def truncate_nested_dicts(list_of_dicts, trunc_size):
outlist = []
outdict = {}
for d in list_of_dicts:
for k,v in list(d.items())[:trunc_size]:
outdict[k]=v
outlist.append(outdict)
outdict={}
return outlist
Something like the following should do the trick. Note that this just keeps the trunc_size number of nested dictionaries. Not specific to entries a and b. It's unclear which was needed, but both answers currently on this question will yield different results if your keys in your dictionaries change.
def truncate_nested_dicts(list_of_dicts, trunc_size):
outlist = []
outdict = {}
for d in list_of_dicts:
for k,v in list(d.items())[:trunc_size]:
outdict[k]=v
outlist.append(outdict)
outdict={}
return outlist
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评论（2）
您可以使用列表理解来执行此操作：
循环在列表上，只需从dict中获取
a值即可。对于myNewlist2，只需做同样的事情，但是有2个值：You can use list comprehension to do this:
Loop over the list and just grab the
avalue from the dict. Formynewlist2, just do the same thing, but with 2 values:像以下内容应该解决问题。请注意，这只会保留
trunc_size嵌套词典的数字。不是条目a和b的特定特定条目。目前尚不清楚需要哪个，但是如果您的词典中的键更改，目前在这个问题上的两个答案都将产生不同的结果。Something like the following should do the trick. Note that this just keeps the
trunc_sizenumber of nested dictionaries. Not specific to entriesaandb. It's unclear which was needed, but both answers currently on this question will yield different results if your keys in your dictionaries change.