我如何計算，而不是在Python中重複相同的字母

Question

這段代碼計算字母，但我不希望它重寫兩次相同的字母。我能做什麼？

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

輸出：

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

我不希望它改寫 '2' 或 'T 2' 兩次

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 

或保持訂單使用OrderedDict從collections模塊：

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 

，或者您也可以使用來自itertools Recipes的unique_everseen配方：

>>> 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

使用裝置然後 - 這允許它只有唯一項目：

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

或者使用collections.Counter計算量可迭代的獨特項目：

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

而且somethig這樣的：

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