Python 使用 numpy 数组调用 lambda
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Calling a lambda with a numpy array
提问by peco
While familiarizing myself with numpy, I noticed an interesting behaviour in numpyarrays:
在熟悉 时numpy，我注意到numpy数组中有一个有趣的行为：
import numpy as np
arr = np.array([1, 2, 3])
scale = lambda x: x * 3
scale(arr) # Gives array([3, 6, 9])
Contrast this with normal Python lists:
将此与普通的 Python 列表进行对比：
arr = [1, 2, 3]
scale = lambda x: x * 3
scale(arr) # Gives [1, 2, 3, 1, 2, 3, 1, 2, 3]
I'm curious as to how this is possible. Does a numpyarray override the multiplication operator or something?
我很好奇这怎么可能。numpy数组是否覆盖乘法运算符或其他什么？
采纳答案by Crispin
回答by Shivkumar kondi
Its all about Overridingoperators in numpy
全部是关于在 numpy 中覆盖运算符
You can learn numpy.arry here
你可以在这里学习 numpy.arry
Let us focus on your lamda function for each;
让我们专注于您的 lamda 函数；
1. numpy array :
1. numpy 数组：
arr = numpy.array([1, 2, 3])
type(arr)
scale = lambda x: x * 3
scale(arr)
this takes each element from array
这从数组中获取每个元素
2. normal list:
2.正常名单：
a =[1,2,3]
type(a)
scale = lambda x: x * 3
scale(a)
this takes full list as x and multiplies the list here itself
这将完整列表作为 x 并在此处乘以列表本身
回答by ibezito
These are two different objects which behaves differently when you use * operator on them.
这是两个不同的对象，当您对它们使用 * 运算符时，它们的行为会有所不同。
In the first case you generate a numpy array. In this case, * operator was overloaded for performing multiplication. i.e. every element will be multiplied by 3.
In the second case you generate a list. In this case the * operator is treated as a repetition operator, and the entire list is repeated 3 times.
在第一种情况下，您生成一个 numpy 数组。在这种情况下，* 运算符被重载以执行乘法。即每个元素都将乘以 3。
在第二种情况下，您生成一个列表。在这种情况下， * 运算符被视为重复运算符，整个列表重复 3 次。
code example:
代码示例：
type(np.array([1,2,3]))
type([1, 2, 3])
result:
结果：
list
numpy.ndarray
