基本python中Numpy.argsort()的等价物？
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Equivalent of Numpy.argsort() in basic python?
提问by Mermoz
采纳答案by Boris Gorelik
I timed the suggestions above and here are my results.
我对上面的建议进行了计时，这是我的结果。
First of all, the functions:
首先是功能：
def f(seq):
# http://stackoverflow.com/questions/3382352/equivalent-of-numpy-argsort-in-basic-python/3383106#3383106
#non-lambda version by Tony Veijalainen
return [i for (v, i) in sorted((v, i) for (i, v) in enumerate(seq))]
def g(seq):
# http://stackoverflow.com/questions/3382352/equivalent-of-numpy-argsort-in-basic-python/3383106#3383106
#lambda version by Tony Veijalainen
return [x for x,y in sorted(enumerate(seq), key = lambda x: x[1])]
def h(seq):
#http://stackoverflow.com/questions/3382352/equivalent-of-numpy-argsort-in-basic-python/3382369#3382369
#by unutbu
return sorted(range(len(seq)), key=seq.__getitem__)
Now, the IPython session:
现在，IPython 会话：
In [16]: seq = rand(10000).tolist()
In [17]: %timeit f(seq)
100 loops, best of 3: 10.5 ms per loop
In [18]: %timeit g(seq)
100 loops, best of 3: 8.83 ms per loop
In [19]: %timeit h(seq)
100 loops, best of 3: 6.44 ms per loop
FWIW
FWIW
回答by unutbu
There is no built-in function, but it's easy to assemble one out of the terrific tools Python makes available:
没有内置函数，但很容易从 Python 提供的极好的工具中组装一个：
def argsort(seq):
# http://stackoverflow.com/questions/3071415/efficient-method-to-calculate-the-rank-vector-of-a-list-in-python
return sorted(range(len(seq)), key=seq.__getitem__)
x = [5,2,1,10]
print(argsort(x))
# [2, 1, 0, 3]
It works on Python array.arrays the same way:
它array.array以相同的方式在 Python 上工作：
import array
x = array.array('d', [5, 2, 1, 10])
print(argsort(x))
# [2, 1, 0, 3]
回答by Tony Veijalainen
My alternative with enumerate:
我的枚举替代方案：
def argsort(seq):
return [x for x,y in sorted(enumerate(seq), key = lambda x: x[1])]
seq=[5,2,1,10]
print(argsort(seq))
# Output:
# [2, 1, 0, 3]
Better though to use answer from https://stackoverflow.com/users/9990/marcelo-cantosanswer to thread python sort without lambda expressions
最好使用https://stackoverflow.com/users/9990/marcelo-cantos 的答案来回答没有 lambda 表达式的线程python 排序
[i for (v, i) in sorted((v, i) for (i, v) in enumerate(seq))]
回答by Jeff M.
Found this question, but needed argsort for a list of objects based on an object property.
找到了这个问题，但需要 argsort 来获取基于对象属性的对象列表。
Extending unutbu's answer, this would be:
扩展 unutbu 的答案，这将是：
sorted(range(len(seq)), key = lambda x: seq[x].sort_property)
