How to return the number of times a string with specific requirements appears in a list?

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

Solution

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]])