基于匹配来自另一个数据帧pandas的值的新列

Question

如果我们在下面的例子中有两个数据帧，如df1和df2 ; 我们如何合并它们来生成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for循环，因为我正在处理大数据集

Answer 1

在使用DataFrame重新创建后检查stack df1的列表列，然后map df2的值

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此外，因为你要求不使用for循环我使用sum ，并且这种情况的sum比*for loop*或itertools慢得多

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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}

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正如我所提到的，我们大多数人都建议，你也可以检查带有熊猫的For循环 - 我什么时候应该关心？

import itertools
d=dict(zip(df2.column3,df2.column4))


l=[set(itertools.chain(*[d[y] for y in x ])) for x in df1.column2.tolist()]
df1['New']=l

Answer 2

你可以这样做：

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)])) )))