在嵌套的Python字典中搜索密钥

Question

I have some Python dictionaries like this:

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

eg

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

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.

This is my try:

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

Answer 1

dpath to the rescue.

http://github.com/akesterson/dpath-python

dpath lets you search by globs, which will get you what you want.

$ easy_install dpath
>>> for (path, value) in dpath.util.search(MY_DICT, '*/11', yielded=True):
>>> ... # 'value' will contain your condition; now do something with it.

It will iterate out all of the conditions in the dictionary, so no special looping constructs required.

See also

• how do i traverse nested dictionaries (python)?
• How to do this - python dictionary traverse and search
• Access nested dictionary items via a list of keys?
• Find all occurrences of a key in nested python dictionaries and lists
• Traverse a nested dictionary and get the path in Python?
• Find all the keys and keys of the keys in a nested dictionary
• Searching for keys in a nested dictionary
• Python: Updating a value in a deeply nested dictionary
• Is there a query language for JSON?
• Chained, nested dict() get calls in python

Answer 2

You're close.

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

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

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

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 .