检查一个数字在python字典中出现的次数？

Question

我试图检查一个数字出现在字典中的次数。

此代码仅适用于输入单个数字的情况。

numbers = input("Enter numbers ") 

d = {} 
d['0'] = 0 
d['1'] = 0 
d['2'] = 0 
d['3'] = 0 
d['4'] = 0 
d['5'] = 0 
d['6'] = 0 
d['7'] = 0 
d['8'] = 0 
d['9'] = 0 

for i in numbers: 
    d[numbers] += 1 

print(d) 

例如，如果我输入8输出将是

{'8': 1, '9': 0, '4': 0, '5': 0, '6': 0, '7': 0, '0': 0, '1': 0, '2': 0, '3': 0} 

但是，如果我输入887655然后它给了我一个builtins.KeyError: '887655'

如果我输入887655输出实际上应该是

{'8': 2, '9': 0, '4': 0, '5': 2, '6': 1, '7': 1, '0': 0, '1': 0, '2': 0, '3': 0} 

Answer 1

你或许应该改变

d[numbers] += 1 

=>

d[i] += 1 

Answer 2

使用collections.Counter代替 - 不需要重新发明轮子。

>>> import collections 
>>> collections.Counter("887655") 
Counter({'8': 2, '5': 2, '6': 1, '7': 1}) 

Answer 3

我想你想要的其实是这样的：

for number in numbers: 
    for digit in str(number): 
     d[digit] += 1 

Answer 4

您应该使用collections.Counter

from collections import Counter 
numbers = input("Enter numbers: ") 
count = Counter(numbers) 
for c in count: 
    print c, "occured", count[c], "times" 

Answer 5

我会建议collections.Counter但这里是你的代码的改进版本：

numbers = input("Enter numbers ") 

d = {} # no need to initialize each key 

for i in numbers: 
    d[i] = d.get(i, 0) + 1 # we can use dict.get for that, default val of 0 

print(d)