如何从字典1中删除基于公共键不在字典2中的键值?

Question

我有两个大字典，两个字典有相同的键(图像名称)和不同的值。

第一个名为train_descriptions的字典，看起来像这样:

{'15970.jpg': 'Turtle Check Men Navy Blue Shirt',
'39386.jpg': 'Peter England Men Party Blue Jeans',
'59263.jpg': 'Titan Women Silver Watch',
....
....
'1855.jpg': 'Inkfruit Mens Chain Reaction T-shirt'}

和第二个名为train_features的字典

{'31973.jpg': array([[0.00125694, 0.        , 0.03409385, ..., 0.00434341, 0.00728011,
0.01451511]], dtype=float32),
'30778.jpg': array([[0.0174035 , 0.04345186, 0.00772929, ..., 0.02230316, 0.        ,
0.03104496]], dtype=float32),
...,
...,
'38246.jpg': array([[0.00403965, 0.03701203, 0.02616892, ..., 0.02296285, 0.00930257,
0.04575242]], dtype=float32)}

两个字典的长度如下:

len(train_descriptions)= 44424,len(train_features)= 44441

可以看到train_description的长度小于train_features的长度。train_features字典比train_descriptions字典有更多的键值。如何从train_features字典中删除不在train_description中的键?使它们的长度相同

Answer 1

使用xor获取字典之间的差异

diff = train_features.keys() ^ train_descriptions.keys()
for k in diff:
del train_features[k]

Answer 2

使用for loop

feat = train_features.keys()
desc = train_description.keys()
common = list(i for i in feat if i not in decc)
for i in common: del train_features[i]

编辑:见下文

以上代码可以工作。但是我们可以通过不将dict_keys转换为list来更有效地做到这一点，如下所示:

for i in train_features.keys() - train_description.keys(): del train_features[i]

当python dict_keys被减去时，它会给出不常见键的dict_keys。第一个代码首先转换为列表，这既不高效也不需要。

Answer 3

如果pop()不存在，则仅为CC_13。

for key in train_descriptions.keys():
if key not in train_features.keys():
train_features.pop(key)