Counting consecutive characters in a string

问题

I need to write a code that slices the string (which is an input), append it to a list, count the number of each letter - and if it is identical to the letter before it, don't put it in the list, but rather increase the appearance number of that letter in the one before.. Well this is how it should look like :

assassin [['a', 1], ['s', 2], ['a', 1], ['s', 2]], ['i', 1], ['n', 1]

the word assassin is just an example of the need.. My code so far goes like this:

userin = raw_input("Please enter a string :")
inputlist = []
inputlist.append(userin)
biglist = []
i=0
count = {}
while i<(len(userin)):
    slicer = inputlist[0][i]
    for s in userin:
        if count.has_key(s):
            count[s] += 1
        else:
            count[s] = 1
    biglist.append([slicer,s])
    i = i+1
print biglist

Thanks!

回答1:

Use Collections.Counter(), dictionary is a better way to store this:

>>> from collections import Counter
>>> strs="assassin"
>>> Counter(strs)
Counter({'s': 4, 'a': 2, 'i': 1, 'n': 1})

or using itertools.groupby():

>>> [[k, len(list(g))] for k, g in groupby(strs)]
[['a', 1], ['s', 2], ['a', 1], ['s', 2], ['i', 1], ['n', 1]]

回答2:

last = ''
results = []
word = 'assassin'
for letter in word:
if letter == last:
    results[-1] = (letter, results[-1][1] +1)
else:
    results.append((letter, 1))
    last = letter

print result # [('a', 1), ('s', 2), ('a', 1), ('s', 2), ('i', 1), ('n', 1)]

回答3:

Using only builtins:

def cnt(s):
    current = [s[0],1]
    out = [current]
    for c in s[1:]:
        if c == current[0]:
            current[1] += 1
        else:
            current = [c, 1]
            out.append(current)
    return out

print cnt('assassin')

来源：https://stackoverflow.com/questions/13197668/counting-consecutive-characters-in-a-string

标签

python

string

count

character