在嵌套的Python词典中搜索键 [英] Search for a key in a nested Python dictionary

问题描述

我有一些这样的Python字典:

I have some Python dictionaries like this:

A = {id: {idnumber: condition},.... 

例如

A = {1: {11 : 567.54}, 2: {14 : 123.13}, .....

我需要搜索字典中是否有任何idnumber == 11，并使用condition进行计算.但是，如果整个字典中没有idnumber == 11，则需要继续使用 next 字典.

I need to search if the dictionary has any idnumber == 11 and calculate something with the condition. But if in the entire dictionary doesn't have any idnumber == 11, I need to continue with the next dictionary.

这是我的尝试:

for id, idnumber in A.iteritems():
    if 11 in idnumber.keys(): 
       calculate = ......
    else:
       break

推荐答案

您已经关闭.

idnum = 11
# The loop and 'if' are good
# You just had the 'break' in the wrong place
for id, idnumber in A.iteritems():
    if idnum in idnumber.keys(): # you can skip '.keys()', it's the default
       calculate = some_function_of(idnumber[idnum])
       break # if we find it we're done looking - leave the loop
    # otherwise we continue to the next dictionary
else:
    # this is the for loop's 'else' clause
    # if we don't find it at all, we end up here
    # because we never broke out of the loop
    calculate = your_default_value
    # or whatever you want to do if you don't find it

如果需要知道内部dict中有多少个11作为键，则可以:

If you need to know how many 11s there are as keys in the inner dicts, you can:

idnum = 11
print sum(idnum in idnumber for idnumber in A.itervalues())

之所以可行，是因为每个dict中只能有一个密钥，因此您只需测试密钥是否退出即可. in返回等于1和0的True或False，因此sum是idnum的出现次数.

This works because a key can only be in each dict once so you just have to test if the key exits. in returns True or False which are equal to 1 and 0, so the sum is the number of occurences of idnum.

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