你近了

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

如果您需要知道11内部dicts中有多少个作为键，则可以：

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

之所以可行，是因为每个密钥只能进入dict一次，因此您只需测试密钥是否退出即可。in返回True或False等于1和0，因此sum是的出现次数idnum。