提问人：George 提问时间：11/1/2023 最后编辑：Goku - stands with PalestineGeorge 更新时间：11/1/2023 访问量：58
搜索两个列表，并按正确的顺序为每对填写“无”
search two lists and fill with None for each pair in correct order
问：
这是此的延续。
现在，结果是：
[{'type': 'POT'}, {'type': 'BOOK'}, {'type': 'GLASS'}]
如果找不到键，我想在结果中插入 a。None
所以，我想要：
[{'type': 'POT'}, {'type': 'BOOK'}, None, None, {'type': 'GLASS'}]
如果我尝试：
stuff_types = {
"spana": {
"type": "BOOK",
},
"geom": {
"type": "GLASS",
},
"hlian": {
"type": "POT",
}
}
a = ['kal', 'khp', 'khp', 'khp', 'geom']
b = ['hlian', 'spana', 'piper', 'meli', 'phin']
the_list = []
for tup in zip(a, b):
for x in tup:
if x in stuff_types:
the_list.append(stuff_types[x])
else:
the_list.append(None)
print(the_list)
只是为了附加，我们收到对中每个元素的：NoneNone
('kal', 'hlian')
('khp', 'spana')
('khp', 'piper')
('khp', 'meli')
('geom', 'phin')
我想在正确的位置接收每对，而不是对中的每个元素。None
答：
1赞
Goku - stands with Palestine
11/1/2023
#1
stuff_types = {
"spana": {
"type": "BOOK",
},
"geom": {
"type": "GLASS",
},
"hlian": {
"type": "POT",
}
}
a = ['kal', 'khp', 'khp', 'khp', 'geom']
b = ['hlian', 'spana', 'piper', 'meli', 'phin']
有一个轻微的修改，你不需要嵌套的for循环：
the_list = []
for k,v in zip(a, b):
if k in stuff_types:
the_list.append(stuff_types[k])
elif v in stuff_types:
the_list.append(stuff_types[v])
else:
the_list.append(None)
print(the_list)
#output
[None, {'type': 'BOOK'}, None, None, {'type': 'GLASS'}]
2赞
quamrana
11/1/2023
#2
您需要测试每个项目，以查看其中任何一个是否具有相关键：tup
stuff_types = {
"spana": {
"type": "BOOK",
},
"geom": {
"type": "GLASS",
},
"hlian": {
"type": "POT",
}
}
def make_value(x,y):
return stuff_types[x] if x in stuff_types else stuff_types[y] if y in stuff_types else None
a = ['kal', 'khp', 'khp', 'khp', 'geom']
b = ['hlian', 'spana', 'piper', 'meli', 'phin']
the_list = []
for (x,y) in zip(a, b):
the_list.append(make_value(x,y))
print(the_list)
按要求输出：
[{'type': 'POT'}, {'type': 'BOOK'}, None, None, {'type': 'GLASS'}]
评论
0赞
JonSG
11/1/2023
简化一下怎么样？我知道这并不是严格意义上的一回事，但给定数据，结果应该是相同的。return stuff_types.get(x) or stuff_types.get(y)
1赞
RomanPerekhrest
11/1/2023
#3
使用单遍历，基于交叉点：set
res = [stuff_types[s.pop()] if (s := set(stuff_types) & set(p)) else None
for p in zip(a, b)]
print(res)
[{'type': 'POT'}, {'type': 'BOOK'}, None, None, {'type': 'GLASS'}]
评论
NoneNonebreakNonebreakdict.get