Python仅枚举反向索引
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Python enumerate reverse index only
提问by rozzy
I am trying to reverse the index given by enumeratewhilst retaining the original order of the list being enumerated.
我试图enumerate在保留被枚举列表的原始顺序的同时反转给定的索引。
Assume I have the following:
假设我有以下内容：
>> range(5)
[0, 1, 2, 3, 4]
If I enumerate this I would get the following:
如果我枚举这个，我会得到以下信息：
>> list(enumerate(range(5)))
[(0, 0), (1, 1), (2, 2), (3, 3), (4, 4)]
However I want to reverse the index provided by enumerate so that I get:
但是我想反转 enumerate 提供的索引，以便我得到：
[(4, 0), (3, 1), (2, 2), (1, 3), (0, 4)]
So far I have the following code:
到目前为止，我有以下代码：
reversed(list(enumerate(reversed(range(5)))))
I was just wondering if there was a neater way to do this?
我只是想知道是否有更简洁的方法来做到这一点？
回答by Netwave
How about using zip instead with a reversed range?
如何使用 zip 而不是反向范围？
>>> zip(range(9, -1, -1), range(10))
[(9, 0), (8, 1), (7, 2), (6, 3), (5, 4), (4, 5), (3, 6), (2, 7), (1, 8), (0, 9)]
>>> def reversedEnumerate(l):
return zip(range(len(l)-1, -1, -1), l)
>>> reversedEnumerate(range(10))
[(9, 0), (8, 1), (7, 2), (6, 3), (5, 4), (4, 5), (3, 6), (2, 7), (1, 8), (0, 9)]
As @julienSpronk suggests, use izipto get a generator, also xrange:
正如@julienSpronk 所建议的，用于izip获取生成器，还有xrange：
import itertools
>>> import itertools
>>> def reversedEnumerate(l):
... return itertools.izip(xrange(len(l)-1, -1, -1), l)
...
>>> reversedEnumerate(range(10))
<itertools.izip object at 0x03749760>
>>> for i in reversedEnumerate(range(10)):
... print i
...
(9, 0)
(8, 1)
(7, 2)
(6, 3)
(5, 4)
(4, 5)
(3, 6)
(2, 7)
(1, 8)
(0, 9)
回答by mandrewcito
I don't know if this solution is better for you, but at least it's shorter:
我不知道这个解决方案是否对你更好，但至少它更短：
>>> [(4 - x, x) for x in range(5)]
[(4, 0), (3, 1), (2, 2), (1, 3), (0, 4)]
回答by RemcoGerlich
Just take the length of your list and subtract the index from that...
只需取列表的长度并从中减去索引...
L = range(5)
for i, n in L:
my_i = len(L) -1 - i
...
Or if you really need a generator:
或者如果你真的需要一个发电机：
def reverse_enumerate(L):
# Only works on things that have a len()
l = len(L)
for i, n in enumerate(L):
yield l-i-1, n
enumerate()can't possibly do this, as it works with generic iterators. For instance, you can pass it infinite iterators, that don't even have a "reverse index".
enumerate()不可能这样做，因为它适用于通用迭代器。例如，您可以将无限迭代器传递给它，甚至没有“反向索引”。
回答by Emil
Assuming your list is not long and you will not run into performance errors, you may use list(enumerate(range(5)[::-1]))[::-1].
假设您的列表不长并且您不会遇到性能错误，您可以使用list(enumerate(range(5)[::-1]))[::-1].
Test:
测试：
>>> list(enumerate(range(5)[::-1]))[::-1]
[(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)]
>>> list(enumerate(range(5)[::-1]))[::-1]
[(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)]
回答by MaThMaX
Actually I'm using the same logic as @RemcoGerlich did, but I use list comprehensiondirectly, which make the code now become 1-liner:
实际上我使用与@RemcoGerlich 相同的逻辑，但我list comprehension直接使用，这使得代码现在变成了 1-liner：
def generatelist(x):
return [(x-1-i,n) for i,n in enumerate(range(x))]
Regarding the dilemma of choosing generatoror list comprehension, hereis the suggested way:
关于选择generatoror的困境list comprehension，这里是建议的方式：
Basically, use a generator expression if all you're doing is iterating once. If you want to store and use the generated results, then you're probably better off with a list comprehension.
基本上，如果您所做的只是迭代一次，请使用生成器表达式。如果您想存储和使用生成的结果，那么最好使用列表理解。
回答by Julien Spronck
If you're going to re-use it several times, you can make your own generator:
如果您要多次重复使用它，您可以制作自己的生成器：
def reverse_enum(lst):
for j, item in enumerate(lst):
yield len(lst)-1-j, item
print list(reverse_enum(range(5)))
# [(4, 0), (3, 1), (2, 2), (1, 3), (0, 4)]
or
或者
def reverse_enum(lst):
return ((len(lst)-1-j, item) for j, item in enumerate(lst))
回答by John
Simply use len(lst)-ieverywhere i is used. or:
只需在使用len(lst)-ii 的地方使用即可。或者：
[(len(range(5)) - x, x) for x in range(5)]
回答by tzot
Python 2
蟒蛇 2
import itertools
def reversed_enumerate(seq):
return itertools.izip(reversed(range(len(seq))), reversed(seq))
Python 3
蟒蛇 3
Substitute zipfor itertools.izip:)
替代zip了itertools.izip:)
回答by Alexander Fedorov
values = 'abcde'
for i, value in zip(reversed(range(len(values))), values):
print(i, value)
Explanation:
解释：
values = 'abcde'
values_len = len(values) # 5
indexes = range(values_len) # [0, 1, 2, 3, 4]
reversed_indexes = reversed(indexes) # [4, 3, 2, 1, 0]
# combine reversed indexes and values
reversed_enumerator = zip(reversed_indexes, values)
for i, value in reversed_enumerator:
print(i, value)
回答by Venfah Nazir
We can use enumerate with len:
我们可以将 enumerate 与 len 一起使用：
$ cat enumerate.py
arr = ['stone', 'cold', 'steve', 'austin']
for i, val in enumerate(arr):
print ("enu {} val {}".format(i, val))
for i, val in enumerate(arr):
print ("enu {} val {}".format(len(arr) - i - 1, val))
$ python enumerate.py
enu 0 val stone
enu 1 val cold
enu 2 val steve
enu 3 val austin
enu 3 val stone
enu 2 val cold
enu 1 val steve
enu 0 val austin
$
