Question

我有以下数组：

var crc32=function(r){for(var a,o=[],c=0;c<256;c++){a=c;for(var f=0;f<8;f++)a=1&a?3988292384^a>>>1:a>>>1;o[c]=a}for(var n=-1,t=0;t<r.length;t++)n=n>>>8^o[255&(n^r.charCodeAt(t))];return(-1^n)>>>0};

console.log(crc32('abc'));

console.log(crc32('abc').toString(16).toUpperCase()); // hex

现在，我想将这些值复制到同一行的n列中。结果应如下所示：

[9.975 9.976 9.977 9.978 9.979 9.98  9.981 9.982 9.983 9.984 9.985 9.986
9.987 9.988 9.989 9.99  9.991 9.992 9.993 9.994]

你知道这是怎么回事吗？

提前致谢。

Answer 1

由于您正在使用numpy，请使用np.repeat + np.reshape：

>>> np.repeat(arr, 3).reshape(-1, 3)
array([[9.975, 9.975, 9.975],
       [9.976, 9.976, 9.976],
       [9.977, 9.977, 9.977],
       [9.978, 9.978, 9.978],
       [9.979, 9.979, 9.979],
       [9.98 , 9.98 , 9.98 ],
       [9.981, 9.981, 9.981],
       [9.982, 9.982, 9.982],
       [9.983, 9.983, 9.983],
       [9.984, 9.984, 9.984],
       [9.985, 9.985, 9.985],
       [9.986, 9.986, 9.986],
       [9.987, 9.987, 9.987],
       [9.988, 9.988, 9.988],
       [9.989, 9.989, 9.989],
       [9.99 , 9.99 , 9.99 ],
       [9.991, 9.991, 9.991],
       [9.992, 9.992, 9.992],
       [9.993, 9.993, 9.993],
       [9.994, 9.994, 9.994]])

Answer 2

试试这个：

a = [ 9.975, 9.976, 9.977, 9.978, 9.979, 9.98,  9.981,
      9.982, 9.983, 9.984, 9.985, 9.986, 9.987, 9.988,
      9.989, 9.99,  9.991, 9.992, 9.993, 9.994 ]
n = 3
print([[x] * n for x in a])

我从你的示例输出中得到这个答案。你的措辞没有明确说明你想要的东西。

如果您使用numpy，我建议使用此解决方案：

a = np.array([ 9.975, 9.976, 9.977, 9.978, 9.979, 9.98,  9.981,
               9.982, 9.983, 9.984, 9.985, 9.986, 9.987, 9.988,
               9.989, 9.99,  9.991, 9.992, 9.993, 9.994 ])
print(np.array([ a ] * 3).transpose())

Answer 3

使用for循环

lst = [9.975, 9.976, 9.977, 9.978, 9.979, 9.98,  9.981, 9.982, 9.983, 9.984, 9.985, 9.986, 9.987, 9.988, 9.989, 9.99,  9.991, 9.992, 9.993, 9.994]
rslt = []
n = 3
for i in lst:
    rslt.append([i for j in range(n)])

编辑：为了使它更加pythonic，虽然可读性稍差：

lst = [9.975, 9.976, 9.977, 9.978, 9.979, 9.98,  9.981, 9.982, 9.983, 9.984, 9.985, 9.986, 9.987, 9.988, 9.989, 9.99,  9.991, 9.992, 9.993, 9.994]
n = 3
rslt = [[i for j in range(n)] for i in lst]

Answer 4

我们可以运行for循环并将数字存储到空列表中。然后我们可以获取该空列表，然后将其转换为数组，以获得您正在寻找的输出。这是一个例子：

import numpy as np



array = np.array([9.975, 9.976, 9.977, 9.978, 9.979, 9.98,  9.981, 9.982, 9.983, 9.984, 9.985, 9.986,
                  9.987, 9.988, 9.989, 9.99,  9.991, 9.992, 9.993, 9.994])
empty_list = []

for number in array:
    num1 = float(number)
    num2 = float(number)
    num3 = float(number)
    empty_list.append(num1)
    empty_list.append(num2)
    empty_list.append(num3)

array = np.array(empty_list).reshape(20, 3)
print(array)

这是你的输出：

[[ 9.975  9.975  9.975]
 [ 9.976  9.976  9.976]
 [ 9.977  9.977  9.977]
 [ 9.978  9.978  9.978]
 [ 9.979  9.979  9.979]
 [ 9.98   9.98   9.98 ]
 [ 9.981  9.981  9.981]
 [ 9.982  9.982  9.982]
 [ 9.983  9.983  9.983]
 [ 9.984  9.984  9.984]
 [ 9.985  9.985  9.985]
 [ 9.986  9.986  9.986]
 [ 9.987  9.987  9.987]
 [ 9.988  9.988  9.988]
 [ 9.989  9.989  9.989]
 [ 9.99   9.99   9.99 ]
 [ 9.991  9.991  9.991]
 [ 9.992  9.992  9.992]
 [ 9.993  9.993  9.993]
 [ 9.994  9.994  9.994]]

简而言之，我们只需对每个数字运行一个for循环，将它存储三次到空列表，创建一个数组，重新整形，然后我们得到你正在寻找的输出。

重复值1D NumPy数组“N”次

4 个答案: