Question

我需要编写一个切割字符串的代码（这是一个输入），将其附加到列表中，计算每个字母的数量 - 如果它与之前的字母相同，则不要将其放入列表，而是增加之前的那个字母的出现次数.. 那么它应该是这样的：

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

刺客这个词只是需要的一个例子.. 到目前为止我的代码是这样的：

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

谢谢！

Answer 1

使用Collections.Counter()，字典是存储它的更好方法：

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

或使用itertools.groupby()：

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

Answer 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)]

Answer 3

仅使用内置：

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')

计算字符串中的连续字符

3 个答案: