Replace NaN value with a median?

问题

So I am trying to use Pandas to replace all NaN values in a table with the median across a particular range. I am working with a larger dataset but for example

np.random.seed(0)
rng = pd.date_range('2020-09-24', periods=20, freq='0.2H')
df = pd.DataFrame({ 'Date': rng, 'Val': np.random.randn(len(rng)), 'Dist' :np.random.randn(len(rng)) }) 
df.Dist[df.Dist<=-0.6] = np.nan
df.Val[df.Val<=-0.5] = np.nan

What I want to do is replace the NaN values for Val and Dist with the median value for each hour for that column. I have managed to get the median values in a separate reference table:

df.set_index('Date', inplace=True)
df = df.assign(Hour = lambda x : x.index.hour)
df_val = df[["Val", "Hour"]].groupby("Hour").median()
df_dist = df[["Dist", "Hour"]].groupby("Hour").median()

But now I have tried all of the below commands in various forms and cannot work out how to fill NaN values.

df[["Val","Hour"]].mask(df['Val'].isna(), df_val.iloc[df.Hour], inplace=True)

df.where(df['Val'].notna(), other=df_val[df.Hour],axis = 0)

df["Val"] = np.where(df['Val'].notna(), df['Val'], df_val(df.Hour))

df.replace({"Val":{np.nan:df_val[df.Hour]}, "Dist":{np.nan:df_dist[df.Hour]}})

回答1:

You can use groupby.transform and fillna:

cols = ['Val','Dist']
df[cols] =  df[cols].fillna(df.groupby(df.Date.dt.floor('H'))
                              [cols].transform('median')
                           )

Output:

                  Date       Val      Dist
0  2020-09-24 00:00:00  1.764052  0.864436
1  2020-09-24 00:12:00  0.400157  0.653619
2  2020-09-24 00:24:00  0.978738  0.864436
3  2020-09-24 00:36:00  2.240893  0.864436
4  2020-09-24 00:48:00  1.867558  2.269755
5  2020-09-24 01:00:00  0.153690  0.757559
6  2020-09-24 01:12:00  0.950088  0.045759
7  2020-09-24 01:24:00 -0.151357 -0.187184
8  2020-09-24 01:36:00 -0.103219  1.532779
9  2020-09-24 01:48:00  0.410599  1.469359
10 2020-09-24 02:00:00  0.144044  0.154947
11 2020-09-24 02:12:00  1.454274  0.378163
12 2020-09-24 02:24:00  0.761038  0.154947
13 2020-09-24 02:36:00  0.121675  0.154947
14 2020-09-24 02:48:00  0.443863 -0.347912
15 2020-09-24 03:00:00  0.333674  0.156349
16 2020-09-24 03:12:00  1.494079  1.230291
17 2020-09-24 03:24:00 -0.205158  1.202380
18 2020-09-24 03:36:00  0.313068 -0.387327
19 2020-09-24 03:48:00  0.323371 -0.302303

回答2:

You can use a groupby -> transform operation, while also utilizing the pd.Grouper class to perform the hourly conversion. This will essentially create a dataframe with the same shape as your original with the hourly medians. Once you have this, you can directly use DataFrame.fillna

hourly_medians = df.groupby(pd.Grouper(key="Date", freq="H")).transform("median")
out = df.fillna(hourly_medians)

print(out)

                  Date       Val      Dist
0  2020-09-24 00:00:00  1.764052  0.864436
1  2020-09-24 00:12:00  0.400157  0.653619
2  2020-09-24 00:24:00  0.978738  0.864436
3  2020-09-24 00:36:00  2.240893  0.864436
4  2020-09-24 00:48:00  1.867558  2.269755
5  2020-09-24 01:00:00  0.153690  0.757559
6  2020-09-24 01:12:00  0.950088  0.045759
7  2020-09-24 01:24:00 -0.151357 -0.187184
8  2020-09-24 01:36:00 -0.103219  1.532779
9  2020-09-24 01:48:00  0.410599  1.469359
10 2020-09-24 02:00:00  0.144044  0.154947
11 2020-09-24 02:12:00  1.454274  0.378163
12 2020-09-24 02:24:00  0.761038  0.154947
13 2020-09-24 02:36:00  0.121675  0.154947
14 2020-09-24 02:48:00  0.443863 -0.347912
15 2020-09-24 03:00:00  0.333674  0.156349
16 2020-09-24 03:12:00  1.494079  1.230291
17 2020-09-24 03:24:00 -0.205158  1.202380
18 2020-09-24 03:36:00  0.313068 -0.387327
19 2020-09-24 03:48:00  0.323371 -0.302303

回答3:

Using what you've done, I'd do this:

df.Val = df.Val.fillna(df.Hour.map(df_val.squeeze()))
df.Dist = df.Val.fillna(df.Hour.map(df_dist.squeeze()))

来源：https://stackoverflow.com/questions/64859946/replace-nan-value-with-a-median

标签

python

pandas

numpy

data-cleaning