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RKIDEV
Asked: 2025-01-08 15:11:11 +0800 CST

Panda 迭代行并将第 n 行的值乘以下一行（n+1）的值

  • 772

我正在尝试迭代多列行并将第 n 行乘以 n+1 行，然后添加列。

我尝试了下面的代码并且运行良好。

还有其他简单的方法可以同时实现减法和乘法部分吗？

import pandas as pd

df = pd.DataFrame({'C': ["Spark","PySpark","Python","pandas","Java"],
                    'F' : [2,4,3,5,4],
                    'D':[3,4,6,5,5]})

df1 = pd.DataFrame({'C': ["Spark","PySpark","Python","pandas","Java"],
                    'F': [1,2,1,2,1],
                    'D':[1,2,2,2,1]})

df = pd.merge(df, df1, on="C")

df['F_x-F_y'] = df['F_x'] - df['F_y']
df['D_x-D_y'] = df['D_x'] - df['D_y']

for index, row in df.iterrows():
    df['F_mul'] = df['F_x-F_y'].mul(df['F_x-F_y'].shift())
    df['D_mul'] = df['D_x-D_y'].mul(df['D_x-D_y'].shift())

df['F+D'] = df['F_mul'] + df['D_mul']

输出 -

         C  F_x  D_x  F_y  D_y  F_x-F_y  D_x-D_y  F_mul  D_mul   F+D
0    Spark    2    3    1    1        1        2    NaN    NaN   NaN
1  PySpark    4    4    2    2        2        2    2.0    4.0   6.0
2   Python    3    6    1    2        2        4    4.0    8.0  12.0
3   pandas    5    5    2    2        3        3    6.0   12.0  18.0
4     Java    4    5    1    1        3        4    9.0   12.0  21.0
python

3 个回答

  1. Best Answer

    首先通过删除迭代iterrows，然后可以通过以下方式简化广义解：

    cols = ['F','D']

for col in cols:
    s = df[f'{col}_x'].sub(df[f'{col}_y'])
    df[f'{col}_mul'] = s.mul(s.shift())

df['+'.join(cols)] = df.filter(like='mul').sum(axis=1, min_count=1)
print (df)
         C  F_x  D_x  F_y  D_y  F_mul  D_mul   F+D
0    Spark    2    3    1    1    NaN    NaN   NaN
1  PySpark    4    4    2    2    2.0    4.0   6.0
2   Python    3    6    1    2    4.0    8.0  12.0
3   pandas    5    5    2    2    6.0   12.0  18.0
4     Java    4    5    1    1    9.0   12.0  21.0

    另一个想法是将所有列一起处理 - 优点是不需要指定要处理的列：

    df1 = (df.filter(like='x').rename(columns=lambda x: x.replace('x','mul'))
         .sub(df.filter(like='y').rename(columns=lambda x: x.replace('y','mul'))))

df2 = df1.mul(df1.shift())
df = df.join(df2)

df['+'.join(x.replace('_mul','') for x in df2.columns)] = df2.sum(axis=1, min_count=1)
print (df)
         C  F_x  D_x  F_y  D_y  F_mul  D_mul   F+D
0    Spark    2    3    1    1    NaN    NaN   NaN
1  PySpark    4    4    2    2    2.0    4.0   6.0
2   Python    3    6    1    2    4.0    8.0  12.0
3   pandas    5    5    2    2    6.0   12.0  18.0
4     Java    4    5    1    1    9.0   12.0  21.0
    • 2

  2. 您可以完全删除遍历行的迭代。您的代码不会在循环内执行任何特定于行的操作iterrows()——也不会index使用任何row值。结果就是，您最终会在每次迭代中重新计算整个列（F_mul 和 D_mul）。

    例子

    ...

# Multiply with shifted values instead of using iterrows()
df['F_mul'] = df['F_x-F_y'] * df['F_x-F_y'].shift()
df['D_mul'] = df['D_x-D_y'] * df['D_x-D_y'].shift()
...
    • 1

  3. 你真的不应该在熊猫中使用循环iterrows。

    您想要做的事情可以通过向量实现。您不应该将键C作为列，而应该将其作为索引，以利用索引的自动对齐：

    # ensure C is the index
df = df.set_index('C')
df1 = df1.set_index('C')

# compute the difference
sub = df.sub(df1)

# combine df/df1/sub and compute the shifted multiplication
out = (pd.concat([df.add_suffix('_x'),
                  df1.add_suffix('_y'),
                  sub.add_suffix('_sub'),
                  sub.mul(sub.shift()).add_suffix('_mul')
                 ], axis=1)
      )

# compute the total sum
out['+'.join(df)] = out.filter(like='_mul').sum(axis=1, min_count=1)

# reset_index
out = out.reset_index()

    你甚至可以将所有内容作为单个表达式（sub 除外）：

    sub = df.sub(df1)

out = (pd.concat([df.add_suffix('_x'),
                  df1.add_suffix('_y'),
                  sub.add_suffix('_sub'),
                  sub.mul(sub.shift()).add_suffix('_mul')
                 ], axis=1)
         .assign(**{'+'.join(df): lambda x: x.filter(like='_mul')
                                             .sum(axis=1, min_count=1)})
         .reset_index()
      )

    输出：

             C  F_x  D_x  F_y  D_y  F_sub  D_sub  F_mul  D_mul   F+D
0    Spark    2    3    1    1      1      2    NaN    NaN   NaN
1  PySpark    4    4    2    2      2      2    2.0    4.0   6.0
2   Python    3    6    1    2      2      4    4.0    8.0  12.0
3   pandas    5    5    2    2      3      3    6.0   12.0  18.0
4     Java    4    5    1    1      3      4    9.0   12.0  21.0

    如果你不需要中间的_sub/_mul列，这会更简单：

    sub = df.sub(df1)

out = pd.concat([df.add_suffix('_x'),
                 df1.add_suffix('_y'),
                 sub.mul(sub.shift())
                    .sum(axis=1, min_count=1).rename('+'.join(df))
                ], axis=1).reset_index()

    或者：

    sub = df.sub(df1)

out = (df.join(df1, lsuffix='_x', rsuffix='_y')
         .assign(**{'+'.join(df): sub.mul(sub.shift())
                                     .sum(axis=1, min_count=1)})
         .reset_index()
      )

    输出：

             C  F_x  D_x  F_y  D_y   F+D
0    Spark    2    3    1    1   NaN
1  PySpark    4    4    2    2   6.0
2   Python    3    6    1    2  12.0
3   pandas    5    5    2    2  18.0
4     Java    4    5    1    1  21.0
    • 0

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