使用lambda时，numpy.piecewise返回错误的答案

Question

I'm trying to "splice" a function by inserting a constant part into it using numpy.piecewise:

import numpy as np

func = lambda x: 20 -x
bid_price = 15.0
bid_power = 1.0
bid_start = 5.0

new_func = lambda x: np.piecewise(x, [0 <= x < bid_start,
              (x>= bid_start) & (x < bid_start + bid_power),  x >= bid_start + bid_power],
              [lambda t: func(t), lambda t : bid_price,
               lambda t: func(t - bid_power)])

While the function gives correct result for x's that match the first condition, any other x gives me a zero:

In[65]: new_func(15.0)
Out[65]: array(0.0)

I looked through the code of numpy. piecewise (can't quite debug it), but it doesn't seem like there's anything there to cause this behavior. Casting x to numpy.array doesn't help. What am I doing wrong here?

Answer 1

Well, it works if you input a non-scalar np.array :

>>> new_func(np.array([15]))
array([6])

With a minor modification (0 <= x) & (x < bid_start) instead of 0 <= x < bid_start it also works on an extended array:

>>> new_func(np.arange(20))
array([20, 19, 18, 17, 16, 15, 15, 14, 13, 12, 11, 10,  9,  8,  7,  6,  5, 4,  3,  2])

I guess np.piecewise is probably designed not to accept scalars (it's a replacement function for arrays after all) or it's a Bug. But if you want to process scalars you should code it in pure python that's way faster:

def new_func(x):
    if x < 0:
        raise ValueError()
    elif x < bid_start:
        return 20 - x
    elif x < bid_start + bid_power:
        return bid_price
    else:
        return 20 - x + bid_power