問題描述

有沒有辦法在 0 和 1 之間步進 0.1?

Is there a way to step between 0 and 1 by 0.1?

我以為我可以這樣做，但它失敗了:

I thought I could do it like the following, but it failed:

for i in range(0, 1, 0.1):
    print i

相反，它說 step 參數不能為零，這是我沒想到的.

Instead, it says that the step argument cannot be zero, which I did not expect.

推薦答案

比起直接使用小數步長，用你想要的點數來表達要安全得多.否則，浮點舍入錯誤很可能會給你一個錯誤的結果.

Rather than using a decimal step directly, it's much safer to express this in terms of how many points you want. Otherwise, floating-point rounding error is likely to give you a wrong result.

您可以使用 linspace 函數NumPy 庫(它不是標準庫的一部分，但相對容易獲得).linspace 需要返回多個點，還可以讓您指定是否包含正確的端點:

You can use the linspace function from the NumPy library (which isn't part of the standard library but is relatively easy to obtain). linspace takes a number of points to return, and also lets you specify whether or not to include the right endpoint:

>>> np.linspace(0,1,11)
array([ 0. ,  0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9,  1. ])
>>> np.linspace(0,1,10,endpoint=False)
array([ 0. ,  0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9])

如果你真的想使用浮點步進值，你可以使用 numpy.arange.

If you really want to use a floating-point step value, you can, with numpy.arange.

>>> import numpy as np
>>> np.arange(0.0, 1.0, 0.1)
array([ 0. ,  0.1,  0.2,  0.3,  0.4,  0.5,  0.6,  0.7,  0.8,  0.9])

浮點舍入誤差會導致問題.這是一個簡單的例子，其中舍入錯誤導致 arange 在它應該只產生 3 個數字時產生一個長度為 4 的數組:

Floating-point rounding error will cause problems, though. Here's a simple case where rounding error causes arange to produce a length-4 array when it should only produce 3 numbers:

>>> numpy.arange(1, 1.3, 0.1)
array([1. , 1.1, 1.2, 1.3])

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