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问题：在Python中，如何将一个列表与另一个列表建立索引？
回答 0
回答 1
回答 2
回答 3
回答 4
回答 5
回答 6

问题：在Python中，如何将一个列表与另一个列表建立索引？

我想用这样的另一个列表索引一个列表

L = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
Idx = [0, 3, 7]
T = L[ Idx ]

T应该最终是一个包含[‘a’，’d’，’h’]的列表。

有没有比这更好的方法

T = []
for i in Idx:
    T.append(L[i])

print T
# Gives result ['a', 'd', 'h']
点击查看英文原文

I would like to index a list with another list like this

L = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
Idx = [0, 3, 7]
T = L[ Idx ]

and T should end up being a list containing [‘a’, ‘d’, ‘h’].

Is there a better way than

T = []
for i in Idx:
    T.append(L[i])

print T
# Gives result ['a', 'd', 'h']

回答 0

T = [L[i] for i in Idx]
点击查看英文原文 
T = [L[i] for i in Idx]

回答 1

如果您使用的是numpy，则可以执行如下扩展切片：

>>> import numpy
>>> a=numpy.array(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'])
>>> Idx = [0, 3, 7]
>>> a[Idx]
array(['a', 'd', 'h'], 
      dtype='|S1')

…并且可能要快得多（如果性能足以影响numpy导入）

点击查看英文原文

If you are using numpy, you can perform extended slicing like that:

>>> import numpy
>>> a=numpy.array(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'])
>>> Idx = [0, 3, 7]
>>> a[Idx]
array(['a', 'd', 'h'], 
      dtype='|S1')

…and is probably much faster (if performance is enough of a concern to to bother with the numpy import)

回答 2

功能方法：

a = [1,"A", 34, -123, "Hello", 12]
b = [0, 2, 5]

from operator import itemgetter

print(list(itemgetter(*b)(a)))
[1, 34, 12]
点击查看英文原文

A functional approach:

a = [1,"A", 34, -123, "Hello", 12]
b = [0, 2, 5]

from operator import itemgetter

print(list(itemgetter(*b)(a)))
[1, 34, 12]

回答 3

T = map(lambda i: L[i], Idx)
点击查看英文原文 
T = map(lambda i: L[i], Idx)

回答 4

我对这些方法都不满意，所以我想出了一个Flexlist类，它允许通过整数，切片或索引列表进行灵活的索引编制：

class Flexlist(list):
    def __getitem__(self, keys):
        if isinstance(keys, (int, slice)): return list.__getitem__(self, keys)
        return [self[k] for k in keys]

例如，您将其用作：

L = Flexlist(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'])
Idx = [0, 3, 7]
T = L[ Idx ]

print(T)  # ['a', 'd', 'h']
点击查看英文原文

I wasn’t happy with any of these approaches, so I came up with a Flexlist class that allows for flexible indexing, either by integer, slice or index-list:

class Flexlist(list):
    def __getitem__(self, keys):
        if isinstance(keys, (int, slice)): return list.__getitem__(self, keys)
        return [self[k] for k in keys]

Which, for your example, you would use as:

L = Flexlist(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'])
Idx = [0, 3, 7]
T = L[ Idx ]

print(T)  # ['a', 'd', 'h']

回答 5

L= {'a':'a','d':'d', 'h':'h'}
index= ['a','d','h'] 
for keys in index:
    print(L[keys])

我会用一个Dict add想要keys的index

点击查看英文原文 
L= {'a':'a','d':'d', 'h':'h'}
index= ['a','d','h'] 
for keys in index:
    print(L[keys])

I would use a Dict add desired keys to index

回答 6

您也可以将__getitem__方法与map以下方法结合使用：

L = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
Idx = [0, 3, 7]
res = list(map(L.__getitem__, Idx))
print(res)
# ['a', 'd', 'h']
点击查看英文原文

You could also use the __getitem__ method combined with map like the following:

L = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
Idx = [0, 3, 7]
res = list(map(L.__getitem__, Idx))
print(res)
# ['a', 'd', 'h']
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