在大熊猫的行和列中对角复制值？

Question

这里是一个示例数据框：

                     datetime temp  T1  T2  T3  T4  T5
115 2020-01-04 02:53:00+00:00   58   0   0   0   0   0
116 2020-01-04 03:53:00+00:00   51   0   0   0   0   0
117 2020-01-04 04:53:00+00:00   49   0   0   0   0   0
118 2020-01-04 05:53:00+00:00   48   0   0   0   0   0
119 2020-01-04 06:00:00+00:00   48   0   0   0   0   0
120 2020-01-04 06:53:00+00:00   47   0   0   0   0   0

这是我想要的输出是：

                     datetime temp  T1  T2  T3  T4  T5
115 2020-01-04 02:53:00+00:00   58   0   0   0   0   0
116 2020-01-04 03:53:00+00:00   51   58  0   0   0   0
117 2020-01-04 04:53:00+00:00   49   51  58  0   0   0
118 2020-01-04 05:53:00+00:00   48   49  51  58  0   0
119 2020-01-04 06:00:00+00:00   48   48  49  51  58  0
120 2020-01-04 06:53:00+00:00   47   48  48  49  51  58

Answer 1

使用Series.shift

Series.shift

我们也可以使用

for col in df.columns[df.columns.str.contains('T')]:
    df[col] = df['temp'].shift(int(col[-1]),fill_value = 0)
print(df)

pd.Index.difference

输出

pd.Index.difference

Answer 2

用途：

for col in df.columns.difference(['datetime','temp']):
        df[col] = df['temp'].shift(int(col[-1]),fill_value = 0)

对于最后的熊猫版本：

                      datetime  temp  T1  T2  T3  T4  T5
115  2020-01-04-02:53:00+00:00    58   0   0   0   0   0
116  2020-01-04-03:53:00+00:00    51  58   0   0   0   0
117  2020-01-04-04:53:00+00:00    49  51  58   0   0   0
118  2020-01-04-05:53:00+00:00    48  49  51  58   0   0
119  2020-01-04-06:00:00+00:00    48  48  49  51  58   0
120  2020-01-04-06:53:00+00:00    47  48  48  49  51  58

---

for i in range(1,len(df)):
    df[f'T{i}'] = df['temp'].shift(i).fillna(0).astype(int)