Question

在下面的示例中，如果我们有两个数据帧，例如df1和df2；我们如何合并它们以生成df3？

import pandas as pd
import numpy as np

data1 = [("a1",["A","B"]),("a2",["A","B","C"]),("a3",["B","C"])]
df1 = pd.DataFrame(data1,columns = ["column1","column2"])
print df1

data2 = [("A",["1","2"]),("B",["1","3","4"]),("C",["5"])]
df2 = pd.DataFrame(data2,columns=["column3","column4"])
print df2

data3 = [("a1",["A","B"],["1","2","3","4"]),("a2",["A","B","C"], 
["1","2","3","4","5"]),("a3",["B","C"],["1","3","4","5"])]
df3 = pd.DataFrame(data3,columns = ["column1","column2","column5"])
print df3

我的目标是不使用循环，因为我正在处理大型数据集

Answer 1

用stack重新创建后，再用DataFrame来检查map df1的列表列

另外，由于您要求不使用for循环，因此我正在使用df2，在这种情况下，sum比sum或*for loop*慢得多

itertools

正如我提到的，我们大多数人都建议，您也可以使用For loops with pandas - When should I care?

进行检查

s=pd.DataFrame(df1.column2.tolist()).stack()
df1['New']=s.map(df2.set_index('column3').column4).sum(level=0).apply(set)
df1
Out[36]: 
  column1    column2              New
0      a1     [A, B]     {2, 4, 3, 1}
1      a2  [A, B, C]  {3, 5, 4, 2, 1}
2      a3     [B, C]     {4, 3, 1, 5}

Answer 2

您可以按照以下步骤进行操作：

df2_dict = {i:j for i,j in zip(df2['column3'].values, df2['column4'].values)}
# print(df2_dict)

def func(val):
    return sorted(list(set(np.concatenate([df2_dict.get(i) for i in val]))))

df1['column5'] = df1['column2'].apply(func)
print(df1)

输出：

  column1    column2          column5
0      a1     [A, B]     [1, 2, 3, 4]
1      a2  [A, B, C]  [1, 2, 3, 4, 5]
2      a3     [B, C]     [1, 3, 4, 5]

Answer 3

这有效：

df1['column2'].apply(lambda x: list(set((np.concatenate([df2.set_index('column3')['column4'][i] for i in list(x)])) )))

基于来自另一个数据框熊猫的匹配值的新列

3 个答案: