如何计算字符串中字符的出现次数（列表）

Question

I'm trying to count the occurrence of each character for any given string input, the occurrences must be output in ascending order( includes numbers and exclamation marks) I have this for my code so far, i am aware of the Counter function, but it does not output the answer in the format I'd like it to, and I do not know how to format Counter. Instead I'm trying to find away to use the count() to count each character. I've also seen the dictionary function , but I'd be hoping that there is a easier way to do it with count()

from collections import Counter

sentence=input("Enter a sentence b'y: ")
lowercase=sentence.lower()

list1=list(lowercase)
list1.sort()

length=len(list1)
list2=list1.count(list1)
print(list2)

p=Counter(list1)
print(p)

Answer 1

Just call .most_common and reverse the output with reversed to get the output from least to most common:

from collections import Counter

sentence= "foobar bar"
lowercase = sentence.lower()
for k, count in  reversed(Counter(lowercase).most_common()):
    print(k,count)

Answer 2

collections.Counter objects provide a most_common() method that returns a list of tuples in decreasing frequency. So, if you want it in ascending frequency, reverse the list:

from collections import Counter

sentence = input("Enter a sentence: ")
c = Counter(sentence.lower())
result = reversed(c.most_common())
print(list(result))

Demo run

Enter a sentence: Here are 3 sentences. This is the first one. Here is the second. The end!
[('a', 1), ('!', 1), ('3', 1), ('f', 1), ('d', 2), ('o', 2), ('c', 2), ('.', 3), ('r', 4), ('i', 4), ('n', 5), ('t', 6), ('h', 6), ('s', 7), (' ', 14), ('e', 14)]

Answer 3

Your best bet is to use Counter (which does work on a string) and then sort on it's output.

from collections import Counter
sentence = input("Enter a sentence b'y: ")
lowercase = sentence.lower()

# Counter will work on strings
p = Counter(lowercase)
count = Counter.items()
# count is now (more or less) equivalent to
#  [('a', 1), ('r', 1), ('b', 1), ('o', 2), ('f', 1)]

# And now you can run your sort
sorted_count = sorted(count)
# Which will sort by the letter. If you wanted to
#  sort by quantity, tell the sort to use the
#  second element of the tuple by setting key:

# sorted_count = sorted(count, key=lambda x:x[1])

for letter, count in sorted_count:
    # will cycle through in order of letters.
    #  format as you wish
    print(letter, count)

Answer 4

If you just want to format the Counter output differently:

for key, value in Counter(list1).items():
    print('%s: %s' % (key, value))

Answer 5

Another way to avoid using Counter.

sentence = 'abc 11 222 a AAnn zzz?? !'
list1 = list(sentence.lower())
#If you want to remove the spaces.
#list1 = list(sentence.replace(" ", ""))

#Removing duplicate characters from the string  
sentence = ''.join(set(list1))
dict = {}
for char in sentence:
    dict[char] = list1.count(char)

for item in sorted(dict.items(), key=lambda x: x[1]):
    print 'Number of Occurences of %s is %d.' % (item[0], item[1])

Output:

Number of Occurences of c is 1.
Number of Occurences of b is 1.
Number of Occurences of ! is 1.
Number of Occurences of n is 2.
Number of Occurences of 1 is 2.
Number of Occurences of ? is 2.
Number of Occurences of 2 is 3.
Number of Occurences of z is 3.
Number of Occurences of a is 4.
Number of Occurences of   is 6.

Answer 6

One way to do this would be by removing instances of your sub string and looking at the length...

def nofsub(s,ss):
    return((len(s)-len(s.replace(ss,"")))/len(ss))

alternatively you could use re or regular expressions,

from re import *
def nofsub(s,ss):
    return(len(findall(compile(ss), s)))

finally you could count them manually,

def nofsub(s,ss):
    return(len([k for n,k in enumerate(s) if s[n:n+len(ss)]==ss]))

Test any of the three with...

>>> nofsub("asdfasdfasdfasdfasdf",'asdf')
5

Now that you can count any given character you can iterate through your string's unique characters and apply a counter for each unique character you find. Then sort and print the result.

def countChars(s):
    s = s.lower()
    d = {}
    for k in set(s):
        d[k]=nofsub(s,k)
    for key, value in sorted(d.iteritems(), key=lambda (k,v): (v,k)):
        print "%s: %s" % (key, value)

Answer 7

You could use the list function to break the words apart`from collections

from collections import Counter

sentence=raw_input("Enter a sentence b'y: ")
lowercase=sentence.lower()

list1=list(lowercase)
list(list1)

length=len(list1)
list2=list1.count(list1)
print(list2)

p=Counter(list1)
print(p)