使用python从列表中随机提取x个项目

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的项目数量（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。例如，假设我想要 50%，输出将是

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

但我想知道是否有更有效的方法来做到这一点，在我的实际用例中，这是从 145,000 个项目中进行子采样。此外，在这个尺度上，randrange 是否足够没有偏差？

谢谢

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格：

 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 Python 2 和 Python 3 中，random.randrange()函数完全消除了偏差（它使用内部_randbelow()进行多次随机选择直到找到无偏差结果的方法）。

在 Python 2 中，随机抽样（）函数略有偏差，但仅限于最后 53 位的舍入。在 Python 3 中，随机抽样（）函数使用内部_randbelow()方法且无偏差。