2 回答

TA贡献1772条经验 获得超5个赞
dict1 = {"AD":{(62,'VMX','Barcelona','ES'): ['123','567','666']}} #dict1 many more rows
dict2 = {"AD":{(151,'CMXS','','ES','62'): ['345','6754']}} #dict 2 many more rows
# create a simplified map of dict2
dict3 = {}
for outer_key, outer_value_dict in dict2.items():
for inner_key_tuple, inner_value_list in outer_value_dict.items():
new_key = (outer_key, inner_key_tuple[4]) # e.g. (AD, 62)
dict3.setdefault(new_key, []).extend(inner_value_list)
# now modify dict1 using dict3 instead of dict2
for outer_key, outer_value in dict1.items():
for inner_key_tuple, inner_value_list in outer_value.items():
lookup_key = (outer_key, str(inner_key_tuple[0]))
# search dict3 using the lookup_key.
# if not present, get will return an empty list
list_to_extend = dict3.get(lookup_key, [])
inner_value_list.extend(list_to_extend)
print(dict1)
输出
/tmp $ python test.py.py
{'AD': {(62, 'VMX', 'Barcelona', 'ES'): ['123', '567', '666', '345', '6754']}}
根据评论,因为您想知道如何避免其中一个循环 - 这是更新版本
for k,v in dict1.items():
a = dict2.get(k)
if a is None or len(v) == 0:
continue
for k2,v2 in v.items():
for c2,a2 in a.items():
if str(k2[0]) == c2[4]:
v2.append(a2)

TA贡献1829条经验 获得超13个赞
只要示例进行,for 循环就会被正确放置。
出乎意料是什么意思?你是这个意思吗?
{'AD': {(62, 'VMX', 'Barcelona', 'ES'): ['123', '567', '666', ['345', '6754']]}}
数组中的那个额外数组?
因为如果是这样,答案就是用 .extend() 切换 .append()
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