如何用 Pandas DataFrame 中的前一个或下一个值替换 NaN? [英] How to replace NaNs by preceding or next values in pandas DataFrame?

问题描述

假设我有一个带有一些 NaN 的 DataFrame:

<预><代码>>>>将熊猫导入为 pd>>>df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])>>>df0 1 20 1 2 31 4 NaN 南2 南 南 9

我需要做的是将每个 NaN 替换为其上方同一列中的第一个非 NaN 值.假设第一行永远不会包含 NaN.所以对于前面的例子，结果是

<代码> 0 1 20 1 2 31 4 2 32 4 2 9

我可以逐列、逐个元素地遍历整个 DataFrame 并直接设置值，但是否有一种简单的(最好是无循环的)方法来实现这一点?

解决方案

您可以使用 fillna 方法在 DataFrame 上并指定方法为 ffill(前向填充):

<预><代码>>>>df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])>>>df.fillna(method='ffill')0 1 20 1 2 31 4 2 32 4 2 9

这个方法...

<块引用>

传播[s]最后一个有效观察结果到下一个有效观察

相反，还有一个 bfill 方法.

此方法不会就地修改 DataFrame - 您需要将返回的 DataFrame 重新绑定到变量或指定 inplace=True:

df.fillna(method='ffill', inplace=True)

Suppose I have a DataFrame with some NaNs:

>>> import pandas as pd
>>> df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
>>> df
    0   1   2
0   1   2   3
1   4 NaN NaN
2 NaN NaN   9

What I need to do is replace every NaN with the first non-NaN value in the same column above it. It is assumed that the first row will never contain a NaN. So for the previous example the result would be

   0  1  2
0  1  2  3
1  4  2  3
2  4  2  9

I can just loop through the whole DataFrame column-by-column, element-by-element and set the values directly, but is there an easy (optimally a loop-free) way of achieving this?

解决方案

You could use the fillna method on the DataFrame and specify the method as ffill (forward fill):

>>> df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
>>> df.fillna(method='ffill')
   0  1  2
0  1  2  3
1  4  2  3
2  4  2  9

This method...

propagate[s] last valid observation forward to next valid

To go the opposite way, there's also a bfill method.

This method doesn't modify the DataFrame inplace - you'll need to rebind the returned DataFrame to a variable or else specify inplace=True:

df.fillna(method='ffill', inplace=True)

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