使用python从列表中随机提取x项

Question

从两个列表开始，例如：

lstOne = [ '1', '2', '3', '4', '5', '6', '7', '8', '9', '10']
lstTwo = [ '1', '2', '3', '4', '5', '6', '7', '8', '9', '10']

我想让用户输入他们想要提取多少项，占整个列表长度的百分比，以及从每个列表中随机提取的相同索引。比如说我想要50%的产量

newLstOne = ['8', '1', '3', '7', '5']
newLstTwo = ['8', '1', '3', '7', '5']

我使用以下代码实现了这一点：

from random import randrange

lstOne = [ '1', '2', '3', '4', '5', '6', '7', '8', '9', '10']
lstTwo = [ '1', '2', '3', '4', '5', '6', '7', '8', '9', '10']

LengthOfList = len(lstOne)
print LengthOfList

PercentageToUse = input("What Percentage Of Reads Do you want to extract? ")
RangeOfListIndices = []

HowManyIndicesToMake = (float(PercentageToUse)/100)*float(LengthOfList)
print HowManyIndicesToMake

for x in lstOne:
    if len(RangeOfListIndices)==int(HowManyIndicesToMake):
        break
    else:
        random_index = randrange(0,LengthOfList)
        RangeOfListIndices.append(random_index)

print RangeOfListIndices


newlstOne = []
newlstTwo = []

for x in RangeOfListIndices:
    newlstOne.append(lstOne[int(x)])
for x in RangeOfListIndices:
    newlstTwo.append(lstTwo[int(x)])

print newlstOne
print newlstTwo

但我想知道是否有一种更有效的方法来实现这一点，在我的实际用例中，这是从145000个项目中进行的子抽样。此外，randrange在这个尺度上是否足够没有偏见？

谢谢你

Answer 1

只需将两个列表放在一起，使用random.sample进行采样，然后再次将zip转换回两个列表。

import random

_zips = random.sample(zip(lstOne,lstTwo), 5)

new_list_1, new_list_2 = zip(*_zips)

演示：

list_1 = range(1,11)
list_2 = list('abcdefghij')

_zips = random.sample(zip(list_1, list_2), 5)

new_list_1, new_list_2 = zip(*_zips)

new_list_1
Out[33]: (3, 1, 9, 8, 10)

new_list_2
Out[34]: ('c', 'a', 'i', 'h', 'j')

Answer 2

我看你这样做还行。

如果要避免多次对同一对象进行采样，可以执行以下操作：

a = len(lstOne)
choose_from = range(a)          #<--- creates a list of ints of size len(lstOne)
random.shuffle(choose_from)
for i in choose_from[:a]:       # selects the desired number of items from both original list
    newlstOne.append(lstOne[i]) # at the same random locations & appends to two newlists in
    newlstTwo.append(lstTwo[i]) # sequence

Answer 3

Q.I want to have the user input how many items they want to extract, as a percentage of the overall list length, and the same indices from each list to be randomly extracted.

A.最直接的方法直接符合您的规范：

 percentage = float(raw_input('What percentage? '))
 k = len(data) * percentage // 100
 indicies = random.sample(xrange(len(data)), k)
 new_list1 = [list1[i] for i in indicies]
 new_list2 = [list2[i] for i in indicies]

Q.in my actual use case this is subsampling from 145,000 items. Furthermore, is randrange sufficiently free of bias at this scale?

A.在Python2和Python3中，random.randrange（）函数完全消除了偏差（它使用内部的方法进行多个随机选择，直到找到无偏差的结果）。

在Python2中，random.sample（）函数稍微有点偏差，但仅在53位的最后一个舍入。在Python 3中，random.sample（）函数使用内部的方法，并且没有偏见。