将 output 作为字符并出现在字符串中

Question

The input I gave is like this

Input- abccdddeffg

and the output I want is character and it's occurrence number

Output- a1b1c2d3e1f2g1

my code

uni = []
string = 'abcccdd'
for i in range(0, len(string)):
    for j in range(i+1, len(string)):
        if (string[i] == string[j]):
            uni.append(string[i])
            
for oc in uni:
    cou= uni.count(oc)
    print(oc,cou)

Thanks in advance

Answer 1

You can use the Counter from collections to get the count of every character in the list. Then use a forloop to generate your result and use set to make sure no character is repeated in the result.

from collections import Counter    
string = "abccdddeffg"
counts = Counter(string)
sets = set()
result = []

for s in string:
    if s not in sets:
        result.append(f"{s}{counts[s]}")
    sets.add(s)

result = ''.join(result)
print(result)

Answer 2

Firstly, your output contains an error, there should be 1 next to e as it was there in first occurring characters.

After clearing this, this is what you need:

import collections

s = "abccdddeffg"  # your string

a = dict((letter,s.count(letter)) for letter in set(s))
a = collections.OrderedDict(sorted(a.items()))

answer = "" # to store the result

for i,j in zip(a.keys(),a.values()):
    answer+= i + str(j)

print(answer)

Answer will return:

'a1b1c2d3e1f2g1'

Answer 3

Here is a simpler approach:

def freq_string(string):
    output, buffer = "", ""
    for letter in string:
        if buffer != letter:
            buffer = letter
            output += f"{letter}{string.count(letter)}"
    return output

Note: this does assume that same characters are in succession rather than spread randomly around the string.

Answer 4

Assuming your input string is sorted and has at least one letter, you can do a simple loop to handle it:

if len(string) == 1:
    # print out the only string and 1 as its occurences
    print(string + '1')
else:
    # initialize first string, its counter, and our result string
    prev = string[0]
    counter = 1
    result = ''
    
    # loop over each letter
    for letter in string[1:]:
        curr = letter
        
        # if current letter is different from previous letter,
        # concat the result and refresh our counter
        # else just increase the counter
        if curr != prev:
            result = result + prev + str(counter)
            counter = 1
        else:
            counter = counter + 1
        prev = curr
        
    # don't forget to handle the last case
    result = result + prev + str(counter)
    print(result)

Answer 5

in the simplest way:

inp = 'abccdddeffg'

l=[]
o=""
for i in inp:
    if i in l:
        pass
    else:
        l.append(i)
        o+="{}{}".format(i,inp.count(i))
print(o)

output

'a1b1c2d3e1f2g1'