Pandas 数据框合并行以删除 NaN [英] Pandas dataframe merging rows to remove NaN

问题描述

我有一个包含一些 NaN 的数据框:

I have a dataframe with some NaNs:

hostname period Teff
51 Peg  4.2293  5773
51 Peg  4.231   NaN
51 Peg  4.23077 NaN
55 Cnc  44.3787 NaN
55 Cnc  44.373  NaN
55 Cnc  44.4175 NaN
55 Cnc  NaN 5234
61 Vir  NaN 5577
61 Vir  38.021  NaN
61 Vir  123.01  NaN

具有相同主机名"的行都指向同一个对象，但正如您所见，某些条目在不同列下具有 NaN.我想合并同一主机名下的所有行，以便保留每列中的第一个有限值(如果所有值都是 NaN，则删除该行).所以结果应该是这样的:

The rows with the same "hostname" all refer to the same object, but as you can see, some entries have NaNs under various columns. I'd like to merge all the rows under the same hostname such that I retain the first finite value in each column (drop the row if all values are NaN). So the result should look like this:

hostname period Teff
51 Peg  4.2293  5773
55 Cnc  44.3787 5234
61 Vir  38.021  5577

你会怎么做?

推荐答案

Use groupby.first;它需要 第一个非 NA 值:

df.groupby('hostname')[['period', 'Teff']].first().reset_index()
#  hostname   period  Teff
#0      Cnc  44.3787  5234
#1      Peg   4.2293  5773
#2      Vir  38.0210  5577

或者使用自定义聚合函数手动执行此操作:

Or manually do this with a custom aggregation function:

df.groupby('hostname')[['period', 'Teff']].agg(lambda x: x.dropna().iat[0]).reset_index()

这要求每组至少有一个非 NA 值.

This requires each group has at least one non NA value.

编写自己的函数来处理边缘情况:

Write your own function to handle the edge case:

def first_(g):
    non_na = g.dropna()
    return non_na.iat[0] if len(non_na) > 0 else pd.np.nan

df.groupby('hostname')[['period', 'Teff']].agg(first_).reset_index()

#  hostname   period  Teff
#0      Cnc  44.3787  5234
#1      Peg   4.2293  5773
#2      Vir  38.0210  5577

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