在熊猫的列之间移动特定字符串答案

【问题标题】：Moving specific strings between columns in pandas在熊猫的列之间移动特定字符串
【发布时间】：2023-02-05 03:43:13
【问题描述】：

我有一个熊猫属性数据框。

**Address**      | **Added on**.       | 
15 Smith Close   |  Added on 17/11/22  |
1 Apple Drive    |  Reduced on 19/11/22|
27 Pride place   |  Added on 18/1//22  |

我想将“Added on”列中所有“reduced on ...”的实例移动到数据框中名为“Reduced on”的另一列。我该怎么做？

非常感谢。

【问题讨论】：

标签： python pandas dataframe

【解决方案1】：

你可以使用pd.DataFrame.where：

df['Reduced on'] = df['Added on'].where(df['Added on'].str.contains('Reduced on'))
df['Added on'] = df['Added on'].where(~ df['Added on'].str.contains('Reduced on'))

df

          Address           Added on           Reduced on
0  15 Smith Close  Added on 17/11/22                  NaN
1   1 Apple Drive                NaN  Reduced on 19/11/22
2  27 Pride place  Added on 18/1//22                  NaN

或者另一种是使用pd.Series.str.extract & pd.DataFrame.concat：

pd.concat([df['Address'], df['Added on'].str.extract('(?P<Added_on>Add.*)|(?P<Reduced_on>Reduced.*)')], axis=1)

          Address           Added_on           Reduced_on
0  15 Smith Close  Added on 17/11/22                  NaN
1   1 Apple Drive                NaN  Reduced on 19/11/22
2  27 Pride place  Added on 18/1//22                  NaN

【讨论】：

谢谢阿努希拉万。就这样排序了。
哈哈，我刚刚注意到 Anoushiravan 的编码方式与我类似。即使 Anoushiravan 是最快的，我也会让我的代码到位
@LaurentB。是的，只要保留它。它经常发生。

【解决方案2】：

建议代码：

import pandas as pd
import numpy as np

# Build Dataframe to work on
df = pd.DataFrame({"**Address** ": ['15 Smith Close' , '1 Apple Drive', '27 Pride place'], 
                   "**Added on**": ['Added on 17/11/22', 'Reduced on 19/11/22', 'Added on 18/1//22']})

# Define the mask m
m = df['**Added on**'].str.contains('Reduced')               

# 1- Move 'Reduced' rows to **New Col**                       
df['**Reduced on**'] = df['**Added on**'].where(m, np.nan)
# 2- Erase 'Reduced' rows from  **Added on**
df['**Added on**'] = df['**Added on**'].where(~m, np.nan) 

print(df)

结果：

     **Address**        **Added on**          **Reduced on**
0  15 Smith Close  Added on 17/11/22                  NaN
1   1 Apple Drive                NaN  Reduced on 19/11/22
2  27 Pride place  Added on 18/1//22                  NaN

【讨论】：

【解决方案3】：

这也应该有效：

(df[['Address']].join(df[['Added on']]
.set_index(df['Added on']
.str.rsplit(n=1)
.str[0]
.rename(None),append=True)['Added on']
.unstack()))

或者

df['Added on'].str.rsplit(' ',n=1).str[0].str.get_dummies().mul(df['Added on'],axis=0)

输出：

          Address           Added on           Reduced on
0  15 Smith Close  Added on 17/11/22                  NaN
1   1 Apple Drive                NaN  Reduced on 19/11/22
2  27 Pride place  Added on 18/1//22                  NaN

【讨论】：