python - 我如何计算而不是在python中重复相同的字母

Question

此代码计算字母，但我不希望它重写同一个字母两次。我能做些什么？

d=list('banttnay')
for letter in d:
    print letter, d.count(letter)

输出：

b 1
a 2
n 2
t 2
t 2
n 2
a 2
y 1

我不希望它重写 'a 2' 或 't 2' 两次

Answer 1

使用set()：

In [40]: d='banttnay'

In [41]: for x in set(d):
   ....:     print x,d.count(x)
   ....:     
a 2
y 1
b 1
t 2
n 2

OrderedDict 或从collections模块维护订单使用：

In [67]: from collections import *

In [68]: c=OrderedDict()

In [69]: for x in d:
    c[x]=c.get(x,0)+1
In [71]: for x in c:
   ....:     print x,c[x]
   ....:     
b 1
a 2
n 2
t 2
y 1

或者您也可以使用以下unique_everseen配方itertools Recipes：

>>> d='banttnay'
>>> from itertools import *
>>> def unique_everseen(iterable, key=None):
...     seen = set()
...     seen_add = seen.add
...     if key is None:
...         for element in ifilterfalse(seen.__contains__, iterable):
...             seen_add(element)
...             yield element
...     else:
...         for element in iterable:
...             k = key(element)
...             if k not in seen:
...                 seen_add(k)
...                 yield element
... 
>>> for x in unique_everseen(d):
...     print x,d.count(x)
... 
b 1
a 2
n 2
t 2
y 1

Answer 2

然后使用集合 - 它只允许其中的唯一项目：

d = 'banttnay'
for letter in set(d):
    print letter, d.count(letter)

或使用collections.Counter计算可迭代项中唯一项的数量：

   from collections import Counter
   cnt = Counter('banttnay')
   print cnt

像这样的东西：

word = 'banttnay'
res = dict(zip(set(word),map(word.count, word)))