我有一个df，如下所示.我已经添加了一个包含所有行的合计的新列和一个包含所有列的合计的新行:
A B C D Total
-------------------
1 2 3 4 10
5 6 7 8 26
6 8 10 12 36
现在我需要再添加一行，该行的第一个元素将是NaN，其余的将是从Total行的前一列中减go 的列.
A B C D Total
1 2 3 4 10
5 6 7 8 26
6 8 10 12 36
NaN 2 2 2 24 <--- new row
谢谢
这将是df.append很少使用的用例之一，但是您可以提取最后一行的iloc[-1]和diff的各个值，然后将其与原始值组合在一起.
Option 1
One method of doing this concatenation would be using pd.concat
df2 = pd.concat([df, df.iloc[-1].diff().to_frame().T])
print (df2)
A B C D Total
0 1.0 2.0 3.0 4.0 10.0
1 5.0 6.0 7.0 8.0 26.0
2 6.0 8.0 10.0 12.0 36.0
2 NaN 2.0 2.0 2.0 24.0
哪里,
df.iloc[-1].diff().to_frame().T # dataframe with 1 row
A B C D Total
2 NaN 2.0 2.0 2.0 24.0
Option 2
An alternative using inplace assignment with loc:
df.loc[len(df.index)] = df.iloc[-1].diff()
print (df)
A B C D Total
0 1.0 2.0 3.0 4.0 10.0
1 5.0 6.0 7.0 8.0 26.0
2 6.0 8.0 10.0 12.0 36.0
3 NaN 2.0 2.0 2.0 24.0
哪里,
df.iloc[-1].diff() # series
A NaN
B 2.0
C 2.0
D 2.0
Total 24.0
Name: 2, dtype: float64
Option 3
Here's an option that has a bit of fun with dictionaries and pd.DataFrame:
pd.DataFrame([*df.to_dict('records'), df.iloc[-1].diff().to_dict()])
A B C D Total
0 1.00 2.00 3.00 4.00 10.00
1 5.00 6.00 7.00 8.00 26.00
2 6.00 8.00 10.00 12.00 36.00
3 NaN 2.00 2.00 2.00 24.00
Option 4 [deprecated]
On older versions (pandas <= 1.4) I would have recommended using append like this:
df2 = df.append(df.iloc[-1].diff(), ignore_index=True)