计算字符串中字符的出现次数

Question

Hey guys I'm trying to get an output of a4b2c1a2d3 with an input of aaaabbcaaddd . I think the problem with my code below is the way I implement the counter for previous. Is there an issue with index > 0: ?

string1 = "aaaabbcaaddd" 
previous = ""
finalstr = ""
finalint = 1
for index, val in enumerate(string1):
    if index > 0:
        previous = string1[index - 1]
        if previous == val:
            finalint += 1
        else:
            finalstr += previous + str(finalint)
            finalint = 1
print(finalstr)

#outputs "a4b2c1a2"

Answer 1

As I said in the comment, you forgot to add the last count after the loop:

finalstr += previous + str(finalint)

However, unless you are doing this as an assigned homework, there is a much more compact way of solving the problem:

from itertools import groupby
''.join(char + str(len(list(group))) for char,group in groupby(string1))
#'a4b2c1a2d3'

Your own code could be rewritten without indexes (they always cause troubles):

cnt = 1
finalstr = ''
for x,y in zip(string1, string1[1:]):
    if x==y:
        cnt += 1
    else:
        finalstr += x + (str(cnt) if cnt > 1 else '')
        cnt = 1
finalstr += x + (str(cnt) if cnt > 1 else '')
finalstr
#'a4b2ca2d3'

Answer 2

This is called "run-length encoding." In the Python library more-itertools there's a function to do this encoding:

>>> from more_itertools import run_length
... 
... string1 = "aaaabbcaaddd"
... list(run_length.encode(string1))
[('a', 4), ('b', 2), ('c', 1), ('a', 2), ('d', 3)]

We can get your desired output by flattening this and string-ifying it.

>>> list(flatten(run_length.encode(string1)))
['a', 4, 'b', 2, 'c', 1, 'a', 2, 'd', 3]

>>> list(map(str, flatten(run_length.encode(string1))))
['a', '4', 'b', '2', 'c', '1', 'a', '2', 'd', '3']

>>> ''.join(map(str, flatten(run_length.encode(string1))))
'a4b2c1a2d3'

Answer 3

If anyone was curious of the solution I got round to (based off DYZ's help)

string1 = "aaaabbcaaddd" 
finalstr = ""
previous = ""
finalint = 0
for c, v in enumerate(string1):
    if c > 0:
        previous = string1[c - 1]
        if previous == v:
            finalint += 1
        else:
            finalstr += previous + str((finalint) if finalint > 1 else '')
            finalint = 1
finalstr += previous + str((finalint) if finalint > 1 else '')
print(finalstr)
#outputs a3b2ca2d3

Answer 4

from itertools import groupby
string1 = "aaaabbcaaddd"
string2 = ''
for c, g in groupby(string1):
    string2 += '%s%d'%(c, len(list(g)))

Answer 5

You could do the same using itertools groupby.

from itertools import groupby
string1 = 'aaaabbcaaddd'
result = [[k, len(list(g))] for k, g in groupby(string1)]
final = "".join([i[0]+str(i[1]) for i in result])
final 

output: #'a4b2c1a2d3'

Answer 6

I am new to programming, writing my first comment, i decided this:

string = "qaabbaaccaqwewdssss"
count = 0
temp = str()

for i in range(len(string) - 1):
    count += 1
    if string[i] != string[i + 1]:
        temp += string[i] + str(count)
        count = 0
    elif i + 1 == len(string) - 1:
        count += 1
        temp += string[i] + str(count)
        count = 0

print(temp)
#output: q1a2b2a2c2a1q1w1e1w1d1s4

Answer 7

Simply by using groupby from itertools this can be done. And there is no need to put everything in one line.

import itertools

s = "aaaabbcaaddd"

iterator = itertools.groupby(s)
f= ""

for key, group in iterator:
    f += str(key) + str(len(list(group)))

print(f)

#output: a4b2c1a2d3