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Dato Dato
Asked:2025-02-07 12:34:00 +0000 UTC2025-02-07 12:34:00 +0000 UTC

如何从一个表中删除另一个表中所存在的行？

  • 772

有两个数据框：

  Reg_x    name BS_name Sector_name Reg_y   tac
0    IR  IR1400  IR1400  IR1400_053    IR  5286
1    IR  IR1397  IR1397  IR1397_053    IR  5286
  Reg    name BS_name Sector_name
0  IR  IR1400  IR1400  IR1400_053
1  IR  IR1400  IR1400  IR1400_052
2  IR  IR1400  IR1400  IR1400_051
3  IR  IR1397  IR1397  IR1397_053
4  IR  IR1397  IR1397  IR1397_052
5  IR  IR1397  IR1397  IR1397_051
3  IR  IR1307  IR1307  IR1307_053
4  IR  IR1307  IR1307  IR1307_052
5  IR  IR1307  IR1307  IR1307_051

我如何从第二个表中保存第一个表中不存在的名称？在这种情况下，您需要获得如下表格：

  Reg    name BS_name Sector_name
0  IR  IR1307  IR1307  IR1307_053
1  IR  IR1307  IR1307  IR1307_052
2  IR  IR1307  IR1307  IR1307_051

我认为可以通过列表来实现这一点，然后比较和过滤它们：

cesList = ces4gNokTable['name'].tolist()
    cesList = list(dict.fromkeys(cesList))
    print("Список (cesList) c незаполненными данными в CES 4G Nokia:")
    print(cesList)
    oldList = old4gTable['name'].tolist()
    oldList = list(dict.fromkeys(oldList))
    print("Список (oldList) c довесами 4G Nokia:")
    print(oldList)

但我认为通过熊猫有一些更好的解决方案

python

2 个回答

  1. Best Answer

    像这样：

    df3 = df2[~df2['name'].isin(df1['name'])]

    解释：

    • df[условие]- 按条件过滤
    • isin- 检查是否包含
    • ~- 条件否定
    • 4 
  2. import pandas as pd
from io import StringIO


data = """Reg_x,name,BS_name,Sector_name,Reg_y,tac
IR,IR1400,IR1400,IR1400_053,IR,5286
IR,IR1397,IR1397,IR1397_053,IR,5286"""
data2 = """Reg,name,BS_name,Sector_name
 IR,IR1400,IR1400,IR1400_053
 IR,IR1400,IR1400,IR1400_052
 IR,IR1400,IR1400,IR1400_051
 IR,IR1397,IR1397,IR1397_053
 IR,IR1397,IR1397,IR1397_052
 IR,IR1397,IR1397,IR1397_051
 IR,IR1307,IR1307,IR1307_053
 IR,IR1307,IR1307,IR1307_052
 IR,IR1307,IR1307,IR1307_051"""

df = pd.read_csv(StringIO(data), header=0)
df2 = pd.read_csv(StringIO(data2), header=0)

udf = df['name'].unique() #Убрать повторяющиеся имена, сохранить результат
print(udf)

df3 = df2[~df2['name'].isin(udf)] #Отфильтровать имена которые не в udf
df3 = df3.reset_index(drop=True)

print(df3)

    ['IR1400' 'IR1397']
   Reg    name BS_name Sector_name
0   IR  IR1307  IR1307  IR1307_053
1   IR  IR1307  IR1307  IR1307_052
2   IR  IR1307  IR1307  IR1307_051

    通过请求的另一种选择：

    df4 = df2.query("name not in @udf")
df4 = df4.reset_index(drop=True)
print(df4)
    • 3

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