments the count by 1 .

  • Otherwise, it does nothing (increments count by 0 ).

• Finally, it returns count .

Answer 2

Your best bet here is to use a list comprehension. A list comprehension has three parts:

• The transformation that you want to perform on each element of the input,
• the input itself, and
• an optional "if" statement that indicates when to produce output

So for example, we can say

[ x * x               # square the number
for x in range(5) ]  # for each number in [0,1,2,3,4]  `

which will produce the list

[0 1 4 9 16]

We can add a third (filtering) line, and get only odd numbers back:

[x * x
for x in range(5) 
if x % 2]     # divide x by 2, take the remainder -- if 1, output the number`

In your particular case, we don't care about the transformation part. We just want to output the word if it fits your criteria:

[word
 for word in word_list
 if len(word) >= 3 and word[0] == word[-1] ]

This will give you a list back. Now you just need to get the length of that list:

len( [word
 for word in word_list
 if len(word) >= 3 and word[0] == word[-1] ]  )

Want to turn that into a function? Here you go:

def count_matching_words(word_list):
    return len([word
                for word in word_list
                if len(word) >= 3 and word[0] == word[-1]])