问题:从列表或元组中明确选择项目
我有以下Python列表(也可以是元组):
myList = ['foo', 'bar', 'baz', 'quux']
我可以说
>>> myList[0:3]
['foo', 'bar', 'baz']
>>> myList[::2]
['foo', 'baz']
>>> myList[1::2]
['bar', 'quux']
如何显式挑选索引没有特定模式的项目?例如,我要选择[0,2,3]
。或者,从1000个很大的清单中,我要选择[87, 342, 217, 998, 500]
。是否有一些Python语法可以做到这一点?看起来像这样:
>>> myBigList[87, 342, 217, 998, 500]
I have the following Python list (can also be a tuple):
myList = ['foo', 'bar', 'baz', 'quux']
I can say
>>> myList[0:3]
['foo', 'bar', 'baz']
>>> myList[::2]
['foo', 'baz']
>>> myList[1::2]
['bar', 'quux']
How do I explicitly pick out items whose indices have no specific patterns? For example, I want to select [0,2,3]
. Or from a very big list of 1000 items, I want to select [87, 342, 217, 998, 500]
. Is there some Python syntax that does that? Something that looks like:
>>> myBigList[87, 342, 217, 998, 500]
回答 0
list( myBigList[i] for i in [87, 342, 217, 998, 500] )
我将答案与python 2.5.2进行了比较:
19.7微秒: [ myBigList[i] for i in [87, 342, 217, 998, 500] ]
20.6 USEC: map(myBigList.__getitem__, (87, 342, 217, 998, 500))
22.7 USEC: itemgetter(87, 342, 217, 998, 500)(myBigList)
24.6 USEC: list( myBigList[i] for i in [87, 342, 217, 998, 500] )
请注意,在Python 3中,第1个已更改为与第4个相同。
另一种选择是以a开头,numpy.array
它允许通过列表或a进行索引numpy.array
:
>>> import numpy
>>> myBigList = numpy.array(range(1000))
>>> myBigList[(87, 342, 217, 998, 500)]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: invalid index
>>> myBigList[[87, 342, 217, 998, 500]]
array([ 87, 342, 217, 998, 500])
>>> myBigList[numpy.array([87, 342, 217, 998, 500])]
array([ 87, 342, 217, 998, 500])
在tuple
不工作方式相同那些片。
list( myBigList[i] for i in [87, 342, 217, 998, 500] )
I compared the answers with python 2.5.2:
19.7 usec: [ myBigList[i] for i in [87, 342, 217, 998, 500] ]
20.6 usec: map(myBigList.__getitem__, (87, 342, 217, 998, 500))
22.7 usec: itemgetter(87, 342, 217, 998, 500)(myBigList)
24.6 usec: list( myBigList[i] for i in [87, 342, 217, 998, 500] )
Note that in Python 3, the 1st was changed to be the same as the 4th.
Another option would be to start out with a numpy.array
which allows indexing via a list or a numpy.array
:
>>> import numpy
>>> myBigList = numpy.array(range(1000))
>>> myBigList[(87, 342, 217, 998, 500)]
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IndexError: invalid index
>>> myBigList[[87, 342, 217, 998, 500]]
array([ 87, 342, 217, 998, 500])
>>> myBigList[numpy.array([87, 342, 217, 998, 500])]
array([ 87, 342, 217, 998, 500])
The tuple
doesn’t work the same way as those are slices.
回答 1
那这个呢:
from operator import itemgetter
itemgetter(0,2,3)(myList)
('foo', 'baz', 'quux')
What about this:
from operator import itemgetter
itemgetter(0,2,3)(myList)
('foo', 'baz', 'quux')
回答 2
它不是内置的,但是如果您愿意,可以创建一个将元组作为“索引”的list的子类:
class MyList(list):
def __getitem__(self, index):
if isinstance(index, tuple):
return [self[i] for i in index]
return super(MyList, self).__getitem__(index)
seq = MyList("foo bar baaz quux mumble".split())
print seq[0]
print seq[2,4]
print seq[1::2]
印刷
foo
['baaz', 'mumble']
['bar', 'quux']
It isn’t built-in, but you can make a subclass of list that takes tuples as “indexes” if you’d like:
class MyList(list):
def __getitem__(self, index):
if isinstance(index, tuple):
return [self[i] for i in index]
return super(MyList, self).__getitem__(index)
seq = MyList("foo bar baaz quux mumble".split())
print seq[0]
print seq[2,4]
print seq[1::2]
printing
foo
['baaz', 'mumble']
['bar', 'quux']
回答 3
也许列表理解是按顺序进行的:
L = ['a', 'b', 'c', 'd', 'e', 'f']
print [ L[index] for index in [1,3,5] ]
生成:
['b', 'd', 'f']
那是您要找的东西吗?
Maybe a list comprehension is in order:
L = ['a', 'b', 'c', 'd', 'e', 'f']
print [ L[index] for index in [1,3,5] ]
Produces:
['b', 'd', 'f']
Is that what you are looking for?
回答 4
>>> map(myList.__getitem__, (2,2,1,3))
('baz', 'baz', 'bar', 'quux')
您也可以创建自己的List
类,该类支持将元组用作__getitem__
要执行的操作的参数myList[(2,2,1,3)]
。
>>> map(myList.__getitem__, (2,2,1,3))
('baz', 'baz', 'bar', 'quux')
You can also create your own List
class which supports tuples as arguments to __getitem__
if you want to be able to do myList[(2,2,1,3)]
.
回答 5
我只想指出,即使itemgetter的语法看起来也很整洁,但是在大型列表上执行时有点慢。
import timeit
from operator import itemgetter
start=timeit.default_timer()
for i in range(1000000):
itemgetter(0,2,3)(myList)
print ("Itemgetter took ", (timeit.default_timer()-start))
物品获取者1.065209062149279
start=timeit.default_timer()
for i in range(1000000):
myList[0],myList[2],myList[3]
print ("Multiple slice took ", (timeit.default_timer()-start))
多个切片花费0.6225321444745759
I just want to point out, even syntax of itemgetter looks really neat, but it’s kinda slow when perform on large list.
import timeit
from operator import itemgetter
start=timeit.default_timer()
for i in range(1000000):
itemgetter(0,2,3)(myList)
print ("Itemgetter took ", (timeit.default_timer()-start))
Itemgetter took 1.065209062149279
start=timeit.default_timer()
for i in range(1000000):
myList[0],myList[2],myList[3]
print ("Multiple slice took ", (timeit.default_timer()-start))
Multiple slice took 0.6225321444745759
回答 6
另一个可能的解决方案:
sek=[]
L=[1,2,3,4,5,6,7,8,9,0]
for i in [2, 4, 7, 0, 3]:
a=[L[i]]
sek=sek+a
print (sek)
Another possible solution:
sek=[]
L=[1,2,3,4,5,6,7,8,9,0]
for i in [2, 4, 7, 0, 3]:
a=[L[i]]
sek=sek+a
print (sek)
回答 7
当你有一个布尔numpy数组时,就像 mask
[mylist[i] for i in np.arange(len(mask), dtype=int)[mask]]
适用于任何序列或np.array的lambda:
subseq = lambda myseq, mask : [myseq[i] for i in np.arange(len(mask), dtype=int)[mask]]
newseq = subseq(myseq, mask)
like often when you have a boolean numpy array like mask
[mylist[i] for i in np.arange(len(mask), dtype=int)[mask]]
A lambda that works for any sequence or np.array:
subseq = lambda myseq, mask : [myseq[i] for i in np.arange(len(mask), dtype=int)[mask]]
newseq = subseq(myseq, mask)