单独检查数组元素的真值

Question

我的代码是：-

import numpy as np
def f(x):
    if x<=0.5:
        return 0
    else:
        return 1
        
X=np.array([0.1,0.2,0.6,0.5,0.01,1])
print(f(X))

此代码给出一个错误。

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

但我不知道a.any()或a.all()对我有什么用处。我要输出的是：-

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

我可以使用for循环，但是是否有任何快捷语法使我只需一次就可以得到所需的输出？

Answer 1

您可以使用vectorize - https://numpy.org/doc/stable/reference/generated/numpy.vectorize.html

In [1]: import numpy as np

In [2]: X=np.array([0.1,0.2,0.6,0.5,0.01,1])

In [3]: f = lambda x: 0 if x<=0.5 else 1

In [4]: func = np.vectorize(f)

In [5]: func(X)
Out[5]: array([0, 0, 1, 0, 0, 1])

或

where - https://numpy.org/doc/stable/reference/generated/numpy.where.html

In [7]: np.where(X<=0.5, 0,1)
Out[7]: array([0, 0, 1, 0, 0, 1])