将 lambda 函数应用于 Pandas 滚动窗口系列 [英] Applying lambda function to a pandas rolling window series

问题描述

我有一个函数，它接受一个数组和一个值，并返回一个值.我想在滚动的基础上将它应用于我的系列 s，因此数组始终是滚动​​窗口.这是我尝试过的(未成功)的最小示例，使用 np.random.choice 代替我的真实函数.我找到了很多用于查找滚动均值和其他内置函数的示例，但无法使其适用于我的任意 lambda 函数.

s = pd.Series([1,2,3,4,5,6,7,8,9])滚动窗口 = s.rolling(3)

第一次尝试:

new_values = s.apply(lambda x: np.random.choice(rolling_window, size=1))

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ValueError: a [a is the first positional param of choice(), so reference to 'rolling_window'] 必须是一维或整数

另一种尝试:

new_values = rolling_window.apply(lambda x: np.random.choice(size=1))

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TypeError: choice() 至少需要 1 个位置参数(0 给定)

...如何对系列中的每个值、系列中的每个滚动数组窗口应用任意 lambda 函数(采用数组和值)?

解决方案

IIUC，如果你想做的只是在窗口上应用一个函数，你的第二次尝试就接近了:

rolling_window.apply(lambda x: np.random.choice(x, size=1))

<小时>

但是，您可以像这样绕过 lambda 的使用:

rolling_window.apply(np.random.choice, kwargs={'size' : 1})0 南1 纳米2 1.03 4.04 4.05 5.06 7.07 7.08 8.0数据类型:float64

传递给函数的附加参数在 args 和 kwargs 中.

I have a function which takes an array and a value, and returns a value. I would like to apply it to my Series s on a rolling basis, so the array is always the rolling window. Here's a minimal example of what I've tried (unsuccessfully), using np.random.choice in place of my real function. I find lots of examples for finding rolling means and other built-in functions, but can't get it to work for my arbitrary lambda function.

s = pd.Series([1,2,3,4,5,6,7,8,9])
rolling_window = s.rolling(3)

First attempt:

new_values = s.apply(lambda x: np.random.choice(rolling_window, size=1)) 

ValueError: a [a is the first positional param of choice(), so refers to 'rolling_window'] must be 1-dimensional or an integer

Another attempt:

new_values = rolling_window.apply(lambda x: np.random.choice(size=1))

TypeError: choice() takes at least 1 positional argument (0 given)

...How do I apply an arbitrary lambda function (taking an array and a value) on each value in my Series, on each rolling-array window in my Series?

解决方案

IIUC, if all you want to do is apply a function on the window, your second attempt comes close:

rolling_window.apply(lambda x: np.random.choice(x, size=1))

---

However, you can circumvent the use of the lambda like this:

rolling_window.apply(np.random.choice, kwargs={'size' : 1})

0    NaN
1    NaN
2    1.0
3    4.0
4    4.0
5    5.0
6    7.0
7    7.0
8    8.0
dtype: float64

Additional arguments to the function you pass go in args and kwargs.

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