[英]Checking how many times a digit appears in a dictionary in python?
I'm trying to check how many times a digit appears in a dictionary. 我正在尝试检查数字在词典中出现了多少次。
This code only works if I input a single digit. 仅当我输入一位数字时,此代码才有效。
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)
For example if I input 8
the output will be 例如,如果我输入8
则输出为
{'8': 1, '9': 0, '4': 0, '5': 0, '6': 0, '7': 0, '0': 0, '1': 0, '2': 0, '3': 0}
But if I input 887655
then it gives me a builtins.KeyError: '887655'
但是,如果我输入887655
那么它会给我一个builtins.KeyError: '887655'
If I input 887655
the output should actually be 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}
Use collections.Counter
for this instead - no need to reinvent the wheel. 为此,请使用collections.Counter
无需重新发明轮子。
>>> import collections
>>> collections.Counter("887655")
Counter({'8': 2, '5': 2, '6': 1, '7': 1})
You should probably change 你应该改变
d[numbers] += 1
=> =>
d[i] += 1
I think what you want is actually this: 我认为您想要的实际上是这样的:
for number in numbers:
for digit in str(number):
d[digit] += 1
You should use collections.Counter
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"
I would recommend collections.Counter
but here's an improved version of your code: 我建议使用collections.Counter
但这是您代码的改进版本:
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)
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