在 Python 中搜索二维元组/列表
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Searching a 2-Dimensional Tuple/List in Python
提问by Koobz
I want to search a tuple of tuplesfor a particular string and return the index of the parent tuple. I seem to run into variations of this kind of search frequently.
我想搜索tuple of tuples特定字符串并返回父元组的索引。我似乎经常遇到这种搜索的变体。
What is the most pythonic way to do this?
什么是最pythonic的方法来做到这一点？
I.E:
IE：
derp = (('Cat','Pet'),('Dog','Pet'),('Spock','Vulcan'))
i = None
for index, item in enumerate(derp):
if item[0] == 'Spock':
i = index
break
>>>print i
2
I could generalize this into a small utility function that takes an iterable, an index (I've hard coded 0in the example) and a search value. It does the trick but I've got this notion that there's probably a one-liner for it ;)
我可以将其概括为一个小的实用程序函数，它接受一个可迭代对象、一个索引（我0在示例中进行了硬编码）和一个搜索值。它可以解决问题，但我有一个想法，即它可能是单行的 ;)
I.E:
IE：
def pluck(iterable, key, value):
for index, item in enumerate(iterable):
if item[key] == value:
return index
return None
采纳答案by HS.
It does the trick but I've got this notion that there's probably a one-liner for it ;)
它可以解决问题，但我有一个想法，即它可能是单行的 ;)
The one-liner is probably notthe pythonic way to do it :)
one-liner 可能不是Pythonic 的方法:)
The method you have used looks fine.
您使用的方法看起来不错。
Edit:
编辑：
If you want to be cute:
如果你想变得可爱：
return next( (i for i,(k,v) in enumerate(items) if k=='Spock'),None)
nexttakes a generator expression and returns the next value or the second argument (in this case None) once the generator has been exhausted.
next接受一个生成器表达式并None在生成器耗尽后返回下一个值或第二个参数（在本例中为）。
回答by Siddharth Srivastava
Or you can do:
或者你可以这样做：
dict(derp)[<key_name>]
eg.
例如。
dict(derp)['Cat']
which will give you 'Pet'
这会给你“宠物”
回答by Siddharth Srivastava
If you're often searching the same tuple, you can build a dict.
如果您经常搜索同一个元组，则可以构建一个 dict。
lookup_table = dict((key, i) for i, (key, unused) in enumerate(derp))
print lookup_table['Spock']
--> 2
回答by Mr Shark
Another way of getting it in one line would be:
在一行中获取它的另一种方法是：
[d[0] for d in derp].index("Spock")
I'm not sure if the iterator evaluates all values before calling index, and therefor being inefficient.
我不确定迭代器是否在调用 index 之前评估所有值，因此效率低下。
回答by ron
Lambdas are fun!
Lambda 很有趣！
return reduce(
lambda x,(i,(a,b)): i,
filter(
lambda (i,(a,b)): a == "Spock",
enumerate(depr)
),
None
)
