Question

我有两个列表（实际上是两个数据帧列）。它们具有相同的元素，但是一个列表是无序的。我想获得与有序列表对应的无序列表的索引。是否有捷径可寻？

即。 list1 [indices] == list2

我需要获取索引变量。

Answer 1

在列表理解中使用list.index()：

l1 = ['a','b','c','d']
l2 = ['c','d','b','a']

[l1.index(x) for x in l2] #[2, 3, 1, 0]

如果您尝试在DataFrame中执行此操作，则可以从np.array转换为list并返回，如：

import numpy as np
import pandas as pd

df = pd.DataFrame({'v1':np.array(l1), 'v2':np.array(l2)})

df['index_of_v2_in_v1'] = np.array([list(df['v1']).index(x) for x in list(df['v2'])])

df
# Result:
#   v1 v2  index_of_v2_in_v1
# 0  a  c                  2
# 1  b  d                  3
# 2  c  b                  1
# 3  d  a                  0

如果您<100>确定已经对列表1进行了排序（如您的问题所示），您可以在列表或数组中使用np.argsort(l2)，如下所示：

np.argsort(df['v2'])
# Returns:
#0    3
#1    2
#2    0
#3    1
#Name: v2, dtype: int64

Answer 2

在此示例中，使用map比列表理解快约3.6倍：

from timeit import timeit

l1 = ['a','b','c','d']
l2 = ['c','d','b','a']

t1 = timeit('map(lambda e: l1.index(e), l2)', globals=globals())
t2 = timeit('[l1.index(x) for x in l2]', globals=globals())
print("t1 = %s, t2 = %s, t2/t1 = %s" % (t1, t2, t2/t1))

结果：

t1 = 0.32407195774213654, t2 = 1.162188749526786, t2/t1 = 3.586205846454439

编辑：其他比较，包括@jbch提出的解决方案：

from timeit import timeit
from random import shuffle

for n in range(10, 70, 10):
    l1 = list(range(n))
    l2 = l1[:]
    shuffle(l2)

    t1 = timeit('indices = {val: i for i, val in enumerate(l1)}; [indices[x] for x in l2]', globals=globals())
    t2 = timeit('[l1.index(x) for x in l2]', globals=globals())
    t3 = timeit('map(lambda e: l1.index(e), l2)', globals=globals())
    print("n = %d, t1 = %g, t2 = %g, t3 = %g" % (n, t1, t2, t3))

结果：

n = 10, t1 = 3.25064, t2 = 3.70473, t3 = 0.339757
n = 20, t1 = 5.01145, t2 = 9.22295, t3 = 0.341116
n = 30, t1 = 7.18546, t2 = 16.6379, t3 = 0.344537
n = 40, t1 = 8.96271, t2 = 26.0522, t3 = 0.336952
n = 50, t1 = 11.0635, t2 = 37.7291, t3 = 0.341935
n = 60, t1 = 12.6453, t2 = 51.1519, t3 = 0.350777

Answer 3

C8H10N42的答案时间复杂度为O（n ^ 2），在大名单上需要很长时间。每次对index（）的调用都是O（n），它被调用n次。

如果您需要更好的性能，可以使用此O（n）解决方案：

from timeit import timeit
from random import shuffle

for n in range(0, 50, 5):
    l1 = list(range(n))
    l2 = l1[:]
    shuffle(l2)

    t1 = timeit('indices = {val: i for i, val in enumerate(l1)}; [indices[x] for x in l2]', 'from __main__ import l1, l2')
    t2 = timeit('[l1.index(x) for x in l2]', 'from __main__ import l1, l2')
    print("n = %s, t1 = %s, t2 = %s, t2/t1 = %s" % (n, t1, t2, t2/t1))

创建字典是O（n），然后您可以用O（1）dict访问替换O（n）index（）调用。所以复杂度是O（n）+ O（n）而不是O（n ^ 2）。

如果你尝试使用不同的列表大小，你会看到列表越大，index（）表现越差：

n = 0, t1 = 0.410041093826, t2 = 0.0470049381256, t2/t1 = 0.114634700847
n = 5, t1 = 1.01210093498, t2 = 0.980098009109, t2/t1 = 0.96837970921
n = 10, t1 = 1.70017004013, t2 = 2.06220698357, t2/t1 = 1.21294160872
n = 15, t1 = 2.12121200562, t2 = 3.28132796288, t2/t1 = 1.54691183823
n = 20, t1 = 2.64426398277, t2 = 4.81948184967, t2/t1 = 1.82261751515
n = 25, t1 = 3.42534303665, t2 = 6.57365703583, t2/t1 = 1.9191237098
n = 30, t1 = 3.95739603043, t2 = 8.52685213089, t2/t1 = 2.15466232475
n = 35, t1 = 4.24842405319, t2 = 10.8080809116, t2/t1 = 2.54402121265
n = 40, t1 = 4.75647592545, t2 = 13.3403339386, t2/t1 = 2.80466760427
n = 45, t1 = 5.33353281021, t2 = 15.6205620766, t2/t1 = 2.92874584865

结果：

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获取无序列表python的索引

3 个答案: