问题
I am trying to count the occurrences of each letter of a word
word = input("Enter a word")
Alphabet=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
for i in range(0,26):
print(word.count(Alphabet[i]))
This currently outputs the number of times each letter occurs including the ones that don't.
How do I list the letters vertically with the frequency alongside it e.g:
word="Hello"
H 1
E 1
L 2
O 1
回答1:
from collections import Counter
counts=Counter(word) # Counter({'l': 2, 'H': 1, 'e': 1, 'o': 1})
for i in word:
print(i,counts[i])
Try using Counter
, which will create a dictionary that contains the frequencies of all items in a collection.
Otherwise, you could do a condition on your current code to print
only if word.count(Alphabet[i])
is greater than 0, though that would slower.
回答2:
def char_frequency(str1):
dict = {}
for n in str1:
keys = dict.keys()
if n in keys:
dict[n] += 1
else:
dict[n] = 1
return dict
print(char_frequency('google.com'))
回答3:
As @Pythonista said, this is a job for collections.Counter:
from collections import Counter
print(Counter('cats on wheels'))
This prints:
{'s': 2, ' ': 2, 'e': 2, 't': 1, 'n': 1, 'l': 1, 'a': 1, 'c': 1, 'w': 1, 'h': 1, 'o': 1}
回答4:
s=input()
t=s.lower()
for i in range(len(s)):
b=t.count(t[i])
print("{} -- {}".format(s[i],b))
回答5:
easy and simple solution without lib.
string=input()
f={}
for i in string:
f[i]=f.get(i,0)+1
print(f)
here is the link for get() https://docs.quantifiedcode.com/python-anti-patterns/correctness/not_using_get_to_return_a_default_value_from_a_dictionary.html
回答6:
Following up what LMc said, your code was already pretty close to functional, you just needed to post-process the result set to remove 'uninteresting' output. Here's one way to make your code work:
#!/usr/bin/env python
word = raw_input("Enter a word: ")
Alphabet = [
'a','b','c','d','e','f','g','h','i','j','k','l','m',
'n','o','p','q','r','s','t','u','v','w','x','y','z'
]
hits = [
(Alphabet[i], word.count(Alphabet[i]))
for i in range(len(Alphabet))
if word.count(Alphabet[i])
]
for letter, frequency in hits:
print letter.upper(), frequency
But the solution using collections.Counter
is much more elegant/Pythonic.
回答7:
For future references: When you have a list with all the words you want, lets say
wordlist
it's pretty simple
for numbers in range(len(wordlist)):
if wordlist[numbers][0] == 'a':
print(wordlist[numbers])
来源:https://stackoverflow.com/questions/40985203/counting-letter-frequency-in-a-string-python