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Merging mulitple rows to one row of a dataframe column
How to merge two dictionaries in a single expression?How to sort a dataframe by multiple column(s)Selecting multiple columns in a pandas dataframeRenaming columns in pandasAdding new column to existing DataFrame in Python pandasDelete column from pandas DataFrameCreating an empty Pandas DataFrame, then filling it?How 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 headers
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
My Current dataframe was like below,
0 1 2
0 HA-567034786 AB-1018724 None
1 AB-6348403 HA-7298656 None
After using the apply(), I just make it like,
def make_dict(row):
s = set(x for x in row if x)
return x: list(s - x) for x in s
result = df.apply(make_dict, axis=1).to_frame(name = 'duplicates')
duplicates
1 'HA-567034786': ['AB-1018724'],'AB-1018724':['HA-567034786']
2 'AB-6348403': ['HA-7298656'],'HA-7298656':['AB-6348403']
Now I'm stucked on to make it a single dimentional dictionary like below,
'HA-567034786': ['AB-1018724'],'AB-1018724':['HA-567034786'],
'AB-6348403': ['HA-7298656'],'HA-7298656':['AB-6348403']
python pandas dataframe
asked Mar 28 at 7:32
Always SunnyAlways Sunny
18.9k3 gold badges32 silver badges56 bronze badges
add a comment |
My Current dataframe was like below,
0 1 2
0 HA-567034786 AB-1018724 None
1 AB-6348403 HA-7298656 None
After using the apply(), I just make it like,
def make_dict(row):
s = set(x for x in row if x)
return x: list(s - x) for x in s
result = df.apply(make_dict, axis=1).to_frame(name = 'duplicates')
duplicates
1 'HA-567034786': ['AB-1018724'],'AB-1018724':['HA-567034786']
2 'AB-6348403': ['HA-7298656'],'HA-7298656':['AB-6348403']
Now I'm stucked on to make it a single dimentional dictionary like below,
'HA-567034786': ['AB-1018724'],'AB-1018724':['HA-567034786'],
'AB-6348403': ['HA-7298656'],'HA-7298656':['AB-6348403']
python pandas dataframe
asked Mar 28 at 7:32
Always SunnyAlways Sunny
18.9k3 gold badges32 silver badges56 bronze badges
add a comment |
My Current dataframe was like below,
0 1 2
0 HA-567034786 AB-1018724 None
1 AB-6348403 HA-7298656 None
After using the apply(), I just make it like,
def make_dict(row):
s = set(x for x in row if x)
return x: list(s - x) for x in s
result = df.apply(make_dict, axis=1).to_frame(name = 'duplicates')
duplicates
1 'HA-567034786': ['AB-1018724'],'AB-1018724':['HA-567034786']
2 'AB-6348403': ['HA-7298656'],'HA-7298656':['AB-6348403']
Now I'm stucked on to make it a single dimentional dictionary like below,
'HA-567034786': ['AB-1018724'],'AB-1018724':['HA-567034786'],
'AB-6348403': ['HA-7298656'],'HA-7298656':['AB-6348403']
python pandas dataframe
asked Mar 28 at 7:32
Always SunnyAlways Sunny
18.9k3 gold badges32 silver badges56 bronze badges
My Current dataframe was like below,
0 1 2
0 HA-567034786 AB-1018724 None
1 AB-6348403 HA-7298656 None
After using the apply(), I just make it like,
def make_dict(row):
s = set(x for x in row if x)
return x: list(s - x) for x in s
result = df.apply(make_dict, axis=1).to_frame(name = 'duplicates')
duplicates
1 'HA-567034786': ['AB-1018724'],'AB-1018724':['HA-567034786']
2 'AB-6348403': ['HA-7298656'],'HA-7298656':['AB-6348403']
Now I'm stucked on to make it a single dimentional dictionary like below,
'HA-567034786': ['AB-1018724'],'AB-1018724':['HA-567034786'],
'AB-6348403': ['HA-7298656'],'HA-7298656':['AB-6348403']
python pandas dataframe
python pandas dataframe
asked Mar 28 at 7:32
Always SunnyAlways Sunny
18.9k3 gold badges32 silver badges56 bronze badges
asked Mar 28 at 7:32
Always SunnyAlways Sunny
18.9k3 gold badges32 silver badges56 bronze badges
asked Mar 28 at 7:32
Always SunnyAlways Sunny
18.9k3 gold badges32 silver badges56 bronze badges
asked Mar 28 at 7:32
Always SunnyAlways Sunny
18.9k3 gold badges32 silver badges56 bronze badges
asked Mar 28 at 7:32
Always SunnyAlways Sunny
18.9k3 gold badges32 silver badges56 bronze badges
18.9k3 gold badges32 silver badges56 bronze badges
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Instead apply use dictionary comprehension with flattening:
print (df)
0 1
0 HA-567034786 AB-1018724
1 AB-6348403 HA-7298656
def make_dict(row):
s = set(x for x in row if x)
return x: list(s - x) for x in s
result = k:v for x in df.values for k, v in make_dict(x).items()
print (result)
'HA-567034786': ['AB-1018724'],
'AB-1018724': ['HA-567034786'],
'HA-7298656': ['AB-6348403'],
'AB-6348403': ['HA-7298656']
Another solution with apply:
result = k:v for x in df.apply(make_dict, axis=1) for k, v in x.items()
answered Mar 28 at 7:38
jezraeljezrael
407k32 gold badges425 silver badges491 bronze badges
1
You're a star mate :)
– Always Sunny
Mar 28 at 9:03
daily I learn a lot from your answers :)
– Always Sunny
Mar 28 at 9:10
add a comment |
Also you can use collections.ChainMap() to group all the dictionaries to one as:
from collections import ChainMap
res =dict(ChainMap(*result['duplicates']))
answered Mar 28 at 9:06
LoochieLoochie
1,0363 silver badges11 bronze badges
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Instead apply use dictionary comprehension with flattening:
print (df)
0 1
0 HA-567034786 AB-1018724
1 AB-6348403 HA-7298656
def make_dict(row):
s = set(x for x in row if x)
return x: list(s - x) for x in s
result = k:v for x in df.values for k, v in make_dict(x).items()
print (result)
'HA-567034786': ['AB-1018724'],
'AB-1018724': ['HA-567034786'],
'HA-7298656': ['AB-6348403'],
'AB-6348403': ['HA-7298656']
Another solution with apply:
result = k:v for x in df.apply(make_dict, axis=1) for k, v in x.items()
answered Mar 28 at 7:38
jezraeljezrael
407k32 gold badges425 silver badges491 bronze badges
1
You're a star mate :)
– Always Sunny
Mar 28 at 9:03
daily I learn a lot from your answers :)
– Always Sunny
Mar 28 at 9:10
add a comment |
Instead apply use dictionary comprehension with flattening:
print (df)
0 1
0 HA-567034786 AB-1018724
1 AB-6348403 HA-7298656
def make_dict(row):
s = set(x for x in row if x)
return x: list(s - x) for x in s
result = k:v for x in df.values for k, v in make_dict(x).items()
print (result)
'HA-567034786': ['AB-1018724'],
'AB-1018724': ['HA-567034786'],
'HA-7298656': ['AB-6348403'],
'AB-6348403': ['HA-7298656']
Another solution with apply:
result = k:v for x in df.apply(make_dict, axis=1) for k, v in x.items()
answered Mar 28 at 7:38
jezraeljezrael
407k32 gold badges425 silver badges491 bronze badges
1
You're a star mate :)
– Always Sunny
Mar 28 at 9:03
daily I learn a lot from your answers :)
– Always Sunny
Mar 28 at 9:10
add a comment |
Instead apply use dictionary comprehension with flattening:
print (df)
0 1
0 HA-567034786 AB-1018724
1 AB-6348403 HA-7298656
def make_dict(row):
s = set(x for x in row if x)
return x: list(s - x) for x in s
result = k:v for x in df.values for k, v in make_dict(x).items()
print (result)
'HA-567034786': ['AB-1018724'],
'AB-1018724': ['HA-567034786'],
'HA-7298656': ['AB-6348403'],
'AB-6348403': ['HA-7298656']
Another solution with apply:
result = k:v for x in df.apply(make_dict, axis=1) for k, v in x.items()
answered Mar 28 at 7:38
jezraeljezrael
407k32 gold badges425 silver badges491 bronze badges
Instead apply use dictionary comprehension with flattening:
print (df)
0 1
0 HA-567034786 AB-1018724
1 AB-6348403 HA-7298656
def make_dict(row):
s = set(x for x in row if x)
return x: list(s - x) for x in s
result = k:v for x in df.values for k, v in make_dict(x).items()
print (result)
'HA-567034786': ['AB-1018724'],
'AB-1018724': ['HA-567034786'],
'HA-7298656': ['AB-6348403'],
'AB-6348403': ['HA-7298656']
Another solution with apply:
result = k:v for x in df.apply(make_dict, axis=1) for k, v in x.items()
answered Mar 28 at 7:38
jezraeljezrael
407k32 gold badges425 silver badges491 bronze badges
edited Mar 28 at 7:46
answered Mar 28 at 7:38
jezraeljezrael
407k32 gold badges425 silver badges491 bronze badges
answered Mar 28 at 7:38
jezraeljezrael
407k32 gold badges425 silver badges491 bronze badges
answered Mar 28 at 7:38
jezraeljezrael
407k32 gold badges425 silver badges491 bronze badges
407k32 gold badges425 silver badges491 bronze badges
1
You're a star mate :)
– Always Sunny
Mar 28 at 9:03
daily I learn a lot from your answers :)
– Always Sunny
Mar 28 at 9:10
add a comment |
1
You're a star mate :)
– Always Sunny
Mar 28 at 9:03
daily I learn a lot from your answers :)
– Always Sunny
Mar 28 at 9:10
1
1
You're a star mate :)
– Always Sunny
Mar 28 at 9:03
You're a star mate :)
– Always Sunny
Mar 28 at 9:03
daily I learn a lot from your answers :)
– Always Sunny
Mar 28 at 9:10
daily I learn a lot from your answers :)
– Always Sunny
Mar 28 at 9:10
add a comment |
Also you can use collections.ChainMap() to group all the dictionaries to one as:
from collections import ChainMap
res =dict(ChainMap(*result['duplicates']))
answered Mar 28 at 9:06
LoochieLoochie
1,0363 silver badges11 bronze badges
add a comment |
Also you can use collections.ChainMap() to group all the dictionaries to one as:
from collections import ChainMap
res =dict(ChainMap(*result['duplicates']))
answered Mar 28 at 9:06
LoochieLoochie
1,0363 silver badges11 bronze badges
add a comment |
Also you can use collections.ChainMap() to group all the dictionaries to one as:
from collections import ChainMap
res =dict(ChainMap(*result['duplicates']))
answered Mar 28 at 9:06
LoochieLoochie
1,0363 silver badges11 bronze badges
Also you can use collections.ChainMap() to group all the dictionaries to one as:
from collections import ChainMap
res =dict(ChainMap(*result['duplicates']))
answered Mar 28 at 9:06
LoochieLoochie
1,0363 silver badges11 bronze badges
answered Mar 28 at 9:06
LoochieLoochie
1,0363 silver badges11 bronze badges
answered Mar 28 at 9:06
LoochieLoochie
1,0363 silver badges11 bronze badges
answered Mar 28 at 9:06
LoochieLoochie
1,0363 silver badges11 bronze badges
1,0363 silver badges11 bronze badges
add a comment |
add a comment |
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