英文:
Map two data frames by common string elements from List column
问题
我想要将两个数据框映射起来，如果两列中的字符串元素匹配，我有一个共同的列，其中的字符串是用逗号分隔的。我尝试过使用映射函数，将其转换为字典，但没有成功。
dfText[Temp,Temp2][Temp4,Temp7,Temp2]ClusterDfLabel Member[Cluster1] [Temp,Temp8][Cluster2] [Temp4,Temp7]
我希望的输出是
dfText Label[Temp,Temp2] [Cluster1][Temp4,Temp7,Temp2] [Cluster2]
英文:
I want to map two data frames if the string element from two columns match, The common column i have is string with comma separated. I tried map function by converting it to dictionary also. But it didn't worked.
dfText[Temp,Temp2][Temp4,Temp7,Temp2]ClusterDfLabel Member[Cluster1] [Temp,Temp8][Cluster2] [Temp4,Temp7]
I want output like
dfText Label[Temp,Temp2] [Cluster1][Temp4,Temp7,Temp2] [Cluster2]
答案1
得分: 1
根据ClusterDf创建字典，然后使用map添加新列，如果没有匹配项则迭代：
d = {v: a[0] for a, b in zip(ClusterDf['Label'], ClusterDf['Member']) for v in b}print (d){'Temp': 'Cluster1', 'Temp8': 'Cluster1', 'Temp4': 'Cluster2', 'Temp7': 'Cluster2'}df['Label'] = df['Text'].map(lambda x: next(iter(d[y] for y in x if y in d), 'no match'))print (df)Text Label0 [Temp, Temp2] Cluster11 [Temp4, Temp7, Temp2] Cluster2
如果需要列表：
df['Label'] = df['Text'].map(lambda x: [next(iter(d[y] for y in x if y in d), 'no match')])print (df)Text Label0 [Temp, Temp2] [Cluster1]1 [Temp4, Temp7, Temp2] [Cluster2]
如果希望存在所有匹配项：
df['Label'] = df['Text'].map(lambda x: [d[y] for y in x if y in d])print (df)Text Label0 [Temp, Temp2] [Cluster1]1 [Temp4, Temp7, Temp2] [Cluster2, Cluster2]
英文:
Create dictionary by ClusterDf and then add new column by map with next and iter if no match:
d = {v: a[0] for a, b in zip(ClusterDf['Label'], ClusterDf['Member']) for v in b}print (d){'Temp': 'Cluster1', 'Temp8': 'Cluster1', 'Temp4': 'Cluster2', 'Temp7': 'Cluster2'}df['Label'] = df['Text'].map(lambda x: next(iter(d[y] for y in x if y in d), 'no match'))print (df)Text Label0 [Temp, Temp2] Cluster11 [Temp4, Temp7, Temp2] Cluster2
If need list:
df['Label'] = df['Text'].map(lambda x: [next(iter(d[y] for y in x if y in d), 'no match')])print (df)Text Label0 [Temp, Temp2] [Cluster1]1 [Temp4, Temp7, Temp2] [Cluster2]
If want all matching if exist:
df['Label'] = df['Text'].map(lambda x: [d[y] for y in x if y in d])print (df)Text Label0 [Temp, Temp2] [Cluster1]1 [Temp4, Temp7, Temp2] [Cluster2, Cluster2]
答案2
得分: 0
感谢 @jezrael，第三种解决方案对我非常有效。非常感谢。
你让我的一天变得美好。
df['Label'] = df['Text'].map(lambda x: [d[y] for y in x if y in d])
英文:
Thanks @jezrael, Third solution worked for me perfectly. Thanks a lot.
You made my day
df['Label'] = df['Text'].map(lambda x: [d[y] for y in x if y in d])
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