blocks|key|1068901|text|一个简单的可读性版本是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1068902|def+find(lst,+key,+value):
++++for+i,+dic+in+enumerate(lst):
++++++++if+dic[key]+==+value:
++++++++++++return+i
++++return+-1|code-block|syntax|javascript|1068903|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|$]]

A simple readable version is

<pre><code>def find(lst, key, value):
 for i, dic in enumerate(lst):
 if dic[key] == value:
 return i
 return -1
</code></pre>

blocks|key|1068925|text|下面是一个查找字典索引位置(如果存在)的函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1068926|dicts+=+[{'id':'1234','name':'Jason'},
+++++++++{'id':'2345','name':'Tom'},
+++++++++{'id':'3456','name':'Art'}]

def+find_index(dicts,+key,+value):
++++class+Null:+pass
++++for+i,+d+in+enumerate(dicts):
++++++++if+d.get(key,+Null)+==+value:
++++++++++++return+i
++++else:
++++++++raise+ValueError('no+dict+with+the+key+and+value+combination+found')

print+find_index(dicts,+'name',+'Tom')
#+1
find_index(dicts,+'name',+'Ensnare')
#+ValueError:+no+dict+with+the+key+and+value+combination+found|code-block|syntax|javascript|1068927|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|$]]

Here's a function that finds the dictionary's index position if it exists.

<pre><code>dicts = [{'id':'1234','name':'Jason'},
 {'id':'2345','name':'Tom'},
 {'id':'3456','name':'Art'}]

def find_index(dicts, key, value):
 class Null: pass
 for i, d in enumerate(dicts):
 if d.get(key, Null) == value:
 return i
 else:
 raise ValueError('no dict with the key and value combination found')

print find_index(dicts, 'name', 'Tom')
# 1
find_index(dicts, 'name', 'Ensnare')
# ValueError: no dict with the key and value combination found
</code></pre>

blocks|key|854533|text|使用过滤器/索引组合似乎是最符合逻辑的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|854534|names=[{},+{'name':+'Tom'},{'name':+'Tony'}]
names.index(next(filter(lambda+n:+n.get('name')+==+'Tom',+names)))
1|code-block|syntax|javascript|854535|如果你认为可能有多个匹配：|854536|[names.index(item)+for+item+in+filter(lambda+n:+n.get('name')+==+'Tom',+names)]
[1]|854537|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Seems most logical to use a filter/index combo:
<pre><code>names=[{}, {'name': 'Tom'},{'name': 'Tony'}]
names.index(next(filter(lambda n: n.get('name') == 'Tom', names)))
1
</code></pre>
And if you think there could be multiple matches:
<pre><code>[names.index(item) for item in filter(lambda n: n.get('name') == 'Tom', names)]
[1]
</code></pre>

blocks|key|854572|text|对于给定的迭代值，more_itertools.locate将生成满足谓词的项的位置。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|854573|import+more_itertools+as+mit


iterable+=+[
++++{"id":+"1234",+"name":+"Jason"},
++++{"id":+"2345",+"name":+"Tom"},
++++{"id":+"3456",+"name":+"Art"}
]

list(mit.locate(iterable,+pred=lambda+d:+d["name"]+==+"Tom"))
#+[1]|code-block|syntax|javascript|854574|more_itertools是一个第三方库，它实现了itertools+recipes以及其他有用的工具。|854575|entityMap|0|LINK|mutability|MUTABLE|url|https://more-itertools.readthedocs.io/en/latest/api.html#more_itertools.locate|1|https://github.com/erikrose/more-itertools|2|https://docs.python.org/3/library/itertools.html#itertools-recipes^0|9|L|9|L|0|0|0|0|E|0|E|1|Q|H|2|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@$9|18|A|19|1|1A]|$9|1B|A|1C|1|1D]]|E|$]]|$1|M|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]|W|$5|P|Q|R|E|$S|X]]]]

For a given iterable, <a href="https://more-itertools.readthedocs.io/en/latest/api.html#more_itertools.locate" rel="nofollow noreferrer"><code>more_itertools.locate</code></a> yields positions of items that satisfy a predicate.

<pre><code>import more_itertools as mit


iterable = [
 {"id": "1234", "name": "Jason"},
 {"id": "2345", "name": "Tom"},
 {"id": "3456", "name": "Art"}
]

list(mit.locate(iterable, pred=lambda d: d["name"] == "Tom"))
# [1]
</code></pre>

<a href="https://github.com/erikrose/more-itertools" rel="nofollow noreferrer"><code>more_itertools</code></a> is a third-party library that implements <a href="https://docs.python.org/3/library/itertools.html#itertools-recipes" rel="nofollow noreferrer">itertools recipes</a> among other useful tools.

blocks|key|854638|text|一个线条！？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|854639|elm+=+([i+for+i+in+mylist+if+i['name']+==+'Tom']+or+[None])[0]|code-block|syntax|javascript|854640|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|$]]

One liner!?

<pre><code>elm = ([i for i in mylist if i['name'] == 'Tom'] or [None])[0]
</code></pre>

blocks|key|854735|text|@faham提供的答案是一个很好的一行代码，但它不会返回包含值的字典的索引。相反，它返回字典本身。下面是一种简单的获取方法:如果有多个索引，则显示一个或多个索引列表；如果没有索引，则显示一个空列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|854736|list+=+[{'id':'1234','name':'Jason'},
++++++++{'id':'2345','name':'Tom'},
++++++++{'id':'3456','name':'Art'}]

[i+for+i,+d+in+enumerate(list)+if+'Tom'+in+d.values()]|code-block|syntax|javascript|854737|输出：|854738|>>>+[1]|854739|我喜欢这种方法的原因是，通过简单的编辑，您可以获得元组形式的索引和字典的列表。这就是我需要解决的问题，并找到了这些答案。在下面的代码中，我在另一个字典中添加了一个重复值，以显示它是如何工作的：|854740|list+=+[{'id':'1234','name':'Jason'},
++++++++{'id':'2345','name':'Tom'},
++++++++{'id':'3456','name':'Art'},
++++++++{'id':'4567','name':'Tom'}]

[(i,+d)+for+i,+d+in+enumerate(list)+if+'Tom'+in+d.values()]|854741|854742|>>>+[(1,+{'id':+'2345',+'name':+'Tom'}),+(3,+{'id':+'4567',+'name':+'Tom'})]|854743|此解决方案查找所有字典的任何值中都包含“Tom”。|854744|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|X|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Z|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|O|3|H|5|6|7|11|8|@]|9|@]|A|$]]|$1|P|3|Q|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|R|3|S|5|6|7|13|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|U|$]]

Answer offered by @faham is a nice one-liner, but it doesn't return the index to the dictionary containing the value. Instead it returns the dictionary itself. Here is a simple way to get: A list of indexes one or more if there are more than one, or an empty list if there are none:

<pre><code>list = [{'id':'1234','name':'Jason'},
 {'id':'2345','name':'Tom'},
 {'id':'3456','name':'Art'}]

[i for i, d in enumerate(list) if 'Tom' in d.values()]
</code></pre>

Output:

<pre><code>&gt;&gt;&gt; [1]
</code></pre>

What I like about this approach is that with a simple edit you can get a list of both the indexes and the dictionaries as tuples. This is the problem I needed to solve and found these answers. In the following, I added a duplicate value in a different dictionary to show how it works:

<pre><code>list = [{'id':'1234','name':'Jason'},
 {'id':'2345','name':'Tom'},
 {'id':'3456','name':'Art'},
 {'id':'4567','name':'Tom'}]

[(i, d) for i, d in enumerate(list) if 'Tom' in d.values()]
</code></pre>

Output:

<pre><code>&gt;&gt;&gt; [(1, {'id': '2345', 'name': 'Tom'}), (3, {'id': '4567', 'name': 'Tom'})]
</code></pre>

This solution finds all dictionaries containing 'Tom' in any of their values.

blocks|key|1069170|text|def+search(itemID,list):
+++++return[i+for+i+in+list+if+i.itemID==itemID]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1069171|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>def search(itemID,list):
 return[i for i in list if i.itemID==itemID]
</code></pre>

blocks|key|854822|text|我需要一个更通用的解决方案来说明列表中有多个字典具有键值的可能性，以及使用列表理解的简单实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|854823|dict_indices+=+[i+for+i,+d+in+enumerate(dict_list)+if+d[dict_key]+==+key_value]+|code-block|syntax|javascript|854824|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|$]]

I needed a more general solution to account for the possibility of multiple dictionaries in the list having the key value, and a straightforward implementation using list comprehension:
<pre><code>dict_indices = [i for i, d in enumerate(dict_list) if d[dict_key] == key_value] 
</code></pre>

blocks|key|1069287|text|下面将返回第一个匹配项的索引：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1069288|['Tom'+in+i['name']+for+i+in+list].index(True)|code-block|syntax|javascript|1069289|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|$]]

The following will return the index for the first matching item:
<pre><code>['Tom' in i['name'] for i in list].index(True)
</code></pre>

blocks|key|854935|text|我的答案最好放在字典里用一下|type|unstyled|depth|inlineStyleRanges|entityRanges|data|854936|food_time_dict+=+{"Lina":+312400,+"Tom":+360054,+"Den":+245800}
print(list(food_time_dict.keys()).index("Lina"))|code-block|syntax|javascript|854937|我从字典中请求关键字，然后翻译列表，如果没有添加，就会出现错误，然后将其用作列表。但是在你的代码中：|854938|lists+=+[{'id':+'1234',+'name':+'Jason'},
+++++++++{'id':+'2345',+'name':+'Tom'},
+++++++++{'id':+'3456',+'name':+'Art'}]
++++
++++
def+dict_in_lists_index(lists,+search):++#+function+for+convenience
++++j+=+0++#+[j][i]
++++for+i+in+lists:
++++++++try:++#+try+our+varible+search+if+not+found+in+list
++++++++++++return+f"[{j}][{list(i.values()).index(search)}]"
++++++++++++#+small+decor
++++++++except+ValueError:+#+error+was+ValueError
++++++++++++pass+#+aa...+what+must+was+what+you+want+to+do
++++++++j+%2B=+1+#+not+found?+ok+j%2B%2B
++++return+"Not+Found"
++++
++++
def+dict_cropped_index(lists,+search):
++++for+i+in+lists:
++++++++try:
++++++++++++return+list(i.values()).index(search)
++++++++except+ValueError:
++++++++++++pass
++++return+"Not+Found"
++++
++++
print(dict_in_lists_index(lists,+'Tom'))+#+and+end
print(dict_cropped_index(lists,+'Tom'))+#+now+for+sure+end|854939|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

my answer is better in one a dictionary to use
<pre class="lang-py prettyprint-override"><code>food_time_dict = {&quot;Lina&quot;: 312400, &quot;Tom&quot;: 360054, &quot;Den&quot;: 245800}
print(list(food_time_dict.keys()).index(&quot;Lina&quot;))
</code></pre>
I request keys from the dictionary, then I translate the list if it is not added, there will be an error then I use it as a list. but on your code:
<pre class="lang-py prettyprint-override"><code>lists = [{'id': '1234', 'name': 'Jason'},
 {'id': '2345', 'name': 'Tom'},
 {'id': '3456', 'name': 'Art'}]
 
 
def dict_in_lists_index(lists, search): # function for convenience
 j = 0 # [j][i]
 for i in lists:
 try: # try our varible search if not found in list
 return f&quot;[{j}][{list(i.values()).index(search)}]&quot;
 # small decor
 except ValueError: # error was ValueError
 pass # aa... what must was what you want to do
 j += 1 # not found? ok j++
 return &quot;Not Found&quot;
 
 
def dict_cropped_index(lists, search):
 for i in lists:
 try:
 return list(i.values()).index(search)
 except ValueError:
 pass
 return &quot;Not Found&quot;
 
 
print(dict_in_lists_index(lists, 'Tom')) # and end
print(dict_cropped_index(lists, 'Tom')) # now for sure end
</code></pre>

I have a list of dicts:

<pre><code>list = [{'id':'1234','name':'Jason'},
 {'id':'2345','name':'Tom'},
 {'id':'3456','name':'Art'}]
</code></pre>

How can I efficiently find the index position [0],[1], or [2] by matching on name = 'Tom'?

If this were a one-dimensional list I could do list.index() but I'm not sure how to proceed by searching the values of the dicts within the list.

Find the index of a dict within a list, by matching the dict's value

我有一个字典列表：list = [{'id':'1234','name':'Jason'},        {'id':'2345','name':'Tom'},        {'id':'3456','name':'Art'}]如何通过匹配name = 'Tom‘有效地找到索引位置1或2？如果这是一个一维列表，我可...

问通过匹配字典的值，在列表中查找字典的索引
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