如何选择列表中最低的 10% 的数字？

Question

我想获得列表中最低的 10% 的数字。

List = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]

从上面的列表中，我希望得到结果。结果 = [1,2] 这是列表中最低的 10%。

Answer 1

result = List[:int(len(List)*0.1)]

Answer 2

如果每个元素都是唯一的，您可以简单地对数据进行排序和切片

l = list(range(1, 21))

number_value_to_get = len(l)//10
print(sorted(l)[:number_value_to_get]

但是，在大多数情况下，这是错误的，你可以使用numpy版本

import numpy as np

l = np.array(range(1, 21))
threshold = np.percentile(l, 10) # calculate the 10th percentile
print(l[l < threshold]) # Filter the list.

注意需要定义这是否包含 10%

import numpy as np

l = np.array([1]*20)
threshold = np.percentile(l, 10)
print(l[l < np.percentile(l, 10)]) # Gives you empty list
print(l[l <= np.percentile(l, 10)]) # Gives you full list

Answer 3

试试这个 -

sorted(lis)[:int((0.1 * len(lis)))]

其中 lis 是您的列表。

Answer 4

 # list of values
lstValues = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]

# get maximum value out of list values
max = max(lstValues)

# calculate 10 percent out of max value
max /= 10

# print all the elements which are under the 10% mark
print([i for i in lstValues if i <= max])

Answer 5

如果你想使用 numpy，有一个内置的百分位数函数：

import numpy

l = numpy.array([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20])

print(l[l < numpy.percentile(l,10)])

Answer 6

给你 =^..^=

import numpy as np

List = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]

percent = 10
values = list(sorted(np.asarray(List, dtype=np.int))[:int(len(List)/(100/percent))])

输出：

[1, 2]

Answer 7

mylist=[int(x) for x in range(1,21)]
mylist.sort()
newlist=[]
for i in range(len(mylist)//10): #just index through the 10% elements with this
    newlist.append(mylist[i])
print(newlist)

如何选择列表中最低的 10% 的数字？

How to choose the numbers which are lowest 10% in the list?

python

min