python中的np.ceil不起作用

Question

我实际上在python中遇到了“np.ceil”的一些问题。

import numpy as np

x_start = 0
y_start = 0
x_end = 2
y_end = 1
x_step = 0.4
y_step = 0.3

x_segment = int(np.ceil((x_end-x_start)/x_step))
y_segment = int(np.ceil((y_end-y_start)/y_step))

print "N. x: " + str(x_segment)
print "N. y: " + str(y_segment)

matrix = np.zeros((y_segment, x_segment), dtype=int)

import matplotlib.pyplot as plt
import matplotlib.patches as patches

def frange(x, y, jump):
  while x < y:
    yield x
    x += jump
for y in frange(y_start, y_end, y_step):
    print "***"
    for x in frange(x_start, x_end, x_step):
        count = 0
        print "(" + str(x) + "; " + str(y) + ")   m[" + str(y_segment - int(np.ceil(y/y_step)) - 1) + "][" + str(int(np.ceil(x/x_step)))+"]" + " | x/x_step: " + str(x/x_step) + " | np.ceil(x/x_step): " + str(int(np.ceil(x/x_step)))

我应该得到类似的东西：

...
***
(0; 0)   m[3][0] | x/x_step: 0.0 | np.ceil(x/x_step): 0
(0.4; 0)   m[3][1] | x/x_step: 1.0 | np.ceil(x/x_step): 1
(0.8; 0)   m[3][2] | x/x_step: 2.0 | np.ceil(x/x_step): 2
(1.2; 0)   m[3][3] | x/x_step: 3.0 | np.ceil(x/x_step): 3
(1.6; 0)   m[3][4] | x/x_step: 4.0 | np.ceil(x/x_step): 4
***
...

但数字“3”被替换为“4”。

***
(0; 0)   m[3][0] | x/x_step: 0.0 | np.ceil(x/x_step): 0
(0.4; 0)   m[3][1] | x/x_step: 1.0 | np.ceil(x/x_step): 1
(0.8; 0)   m[3][2] | x/x_step: 2.0 | np.ceil(x/x_step): 2
(1.2; 0)   m[3][4] | x/x_step: 3.0 | np.ceil(x/x_step): 4
(1.6; 0)   m[3][4] | x/x_step: 4.0 | np.ceil(x/x_step): 4
***

你知道为什么吗？我该如何修复我的代码？谢谢！

Answer 1

Short Answer: Float imprecision issue

答案很长：

frange似乎有一些浮动精度问题。看看当你通过frange时会发生什么：

如果你只是打印原始浮动：

>>> [x for x in frange(x_start, x_end, x_step)]
[0, 0.4, 0.8, 1.2000000000000002, 1.6]

由于某种原因，1.2不完全是1.2。当你这样做时：

np.ceil(1.2000000000000002/x_step)

你得到4.0（换句话说，np.ceil工作正常）。

你想要的基本上是np.ceil(1.2/x_step)，它等于3.0

在使用np.round()之前，我建议使用x或类似的东西来围绕你的np.ceil()值