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Outer join within a Pandas table
Performant cartesian product (CROSS JOIN) with pandasPython join: why is it string.join(list) instead of list.join(string)?Selecting multiple columns in a pandas dataframeRenaming columns in pandasAdding new column to existing DataFrame in Python pandasDelete column from pandas DataFrame by column name“Large data” work flows using pandasHow to iterate over rows in a DataFrame in Pandas?Select rows from a DataFrame based on values in a column in pandasGet list from pandas DataFrame column headerspandas three-way joining multiple dataframes on columns
I have a PANDAS dataframe with three string columns that looks something like this:
Name Surname MiddleName
James Bond A
Maggie Sweenie B
I want to create a kind of outer join within the table so that every possible combination of Name, Surname and MiddleName is output. The output that I am looking for is:
Name Surname MiddleName
James Bond A
Maggie Sweenie B
James Sweenie A
James Sweenie B
Maggie Bond A
Maggie Bond B
Any ideas what's the most efficient way to do this ?
python pandas
asked Mar 8 at 20:42
Number LogicNumber Logic
17010
add a comment |
I have a PANDAS dataframe with three string columns that looks something like this:
Name Surname MiddleName
James Bond A
Maggie Sweenie B
I want to create a kind of outer join within the table so that every possible combination of Name, Surname and MiddleName is output. The output that I am looking for is:
Name Surname MiddleName
James Bond A
Maggie Sweenie B
James Sweenie A
James Sweenie B
Maggie Bond A
Maggie Bond B
Any ideas what's the most efficient way to do this ?
python pandas
asked Mar 8 at 20:42
Number LogicNumber Logic
17010
Sorry it is different , I reopen it .
– Wen-Ben
Mar 8 at 20:45
add a comment |
I have a PANDAS dataframe with three string columns that looks something like this:
Name Surname MiddleName
James Bond A
Maggie Sweenie B
I want to create a kind of outer join within the table so that every possible combination of Name, Surname and MiddleName is output. The output that I am looking for is:
Name Surname MiddleName
James Bond A
Maggie Sweenie B
James Sweenie A
James Sweenie B
Maggie Bond A
Maggie Bond B
Any ideas what's the most efficient way to do this ?
python pandas
asked Mar 8 at 20:42
Number LogicNumber Logic
17010
I have a PANDAS dataframe with three string columns that looks something like this:
Name Surname MiddleName
James Bond A
Maggie Sweenie B
I want to create a kind of outer join within the table so that every possible combination of Name, Surname and MiddleName is output. The output that I am looking for is:
Name Surname MiddleName
James Bond A
Maggie Sweenie B
James Sweenie A
James Sweenie B
Maggie Bond A
Maggie Bond B
Any ideas what's the most efficient way to do this ?
python pandas
python pandas
asked Mar 8 at 20:42
Number LogicNumber Logic
17010
asked Mar 8 at 20:42
Number LogicNumber Logic
17010
asked Mar 8 at 20:42
Number LogicNumber Logic
17010
asked Mar 8 at 20:42
Number LogicNumber Logic
17010
asked Mar 8 at 20:42
Number LogicNumber Logic
17010
17010
Sorry it is different , I reopen it .
– Wen-Ben
Mar 8 at 20:45
add a comment |
Sorry it is different , I reopen it .
– Wen-Ben
Mar 8 at 20:45
Sorry it is different , I reopen it .
– Wen-Ben
Mar 8 at 20:45
Sorry it is different , I reopen it .
– Wen-Ben
Mar 8 at 20:45
add a comment |
3 Answers
3
active
oldest
votes
IIUC using product
import itertools
yourdf=pd.DataFrame(list(itertools.product(*df.values.T.tolist())),columns=df.columns)
yourdf
Out[937]:
Name Surname MiddleName
0 James Bond A
1 James Bond B
2 James Sweenie A
3 James Sweenie B
4 Maggie Bond A
5 Maggie Bond B
6 Maggie Sweenie A
7 Maggie Sweenie B
answered Mar 8 at 20:48
Wen-BenWen-Ben
123k83671
add a comment |
You are looking for a kind of expand_grid functionality, which can be implemented with itertools.product(). From the pandas documentation , you can define expand_grid:
import itertools
def expand_grid(data_dict):
rows = itertools.product(*data_dict.values())
return pd.DataFrame.from_records(rows, columns=data_dict.keys())
expand_grid(df.to_dict('list'))
Out[38]:
Name Surname MidName
0 James Bond A
1 James Bond B
2 James Sweenie A
3 James Sweenie B
4 Maggie Bond A
5 Maggie Bond B
6 Maggie Sweenie A
7 Maggie Sweenie B
answered Mar 8 at 20:55
Xinbin H.Xinbin H.
305
add a comment |
Without itertools:
pd.MultiIndex.from_product(df.T.values.tolist()).to_frame(index=False)
answered Mar 8 at 22:28
JoergVanAkenJoergVanAken
88667
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
IIUC using product
import itertools
yourdf=pd.DataFrame(list(itertools.product(*df.values.T.tolist())),columns=df.columns)
yourdf
Out[937]:
Name Surname MiddleName
0 James Bond A
1 James Bond B
2 James Sweenie A
3 James Sweenie B
4 Maggie Bond A
5 Maggie Bond B
6 Maggie Sweenie A
7 Maggie Sweenie B
answered Mar 8 at 20:48
Wen-BenWen-Ben
123k83671
add a comment |
IIUC using product
import itertools
yourdf=pd.DataFrame(list(itertools.product(*df.values.T.tolist())),columns=df.columns)
yourdf
Out[937]:
Name Surname MiddleName
0 James Bond A
1 James Bond B
2 James Sweenie A
3 James Sweenie B
4 Maggie Bond A
5 Maggie Bond B
6 Maggie Sweenie A
7 Maggie Sweenie B
answered Mar 8 at 20:48
Wen-BenWen-Ben
123k83671
add a comment |
IIUC using product
import itertools
yourdf=pd.DataFrame(list(itertools.product(*df.values.T.tolist())),columns=df.columns)
yourdf
Out[937]:
Name Surname MiddleName
0 James Bond A
1 James Bond B
2 James Sweenie A
3 James Sweenie B
4 Maggie Bond A
5 Maggie Bond B
6 Maggie Sweenie A
7 Maggie Sweenie B
answered Mar 8 at 20:48
Wen-BenWen-Ben
123k83671
IIUC using product
import itertools
yourdf=pd.DataFrame(list(itertools.product(*df.values.T.tolist())),columns=df.columns)
yourdf
Out[937]:
Name Surname MiddleName
0 James Bond A
1 James Bond B
2 James Sweenie A
3 James Sweenie B
4 Maggie Bond A
5 Maggie Bond B
6 Maggie Sweenie A
7 Maggie Sweenie B
answered Mar 8 at 20:48
Wen-BenWen-Ben
123k83671
answered Mar 8 at 20:48
Wen-BenWen-Ben
123k83671
answered Mar 8 at 20:48
Wen-BenWen-Ben
123k83671
answered Mar 8 at 20:48
Wen-BenWen-Ben
123k83671
123k83671
add a comment |
add a comment |
You are looking for a kind of expand_grid functionality, which can be implemented with itertools.product(). From the pandas documentation , you can define expand_grid:
import itertools
def expand_grid(data_dict):
rows = itertools.product(*data_dict.values())
return pd.DataFrame.from_records(rows, columns=data_dict.keys())
expand_grid(df.to_dict('list'))
Out[38]:
Name Surname MidName
0 James Bond A
1 James Bond B
2 James Sweenie A
3 James Sweenie B
4 Maggie Bond A
5 Maggie Bond B
6 Maggie Sweenie A
7 Maggie Sweenie B
answered Mar 8 at 20:55
Xinbin H.Xinbin H.
305
add a comment |
You are looking for a kind of expand_grid functionality, which can be implemented with itertools.product(). From the pandas documentation , you can define expand_grid:
import itertools
def expand_grid(data_dict):
rows = itertools.product(*data_dict.values())
return pd.DataFrame.from_records(rows, columns=data_dict.keys())
expand_grid(df.to_dict('list'))
Out[38]:
Name Surname MidName
0 James Bond A
1 James Bond B
2 James Sweenie A
3 James Sweenie B
4 Maggie Bond A
5 Maggie Bond B
6 Maggie Sweenie A
7 Maggie Sweenie B
answered Mar 8 at 20:55
Xinbin H.Xinbin H.
305
add a comment |
You are looking for a kind of expand_grid functionality, which can be implemented with itertools.product(). From the pandas documentation , you can define expand_grid:
import itertools
def expand_grid(data_dict):
rows = itertools.product(*data_dict.values())
return pd.DataFrame.from_records(rows, columns=data_dict.keys())
expand_grid(df.to_dict('list'))
Out[38]:
Name Surname MidName
0 James Bond A
1 James Bond B
2 James Sweenie A
3 James Sweenie B
4 Maggie Bond A
5 Maggie Bond B
6 Maggie Sweenie A
7 Maggie Sweenie B
answered Mar 8 at 20:55
Xinbin H.Xinbin H.
305
You are looking for a kind of expand_grid functionality, which can be implemented with itertools.product(). From the pandas documentation , you can define expand_grid:
import itertools
def expand_grid(data_dict):
rows = itertools.product(*data_dict.values())
return pd.DataFrame.from_records(rows, columns=data_dict.keys())
expand_grid(df.to_dict('list'))
Out[38]:
Name Surname MidName
0 James Bond A
1 James Bond B
2 James Sweenie A
3 James Sweenie B
4 Maggie Bond A
5 Maggie Bond B
6 Maggie Sweenie A
7 Maggie Sweenie B
answered Mar 8 at 20:55
Xinbin H.Xinbin H.
305
answered Mar 8 at 20:55
Xinbin H.Xinbin H.
305
answered Mar 8 at 20:55
Xinbin H.Xinbin H.
305
answered Mar 8 at 20:55
Xinbin H.Xinbin H.
305
305
add a comment |
add a comment |
Without itertools:
pd.MultiIndex.from_product(df.T.values.tolist()).to_frame(index=False)
answered Mar 8 at 22:28
JoergVanAkenJoergVanAken
88667
add a comment |
Without itertools:
pd.MultiIndex.from_product(df.T.values.tolist()).to_frame(index=False)
answered Mar 8 at 22:28
JoergVanAkenJoergVanAken
88667
add a comment |
Without itertools:
pd.MultiIndex.from_product(df.T.values.tolist()).to_frame(index=False)
answered Mar 8 at 22:28
JoergVanAkenJoergVanAken
88667
Without itertools:
pd.MultiIndex.from_product(df.T.values.tolist()).to_frame(index=False)
answered Mar 8 at 22:28
JoergVanAkenJoergVanAken
88667
answered Mar 8 at 22:28
JoergVanAkenJoergVanAken
88667
answered Mar 8 at 22:28
JoergVanAkenJoergVanAken
88667
answered Mar 8 at 22:28
JoergVanAkenJoergVanAken
88667
88667
add a comment |
add a comment |
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Sorry it is different , I reopen it .
– Wen-Ben
Mar 8 at 20:45