不使用try填充字典 [英] fill a dictionary without using try except

问题描述

假设我有字典，我想用一些键和值来填充它，第一个字典为空，并假设我需要这个字典作为计数器，例如用这种方式对字符串中的一些键进行计数:

Suppose I have dictionary and I want to fill that with some keys and values , first dictionary is empty and suppose I need this dictionary for a counter for example count some keys in a string I have this way:

myDic = {}
try :
    myDic[desiredKey] += 1
except KeyError:
    myDic[desiredKey] = 1

或者也许某些时候值应该是一个列表，而我需要以这种方式将一些值附加到值列表中:

Or maybe some times values should be a list and I need to append some values to a list of values I have this way:

myDic = {}
try:
    myDic[desiredKey].append(desiredValue)
except KeyError:
    myDic[desiredKey] = []
    myDic[desiredKey].append(desiredValue)

对于不使用try except部分的作品(两者)是否都有更好的替代?

Is there better alternarive for this works (both of them) that dont uses try except section?

推荐答案

您可以使用collections.defaultdict()，它允许您提供一个函数，该函数在每次访问丢失的键时都会被调用.

You can use collections.defaultdict() which allows you to supply a function that will be called each time a missing key is accessed.

对于第一个示例，您可以将int作为缺少的函数传递给它，该函数在首次访问时将被评估为0.

And for your first example you can pass the int to it as the missing function which will be evaluated as 0 at first time that it gets accessed.

第二个可以通过list.

示例:

from collections import defaultdict 
my_dict = defaultdict(int)

>>> lst = [1,2,2,2,3] 
>>> for i in lst:
...      my_dict[i]+=1
... 
>>> 
>>> my_dict
defaultdict(<type 'int'>, {1: 1, 2: 3, 3: 1})
>>> my_dict = defaultdict(list)
>>> 
>>> for i,j in enumerate(lst):
...     my_dict[j].append(i)
... 
>>> my_dict
defaultdict(<type 'list'>, {1: [0], 2: [1, 2, 3], 3: [4]})

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