Python 合并数据帧时如何合并两个列表列？
我有两个数据帧：Python 合并数据帧时如何合并两个列表列？,python,pandas,merge,Python,Pandas,Merge,我有两个数据帧： date ids_x ids_y 0 2015-10-13 [978] [978, 12] 1 2015-10-14 [978, 121] [2, 1] df1: date ids 0 2015-10-13 [978] 1 2015-10-14 [978, 121] date ids 0
date ids_x ids_y
0 2015-10-13 [978] [978, 12]
1 2015-10-14 [978, 121] [2, 1]
date ids
0 2015-10-13 [978]
1 2015-10-14 [978, 121]
date ids
0 2015-10-13 [978, 12]
1 2015-10-14 [2, 1]
date ids
0 2015-10-13 [978]
1 2015-10-14 [978, 121]
date ids
0 2015-10-13 [978, 12]
1 2015-10-14 [2, 1]
日期将它们合并时，如下所示：
df = pandas.merge(df1, df2, on='date', sort=False)
我将使用下面的数据帧
：
date ids_x ids_y
0 2015-10-13 [978] [978, 12]
1 2015-10-14 [978, 121] [2, 1]
我想从两个列表中合并一个
列，如[978，978，12]
，或者最好删除重复项，并有类似[978，12]
的解决方案：
df = pandas.merge(df1, df2, on='date', sort=False)
df['ids'] = df['ids_x'] + df['ids_y']
df = df.drop(['ids_x','ids_y'], 1)
要合并两个列表，请使用apply
功能：
df['ids'] = df.apply(lambda row: list(set(row['ids'])), axis=1)
您可以将这两列添加到一起以获得要查找的列表，然后使用df.drop（）
和axis=1
删除ids\u x
和ids\u y
列。范例-
df = pd.merge(df1, df2, on='date', sort=False)
df['ids'] = df['ids_x'] + df['ids_y']
df = df.drop(['ids_x','ids_y'],axis=1)
df['ids'] = df['ids'].apply(lambda x: list(set(x)))
演示-
In [65]: df
Out[65]:
date ids_x ids_y
0 2015-10-13 [978] [978, 12]
1 2015-10-14 [978, 121] [2, 1]
In [67]: df['ids'] = df['ids_x'] + df['ids_y']
In [68]: df
Out[68]:
date ids_x ids_y ids
0 2015-10-13 [978] [978, 12] [978, 978, 12]
1 2015-10-14 [978, 121] [2, 1] [978, 121, 2, 1]
In [70]: df = df.drop(['ids_x','ids_y'],axis=1)
In [71]: df
Out[71]:
date ids
0 2015-10-13 [978, 978, 12]
1 2015-10-14 [978, 121, 2, 1]
In [72]: df['ids'] = df['ids'].apply(lambda x: list(set(x)))
In [73]: df
Out[73]:
date ids
0 2015-10-13 [978, 12]
1 2015-10-14 [121, 978, 2, 1]
In [79]: df['ids'] = df['ids'].apply(lambda x: np.unique(x))
In [80]: df
Out[80]:
date ids
0 2015-10-13 [12, 978]
1 2015-10-14 [1, 2, 121, 978]
如果您还想删除重复的值，并且不关心顺序，则可以使用系列。应用，然后将列表转换为集
，然后返回到列表
。范例-
df = pd.merge(df1, df2, on='date', sort=False)
df['ids'] = df['ids_x'] + df['ids_y']
df = df.drop(['ids_x','ids_y'],axis=1)
df['ids'] = df['ids'].apply(lambda x: list(set(x)))
演示-
In [65]: df
Out[65]:
date ids_x ids_y
0 2015-10-13 [978] [978, 12]
1 2015-10-14 [978, 121] [2, 1]
In [67]: df['ids'] = df['ids_x'] + df['ids_y']
In [68]: df
Out[68]:
date ids_x ids_y ids
0 2015-10-13 [978] [978, 12] [978, 978, 12]
1 2015-10-14 [978, 121] [2, 1] [978, 121, 2, 1]
In [70]: df = df.drop(['ids_x','ids_y'],axis=1)
In [71]: df
Out[71]:
date ids
0 2015-10-13 [978, 978, 12]
1 2015-10-14 [978, 121, 2, 1]
In [72]: df['ids'] = df['ids'].apply(lambda x: list(set(x)))
In [73]: df
Out[73]:
date ids
0 2015-10-13 [978, 12]
1 2015-10-14 [121, 978, 2, 1]
In [79]: df['ids'] = df['ids'].apply(lambda x: np.unique(x))
In [80]: df
Out[80]:
date ids
0 2015-10-13 [12, 978]
1 2015-10-14 [1, 2, 121, 978]
或者，如注释中所述，如果要使用numpy.unique（）
，您可以将其与系列一起使用。也可以应用-
import numpy as np
df['ids'] = df['ids'].apply(lambda x: np.unique(x))
演示-
In [65]: df
Out[65]:
date ids_x ids_y
0 2015-10-13 [978] [978, 12]
1 2015-10-14 [978, 121] [2, 1]
In [67]: df['ids'] = df['ids_x'] + df['ids_y']
In [68]: df
Out[68]:
date ids_x ids_y ids
0 2015-10-13 [978] [978, 12] [978, 978, 12]
1 2015-10-14 [978, 121] [2, 1] [978, 121, 2, 1]
In [70]: df = df.drop(['ids_x','ids_y'],axis=1)
In [71]: df
Out[71]:
date ids
0 2015-10-13 [978, 978, 12]
1 2015-10-14 [978, 121, 2, 1]
In [72]: df['ids'] = df['ids'].apply(lambda x: list(set(x)))
In [73]: df
Out[73]:
date ids
0 2015-10-13 [978, 12]
1 2015-10-14 [121, 978, 2, 1]
In [79]: df['ids'] = df['ids'].apply(lambda x: np.unique(x))
In [80]: df
Out[80]:
date ids
0 2015-10-13 [12, 978]
1 2015-10-14 [1, 2, 121, 978]
使用numpy.unique（）。谢谢你抽出时间+1@AlirezaHos请注意，根据我的测试，set
比numpy.unique
快一点。你说得对Numpy.unique
花费了0.16
秒，而set
花费了0.05
秒，用于30000个ID。很快。谢谢你的回答+1.