How do I count the frequency of letters in a string and sort by alphabetical orders?, if their frequ-CodePudding

Given string S consisting of only lowercase letters (a-z). Print to the screen all the letters that appear more than once in the string S and their number in lexicographical order.

a = input()
for i in range(len(a)):
    if a.count(a[i])!=1:
        print(a[i], a.count(a[i]))

the problem here is it does print the frequency, but doesn't sort in alphabetical order, and some letter and their frequencies it print out more than 1

example input

thequickbrownfoxjumpsoverthelazydog

example output

e 3
h 2
o 4
r 2
t 2
u 2

How do I do this pls help

CodePudding user response：

Make a mapping of letters to counts. One quick way with a dictionary would be:

word = input()
letters = {}
for letter in word:
    if letter not in letters:
        letters[letter] = 0
    letters[letter]  = 1

You can simplify this using a defaultdict:

from collections import defaultdict

word = input()
letters = defaultdict(int)
for letter in word:
    letters[letter]  = 1

Or even simpler with a counter:

from collections import Counter

word = input()
letters = Counter(word)

Now that you have your letter counts, you just need to sort and print the items:

for letter, count in sorted(letters.items()):
    if count > 1:
        print(letter, count)

CodePudding user response：

import collections

s = "thequickbrownfoxjumpsoverthelazydog"

d = {}
for c in s:
    if c in d:
        d[c]  = 1
    else:
        d[c] = 1

od = collections.OrderedDict(sorted(d.items()))

for k, v in od.items():
    if v > 1:
        print(k, v)

store the letters and their appearance in a dictionary, sort it and print the key and value if the value is greater than 1.