如何使用pyhton知道以txt文件的每个字母开头的名称数量

Question

I need to know how I can calculate the number of words in the list that start with the letter A, B, C .. Z.

Here I leave the reading part of the txt file

#!/usr/bin/python


def main():
  lines = []
  xs = []

  try:
    with open("bin-nombres.txt", 'r') as fp:
      lines = [lines.strip() for lines in fp]

    for i in lines:
      print(i[0])
      xs = counterByLetter(i[0])

    print(xs)
  except EOFError as e:
    print(e)

  finally:
    pass

def counterByLetter(data):
  return [(k, v) for k, v in {v: data.count(v) for v in 'abcdefghijklmnopqrstuvwxyz'}.items()]  

if __name__ == "__main__":
  main()

I must calculate the number of words that begin with [A ... Z]. For examples.

• There are 3 words that start with A.
• There are 20 words that start with B.
• etc..

Here I leave the solution to the problem. Thanking those who helped me !!

import string


def main():
  try:
    # this initiates the counter with 0 for each letter
    letter_count = {letter: 0 for letter in list(string.ascii_lowercase)}
    with open("bin-nombres.txt", 'r') as fp:
      for line in fp:
        line = line.strip()
        initial = line[0].lower()
        letter_count[initial] += 1  # and here I increment per word


    #iterating over the dictionary to get the key and the value.
    #In the iteration process the values will be added to know the amount of words.
    size = 0
    for key , value in letter_count.items():
      size += value
      print("Names that start with the letter '{}' have {} numbers.".format(key , value))

    print("Total names in the file: {}".format(size))


  except EOFError as e:
    print(e)


if __name__ == "__main__":
  main()

Answer 1

Assume that, there have a list name list which have 3 elements:

list = ["Geeks", "For", "Triks"]

And have a array which have 26 elements.

array = ["0", "0", ......"0", "0"......"0","0"]

array[0] represent the number of words start with A . .................. .................. array[25] represent the number of words start with Z . Then, if list[n][0] start with A then you need to increment array[0] by 1.

if array[5] = 7 then it's mean that there are 7 words start with F . This is the straightforward logic for find the result.

Answer 2

So, according to your updated answer (1 word per line, already alphabetically sorted), something like this should work:

import string


def main():
  try:
    # this initiates our counter with 0 for each letter
    letter_count = {letter: 0 for letter in list(string.ascii_lowercase)}
    with open("words.txt", 'r') as fp:
      for line in fp:
        line = line.strip()
        initial = line[0].lower()
        letter_count[initial] += 1  # and here we increment per word

    print(letter_count)

  except EOFError as e:
    print(e)


if __name__ == "__main__":
  main()

UPDATE:

It's good that you don't just want a readymade solution, but your code has a few issues and some points are not super pythonic, that's why I suggested to do it as above. If you really want to go with your solution, you need to fix your counterByLetter function. The problem with it is that you're not actually storing the results anywhere, you're always returning a new array of results for each word. You probably have a word starting with 'z' as the last word of the file, hence the result having 0 as the count for all letters, except 'z', which has one. You need to update your values for the current letter in that function, instead of calculating the whole array at once.

Answer 3

I'd suggest to change a bit your code like this.

Use collection.defaultdic t set to int as value: using the first letter as key of the dictionary you are able to increment its value each there is a match. So:

from collections import defaultdict

Set xs as xs = defaultdict(int)

Change the for i in lines: body to

for i in lines:
  xs[i[0]] += 1

If you print xs at the end of the for loop you'll get something like:

defaultdict(<class 'int'>, {'P': 3, 'G': 2, 'R': 2})

Keys in dict are case sensitive, so, take care of transforming the case, if required.

You don't need an external method to do the counting job.