为什么一维数组的形状未将行数显示为1? [英] Why does the shape of a 1D array not show the number of rows as 1?

问题描述

我知道numpy数组有一个叫做shape的方法，该方法返回[行数，列数]，shape [0]给出行数，shape [1]给出列数.

I know that numpy array has a method called shape that returns [No.of rows, No.of columns], and shape[0] gives you the number of rows, shape[1] gives you the number of columns.

a = numpy.array([[1,2,3,4], [2,3,4,5]])
a.shape
>> [2,4]
a.shape[0]
>> 2
a.shape[1]
>> 4

但是，如果我的数组只有一行，那么它将返回[列数].并且shape [1]将不在索引中.例如

However, if my array only have one row, then it returns [No.of columns, ]. And shape[1] will be out of the index. For example

a = numpy.array([1,2,3,4])
a.shape
>> [4,]
a.shape[0]
>> 4    //this is the number of column
a.shape[1]
>> Error out of index

现在，如果该数组可能只有一行，如何获取numpy数组的行数?

Now how do I get the number of rows of an numpy array if the array may have only one row?

谢谢

推荐答案

当您拥有2D数组时，行和列的概念适用.但是，数组numpy.array([1,2,3,4])是一维数组，因此只有一个维度，因此shape正确地返回一个可迭代的单值.

The concept of rows and columns applies when you have a 2D array. However, the array numpy.array([1,2,3,4]) is a 1D array and so has only one dimension, therefore shape rightly returns a single valued iterable.

对于同一阵列的2D版本，请考虑以下内容:

For a 2D version of the same array, consider the following instead:

>>> a = numpy.array([[1,2,3,4]]) # notice the extra square braces
>>> a.shape
(1, 4)

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