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[英]How to join unique strings from a column in dataframe based on another column
[英]How to join a column of lists in one dataframe with a column of strings in another dataframe?
我有兩個數據框。 第一個(稱為A)稱為列(稱為col1),其元素為字符串列表。 另一個(稱為B)稱為列(稱為“ col2”),其元素為字符串。 我想在B.col2在A.col1中的列表中的這兩個數據幀之間進行聯接。 這是一對多聯接。
另外,我需要該解決方案具有可伸縮性,因為我想將兩個數據幀與成千上萬的行連接在一起。
我嘗試串聯A.col1中的值並創建一個新列(我們將其稱為“ col3”)並加入以下條件:A.col3.contains(B.col2)。 但是,我的理解是,這種情況觸發了兩個數據框之間的笛卡爾積,考慮到數據框的大小,我無法承受。
def joinIds(IdList):
return "__".join(IdList)
joinIds_udf = udf(joinIds)
pnr_corr = pnr_corr.withColumn('joinedIds', joinIds_udf(pnr_corr.pnrCorrelations.correlationPnrSchedule.scheduleIds)
pnr_corr_skd = pnr_corr.join(skd, pnr_corr.joinedIds.contains(skd.id), how='inner')
這是我想到的聯接示例:
dataframe A:
listColumn
["a","b","c"]
["a","b"]
["d","e"]
dataframe B:
valueColumn
a
b
d
output:
listColumn valueColumn
["a","b","c"] a
["a","b","c"] b
["a","b"] a
["a","b"] b
["d","e"] d
我不知道是否有一種有效的方法來做到這一點,但這給出了正確的輸出:
import pandas as pd
from itertools import chain
df1 = pd.Series([["a","b","c"],["a","b"],["d","e"]])
df2 = pd.Series(["a","b","d"])
result = [ [ [el2,list1] for el2 in df2.values if el2 in list1 ]
for list1 in df1.values ]
result_flat = list(chain(*result))
result_df = pd.DataFrame(result_flat)
你得到:
In [26]: result_df
Out[26]:
0 1
0 a [a, b, c]
1 b [a, b, c]
2 a [a, b]
3 b [a, b]
4 d [d, e]
另一種方法是使用pandas> = 0.25中的新explode()
方法並像這樣合並:
import pandas as pd
df1 = pd.DataFrame({'col1': [["a","b","c"],["a","b"],["d","e"]]})
df2 = pd.DataFrame({'col2': ["a","b","d"]})
df1_flat = df1.col1.explode().reset_index()
df_merged = pd.merge(df1_flat,df2,left_on='col1',right_on='col2')
df_merged['col2'] = df1.loc[df_merged['index']].values
df_merged.drop('index',axis=1, inplace=True)
這給出了相同的結果:
col1 col2
0 a [a, b, c]
1 a [a, b]
2 b [a, b, c]
3 b [a, b]
4 d [d, e]
怎么樣:
df['col1'] = [df['col1'].values[i] + [df['col2'].values[i]] for i in range(len(df))]
其中“ col1”是字符串列表,“ col2”是字符串。
如果您不再希望使用以下命令,也可以刪除“ col2”:
df = df.drop('col2',axis=1)
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