blocks|key|1255847|text|for+key,+value+in+your_dict.items():
++++if+key+not+in+your_blacklisted_set:
++++++++print+value|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1255848|美妙之处在于这个伪代码示例是有效的python代码。|unstyled|1255849|它也可以表示为列表理解：|1255850|resultset+=+[value+for+key,+value+in+your_dict.items()+if+key+not+in+your_blacklisted_set]|1255851|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$B|C]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>for key, value in your_dict.items():
 if key not in your_blacklisted_set:
 print value
</code></pre>

the beauty is that this pseudocode example is valid python code.

it can also be expressed as a list comprehension:

<pre><code>resultset = [value for key, value in your_dict.items() if key not in your_blacklisted_set]
</code></pre>

blocks|key|4053672|text|如果您的目标是返回一个新字典，其中包含除一个或几个键/值之外的所有键/值，请使用以下方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4053673|exclude_keys+=+['exclude',+'exclude2']
new_d+=+{k:+d[k]+for+k+in+set(list(d.keys()))+-+set(exclude_keys)}|code-block|syntax|javascript|4053674|其中'exclude'可以由应该被排除的(一个或多个)键(列表)替换。|offset|length|style|CODE|4053675|entityMap^0|0|0|2|9|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If your goal is to return a new dictionary, with all key/values except one or a few, use the following:

<pre><code>exclude_keys = ['exclude', 'exclude2']
new_d = {k: d[k] for k in set(list(d.keys())) - set(exclude_keys)}
</code></pre>

where <code>'exclude'</code> can be replaced by (a list of) key(s) which should be excluded.

blocks|key|4053616|text|只是为了玩玩布景|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4053617|keys+=+set(dict.keys())
excludes+=+set([...])

for+key+in+keys.difference(excludes):
++++print+dict[key]|code-block|syntax|javascript|4053618|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Just for fun with sets

<pre><code>keys = set(dict.keys())
excludes = set([...])

for key in keys.difference(excludes):
 print dict[key]
</code></pre>

blocks|key|4053634|text|keys+=+['a',+'b']
a_dict+=+{'a':1,+'b':2,+'c':3,+'d':4}
[a_dict.pop(key)+for+key+in+keys]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4053635|在取出要丢弃的密钥后，a_dict将保留您要丢弃的密钥。|unstyled|4053636|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>keys = ['a', 'b']
a_dict = {'a':1, 'b':2, 'c':3, 'd':4}
[a_dict.pop(key) for key in keys]
</code></pre>

After popping out the keys to be discarded, a_dict will retain the ones you're after.

blocks|key|12988|text|给出一本字典说|type|unstyled|depth|inlineStyleRanges|entityRanges|data|12989|d+=+{
+++++2:+2,+5:+16,+6:+5,
+++++7:+6,+11:+17,+12:+9,
+++++15:+18,+16:+1,+18:+16,
+++++19:+17,+20:+10
+++++}|code-block|syntax|javascript|12990|然后，这个简单的理解示例就会达到您所希望的效果|12991|[v+for+k,v+in+d.iteritems()+if+k+not+in+(2,5)]|12992|此示例列出了不带键{2,5}的所有值|12993|例如，上述理解的O/P为|12994|[5,+6,+1,+17,+9,+18,+1,+16,+17,+10]|12995|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Given a dictionary say

<pre><code>d = {
 2: 2, 5: 16, 6: 5,
 7: 6, 11: 17, 12: 9,
 15: 18, 16: 1, 18: 16,
 19: 17, 20: 10
 }
</code></pre>

then the simple comprehension example would attain what you possibly desire

<pre><code>[v for k,v in d.iteritems() if k not in (2,5)]
</code></pre>

This example lists all values not with keys {2,5}

for example the O/P of the above comprehension is

<pre><code>[5, 6, 1, 17, 9, 18, 1, 16, 17, 10]
</code></pre>

blocks|key|12976|text|此外，作为使用集合的列表理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|12977|d+=+dict(zip(range(9),"abcdefghi"))
blacklisted+=+[2,5]
outputs+=+[d[k]+for+k+in+set(d.keys())-set(blacklisted)]|code-block|syntax|javascript|12978|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Also, as a list comprehension using sets:

<pre><code>d = dict(zip(range(9),"abcdefghi"))
blacklisted = [2,5]
outputs = [d[k] for k in set(d.keys())-set(blacklisted)]
</code></pre>

blocks|key|4053554|text|下面的内容如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4053555|In+[7]:+d+=+dict((i,i%2B100)+for+i+in+xrange(10))

In+[8]:+d
Out[8]:+
{0:+100,
+1:+101,
+2:+102,
+3:+103,
+4:+104,
+5:+105,
+6:+106,
+7:+107,
+8:+108,
+9:+109}

In+[9]:+exc+=+set((2,+5))

In+[10]:+for+k,+v+in+d.items():
+++....:+++++if+k+not+in+exc:
+++....:+++++++++print+v
+++....:+++++++++
+++....:+++++++++
100
101
103
104
106
107
108
109|code-block|syntax|javascript|4053556|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

How about something along the following lines:

<pre><code>In [7]: d = dict((i,i+100) for i in xrange(10))

In [8]: d
Out[8]: 
{0: 100,
 1: 101,
 2: 102,
 3: 103,
 4: 104,
 5: 105,
 6: 106,
 7: 107,
 8: 108,
 9: 109}

In [9]: exc = set((2, 5))

In [10]: for k, v in d.items():
 ....: if k not in exc:
 ....: print v
 ....: 
 ....: 
100
101
103
104
106
107
108
109
</code></pre>

blocks|key|4053705|text|使用异常处理|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4053706|facebook_posts+=+[
++++{'Likes':+21,+'Comments':+2},+
++++{'Likes':+13,+'Comments':+2,+'Shares':+1},+
++++{'Likes':+33,+'Comments':+8,+'Shares':+3},+
++++{'Comments':+4,+'Shares':+2},+
++++{'Comments':+1,+'Shares':+1},+
++++{'Likes':+19,+'Comments':+3}
]

total_likes+=+0

for+post+in+facebook_posts:
++++try:
++++++++total_likes+=+total_likes+%2B+post['Likes']
++++except+KeyError:
++++++++pass
print(total_likes)|code-block|syntax|javascript|4053707|您还可以使用：|4053708|except+KeyError:+'My+Key'|4053709|但可能只有一次好用|4053710|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

using exception handling
<pre><code>facebook_posts = [
 {'Likes': 21, 'Comments': 2}, 
 {'Likes': 13, 'Comments': 2, 'Shares': 1}, 
 {'Likes': 33, 'Comments': 8, 'Shares': 3}, 
 {'Comments': 4, 'Shares': 2}, 
 {'Comments': 1, 'Shares': 1}, 
 {'Likes': 19, 'Comments': 3}
]

total_likes = 0

for post in facebook_posts:
 try:
 total_likes = total_likes + post['Likes']
 except KeyError:
 pass
print(total_likes)
</code></pre>
you can also use:
<pre><code>except KeyError: 'My Key'
</code></pre>
but probably only good for once off use

blocks|key|4053738|text|如果您打算修改结果字典，请注意使用浅层副本。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4053739|import+copy

def+copy_dict_except_keys_shallow(d:dict,+exclude_keys):
++return+{k:+d[k]+for+k+in+set(list(d.keys()))+-+set(exclude_keys)}

def+copy_dict_except_keys_deep(d:dict,+exclude_keys):
++return+{k:+copy.deepcopy(d[k])+for+k+in+set(list(d.keys()))+-+set(exclude_keys)}

original+=+{'a':+1,+'b':+[1,+2,+3]}
deep+=+copy_dict_except_keys_deep(original,+['a'])
deep['b'].append(4)

print("Modifying+the+deep+copy+preserves+the+original")
print(f"original:+{original}")
print(f"deep+copy:+{deep}")
print()

shallow+=+copy_dict_except_keys_shallow(original,+'a')
shallow['b'].append(4)

print("Modifying+the+shallow+copy+changes+the+original")
print(f"original:+{original}")
print(f"shallow+copy:+{shallow}")|code-block|syntax|javascript|4053740|Modifying+the+deep+copy+preserves+the+original
original:+{'a':+1,+'b':+[1,+2,+3]}
deep+copy:+{'b':+[1,+2,+3,+4]}

Modifying+the+shallow+copy+changes+the+original
original:+{'a':+1,+'b':+[1,+2,+3,+4]}
shallow+copy:+{'b':+[1,+2,+3,+4]}|4053741|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|D|7|M|8|@]|9|@]|A|$E|F]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Be careful of using a shallow copy if you intend to modify the resulting dict.
<pre><code>import copy

def copy_dict_except_keys_shallow(d:dict, exclude_keys):
 return {k: d[k] for k in set(list(d.keys())) - set(exclude_keys)}

def copy_dict_except_keys_deep(d:dict, exclude_keys):
 return {k: copy.deepcopy(d[k]) for k in set(list(d.keys())) - set(exclude_keys)}

original = {'a': 1, 'b': [1, 2, 3]}
deep = copy_dict_except_keys_deep(original, ['a'])
deep['b'].append(4)

print(&quot;Modifying the deep copy preserves the original&quot;)
print(f&quot;original: {original}&quot;)
print(f&quot;deep copy: {deep}&quot;)
print()

shallow = copy_dict_except_keys_shallow(original, 'a')
shallow['b'].append(4)

print(&quot;Modifying the shallow copy changes the original&quot;)
print(f&quot;original: {original}&quot;)
print(f&quot;shallow copy: {shallow}&quot;)
</code></pre>
<pre><code>Modifying the deep copy preserves the original
original: {'a': 1, 'b': [1, 2, 3]}
deep copy: {'b': [1, 2, 3, 4]}

Modifying the shallow copy changes the original
original: {'a': 1, 'b': [1, 2, 3, 4]}
shallow copy: {'b': [1, 2, 3, 4]}
</code></pre>

I have a python dictionary. Just to give out context, I am trying to write my own simple cross validation unit.

So basically what I want is to get all the values except for the given keys.
And depending on the input, it returns all the values from a dictionary except to those what has been given.

So if the input is 2 and 5 then the output values doesn't have the values from the keys 2 and 5?

Retain all entries except for one key python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一本蟒蛇字典。为了给出上下文，我正在尝试编写自己的简单的交叉验证单元。所以基本上我想要的是获得除了给定键之外的所有值。根据输入，它将返回字典中除已给定值以外的所有值。那么，如果输入是2和5，那么输出值就没有来自键2和5的值了？

问保留除一个密钥python之外的所有条目
EN

回答 9

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