Python函数，用于标识列表或数组中的数字是否接近0或1

Question

我有一个数字的numpy数组。下面是一个例子：

[[-2.10044520e-04  1.72314372e-04  1.77235336e-04 -1.06613465e-04
6.76617611e-07  2.71623057e-03 -3.32789944e-05  1.44899758e-05
5.79249863e-05  4.06502549e-04 -1.35823707e-05 -4.13955189e-04
5.29862793e-05 -1.98286005e-04 -2.22829175e-04 -8.88758230e-04
5.62228710e-05  1.36249752e-05 -2.00474996e-05 -2.10090068e-05
1.00007518e+00  1.00007569e+00 -4.44597417e-05 -2.93724453e-04
1.00007513e+00  1.00007496e+00  1.00007532e+00 -1.22357142e-03
3.27903892e-06  1.00007592e+00  1.00007468e+00  1.00007558e+00
2.09869172e-05 -1.97610235e-05  1.00007529e+00  1.00007530e+00
1.00007503e+00 -2.68725642e-05 -3.00372853e-03  1.00007386e+00
1.00007443e+00  1.00007388e+00  5.86993822e-05 -8.69989983e-06
1.00007590e+00  1.00007488e+00  1.00007515e+00  8.81850779e-04
2.03875532e-05  1.00007480e+00  1.00007425e+00  1.00007517e+00
-2.44678912e-05 -4.36556267e-08  1.00007436e+00  1.00007558e+00
1.00007571e+00 -5.42990711e-04  1.45517859e-04  1.00007522e+00
1.00007469e+00  1.00007575e+00 -2.52271817e-05 -7.46339417e-05
1.00007427e+00]]

我想知道每个数字是接近0还是接近1。Python中有没有一个函数可以做到这一点，或者我必须手动完成这一点？

Answer 1

一种简单的方法：

lst=[0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9]
closerTo1 = [x >= 0.5 for x in lst]

或者你可以使用np:

import numpy as np
lst=[0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9]
arr = np.array(lst)
closerTo1 = arr >= 0.5

注意，>= 0.5可以更改为> 0.5，但您可以选择进行处理。

Answer 2

numpy.rint是一个ufunc，它将数组的元素四舍五入到最近的整数。

>>> a = np.arange(0, 1.1, 0.1)
>>> a
array([0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1. ])
>>> np.rint(a)
array([0., 0., 0., 0., 0., 0., 1., 1., 1., 1., 1.])

如果数字不必在0和1之间怎么办？

在这种情况下，我会使用numpy.where。

>>> a = np.arange(-2, 2.1, 0.1)
>>> a
array([-2.00000000e+00, -1.90000000e+00, -1.80000000e+00, -1.70000000e+00,
-1.60000000e+00, -1.50000000e+00, -1.40000000e+00, -1.30000000e+00,
-1.20000000e+00, -1.10000000e+00, -1.00000000e+00, -9.00000000e-01,
-8.00000000e-01, -7.00000000e-01, -6.00000000e-01, -5.00000000e-01,
-4.00000000e-01, -3.00000000e-01, -2.00000000e-01, -1.00000000e-01,
1.77635684e-15,  1.00000000e-01,  2.00000000e-01,  3.00000000e-01,
4.00000000e-01,  5.00000000e-01,  6.00000000e-01,  7.00000000e-01,
8.00000000e-01,  9.00000000e-01,  1.00000000e+00,  1.10000000e+00,
1.20000000e+00,  1.30000000e+00,  1.40000000e+00,  1.50000000e+00,
1.60000000e+00,  1.70000000e+00,  1.80000000e+00,  1.90000000e+00,
2.00000000e+00])
>>> np.where(a <= 0.5, 0, 1)
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1])

Answer 3

这里有一个简单的方法：

>>> a = np.arange(-2, 2.1, 0.1)
>>> (a >= .5).astype(np.float)
array([ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,
0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  1.,
1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,
1.,  1.])

(如果需要整数，请将np.float更改为np.int。(

Answer 4

您可以使用numpy.where:

import numpy as np
arr = np.array([0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 2.0])
result = np.where(arr >= 0.5, 1, 0)
print(result)

输出

[0 0 0 0 1 1 1 1 1 1]

请注意，对于1以上的数字(例如2(，这将返回1。

Answer 5

您可以使用abs()来测量数字与0和1之间的距离，并检查哪个更短。

x = [[-2.10044520e-04,  1.72314372e-04,  1.77235336e-04, -1.06613465e-04,
6.76617611e-07,  2.71623057e-03, -3.32789944e-05,  1.44899758e-05,
5.79249863e-05,  4.06502549e-04, -1.35823707e-05, -4.13955189e-04,
5.29862793e-05, -1.98286005e-04, -2.22829175e-04, -8.88758230e-04,
5.62228710e-05,  1.36249752e-05, -2.00474996e-05, -2.10090068e-05,
1.00007518e+00,  1.00007569e+00, -4.44597417e-05, -2.93724453e-04,
1.00007513e+00,  1.00007496e+00,  1.00007532e+00, -1.22357142e-03,
3.27903892e-06,  1.00007592e+00,  1.00007468e+00,  1.00007558e+00,
2.09869172e-05, -1.97610235e-05,  1.00007529e+00,  1.00007530e+00,
1.00007503e+00, -2.68725642e-05, -3.00372853e-03,  1.00007386e+00,
1.00007443e+00,  1.00007388e+00,  5.86993822e-05, -8.69989983e-06,
1.00007590e+00,  1.00007488e+00,  1.00007515e+00,  8.81850779e-04,
2.03875532e-05,  1.00007480e+00,  1.00007425e+00,  1.00007517e+00,
-2.44678912e-05, -4.36556267e-08,  1.00007436e+00,  1.00007558e+00,
1.00007571e+00, -5.42990711e-04,  1.45517859e-04,  1.00007522e+00,
1.00007469e+00,  1.00007575e+00, -2.52271817e-05, -7.46339417e-05,
1.00007427e+00]]
rounded_x = [0 if abs(i) < abs(1-i) else 1 for i in x[0]]
print(rounded_x)

输出：

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1]

Answer 6

这里有一个对任意数字a和b的简单概括，而不仅仅是0和1：

def closerab(l, a=0, b=1):
l = np.asarray(l)
boolarr = (np.abs(l - b) > np.abs(l - a))
# returns two lists of indices, one for numbers closer to a and one for numbers closer to b
return boolarr.nonzero()[0], (boolarr==0).nonzero()[0]

这将返回两个列表，一个列表中的数字索引更接近a，另一个列表的数字索引接近b。

测试：

l = [
-2.10044520e-04, 1.72314372e-04, 1.77235336e-04, 1.06613465e-04,
6.76617611e-07, 2.71623057e-03, 3.32789944e-05, 1.44899758e-05,
5.79249863e-05, 4.06502549e-04, 1.35823707e-05, 4.13955189e-04,
5.29862793e-05, 1.98286005e-04, 2.22829175e-04, 8.88758230e-04,
5.62228710e-05, 1.36249752e-05, 2.00474996e-05, 2.10090068e-05,
1.00007518e+00, 1.00007569e+00, 4.44597417e-05, 2.93724453e-04,
1.00007513e+00, 1.00007496e+00, 1.00007532e+00, 1.22357142e-03,
3.27903892e-06, 1.00007592e+00, 1.00007468e+00, 1.00007558e+00,
2.09869172e-05, 1.97610235e-05, 1.00007529e+00, 1.00007530e+00,
1.00007503e+00, 2.68725642e-05, 3.00372853e-03, 1.00007386e+00,
1.00007443e+00, 1.00007388e+00, 5.86993822e-05, 8.69989983e-06,
1.00007590e+00, 1.00007488e+00, 1.00007515e+00, 8.81850779e-04,
2.03875532e-05, 1.00007480e+00, 1.00007425e+00, 1.00007517e+00,
-2.44678912e-05, 4.36556267e-08, 1.00007436e+00, 1.00007558e+00,
1.00007571e+00, 5.42990711e-04, 1.45517859e-04, 1.00007522e+00,
1.00007469e+00, 1.00007575e+00, 2.52271817e-05, 7.46339417e-05,
1.00007427e+00
]
print(closerab(l, 0, 1))

该输出：

(array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 22, 23, 27, 28, 32, 33, 37, 38, 42, 43, 47, 48, 52, 53,
57, 58, 62, 63]),
array([20, 21, 24, 25, 26, 29, 30, 31, 34, 35, 36, 39, 40, 41, 44, 45, 46,
49, 50, 51, 54, 55, 56, 59, 60, 61, 64]))

Answer 7

或者，您可以使用三元运算符。

x = [-0.2, 0.1, 1.1, 0.75, 0.4, 0.2, 1.5, 0.9]
a = 0
b = 1
[a if i <= (a+b)/2 else b for i in x]

Answer 8

来自Python内置函数文档round(number[, ndigits]):

返回浮点值数字，四舍五入到小数点后的两位数。如果省略ndigits，则默认为零。结果是一个浮点数字。值四舍五入到10的最接近倍数减去ndigits的幂；如果两个倍数相等，则从0开始取整(例如，round(0.5)是1.0，round(-0.5)是-1.0(。

特别是对于numpy数组，可以使用numpy.round_函数。

Answer 9

your_list=[[-2.10044520e-04, 1.72314372e-04, 1.77235336e-04, 1.06613465e-04,
6.76617611e-07, 2.71623057e-03, 3.32789944e-05, 1.44899758e-05,
5.79249863e-05, 4.06502549e-04, 1.35823707e-05, 4.13955189e-04,
5.29862793e-05, 1.98286005e-04, 2.22829175e-04, 8.88758230e-04,
5.62228710e-05, 1.36249752e-05, 2.00474996e-05, 2.10090068e-05,
1.00007518e+00, 1.00007569e+00, 4.44597417e-05, 2.93724453e-04,
1.00007513e+00, 1.00007496e+00, 1.00007532e+00, 1.22357142e-03,
3.27903892e-06, 1.00007592e+00, 1.00007468e+00, 1.00007558e+00,
2.09869172e-05, 1.97610235e-05, 1.00007529e+00, 1.00007530e+00,
1.00007503e+00, 2.68725642e-05, 3.00372853e-03, 1.00007386e+00,
1.00007443e+00, 1.00007388e+00, 5.86993822e-05, 8.69989983e-06,
1.00007590e+00, 1.00007488e+00, 1.00007515e+00, 8.81850779e-04,
2.03875532e-05, 1.00007480e+00, 1.00007425e+00, 1.00007517e+00,
-2.44678912e-05, 4.36556267e-08, 1.00007436e+00, 1.00007558e+00,
1.00007571e+00, 5.42990711e-04, 1.45517859e-04, 1.00007522e+00,
1.00007469e+00, 1.00007575e+00, 2.52271817e-05, 7.46339417e-05,
1.00007427e+00]]
close_to_one_or_zero=[1 if x > 0.5 else 0 for x in your_list[0]]
close_to_one_or_zero
[0, 0, 0, 0, 0,....... 1, 1, 1, 0, 0, 1]

Answer 10

您可以使用round:

[round(i) for i in [0.1,0.2,0.3,0.8,0.9]]