将数据帧中的某些列相乘，而不删除其他现有列

Question

我正在努力用Python（pandas）创建一个新的数据帧，它保留数据帧中的所有现有列，同时只将某些列乘以100000

我使用的代码是：

df[['t1','t2','t3','t4','t5','t6']].multiply(100000)

此代码保留t1-t6的列，但删除其他所有内容。（大约10个其他列）运行此代码时，是否有方法保留我的其他现有列

Answer 1

可以考虑使用{{CD1>}，其中要处理

的^ {< CD2>}信息

import pandas as pd

d={'t1':[2,3,1,4,5],
   't2':[5,4,3,2,1],
   'other1':[1,9,9,8,7],
   'other2':[3,3,4,5,6]}
df=pd.DataFrame(d)
print(df)

target_columns=['t1','t2']
for one_column in target_columns:
  df[one_column]=df[one_column].map(lambda x:x*100000)
print(df)

Answer 2

许多选择

1. 重新分配给指定的列：

df[['t1', 't2', 't3', 't4', 't5', 't6']] = \
    df[['t1', 't2', 't3', 't4', 't5', 't6']].multiply(100000)

1. ^{}+multiply：

df.loc[:, ['t1', 't2', 't3', 't4', 't5', 't6']] = \
    df[['t1', 't2', 't3', 't4', 't5', 't6']].multiply(100000)

1. *=：

df[['t1', 't2', 't3', 't4', 't5', 't6']] *= 100000

1. ^{}+*=：

df.loc[:, ['t1', 't2', 't3', 't4', 't5', 't6']] *= 100000

1. ^{}：

df.update(df[['t1', 't2', 't3', 't4', 't5', 't6']].multiply(100000))

测试框架：

df = pd.DataFrame({
    't1': {0: 1}, 't2': {0: 2}, 't3': {0: 3}, 't4': {0: 4}, 't5': {0: 5},
    't6': {0: 6}, 't7': {0: 7}, 't8': {0: 8}, 't9': {0: 9}, 't10': {0: 10},
    't11': {0: 11}
})

输出：

       t1      t2      t3      t4      t5      t6  t7  t8  t9  t10  t11
0  100000  200000  300000  400000  500000  600000   7   8   9   10   11

---

通过perfplot的定时信息：

^{tb1}$

import numpy as np
import pandas as pd
import perfplot


def gen_data(n):
    return pd.DataFrame({f't{i}': np.arange(0, n) for i in range(1, 12)})


def assign_back_multiply(df):
    df[['t1', 't2', 't3', 't4', 't5', 't6']] = \
        df[['t1', 't2', 't3', 't4', 't5', 't6']].multiply(100000)
    return df


def assign_back_with_loc_multiply(df):
    df.loc[:, ['t1', 't2', 't3', 't4', 't5', 't6']] = \
        df[['t1', 't2', 't3', 't4', 't5', 't6']].multiply(100000)
    return df


def times_equals(df):
    df[['t1', 't2', 't3', 't4', 't5', 't6']] *= 100000
    return df


def times_equals_with_loc(df):
    df.loc[:, ['t1', 't2', 't3', 't4', 't5', 't6']] *= 100000
    return df


def df_update(df):
    df.update(df[['t1', 't2', 't3', 't4', 't5', 't6']].multiply(100000))
    return df


# @YoungMin Park's Answer
def for_loop_map(df):
    target_columns = ['t1', 't2', 't3', 't4', 't5', 't6']
    for one_column in target_columns:
        df[one_column] = df[one_column].map(lambda x: x * 100000)
    return df


if __name__ == '__main__':
    out = perfplot.bench(
        setup=gen_data,
        kernels=[
            assign_back_multiply,
            assign_back_with_loc_multiply,
            times_equals,
            times_equals_with_loc,
            df_update,
            for_loop_map
        ],
        labels=[
            'assign_back_multiply',
            'assign_back_with_loc_multiply',
            'times_equals',
            'times_equals_with_loc',
            'df_update',
            'for_loop_map @YoungMin Park'
        ],
        n_range=[2 ** k for k in range(25)],
        equality_check=None
    )
    out.save('perfplot_results.png', transparent=False)