Filter and Generate Sub-DataFrame From a List
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I have a set:
CompanyList={'Apple','LG','Samsung'}
and a pandas DataFrame:
sales=[{'name':'Samsung Korea','model':'S1'},
{'name':'Samsung Vienam','model':'J1'},
{'name':'LG America','model':'L1'}
]
df=pd.DataFrame(sales)
I'd like to go through the CompanyList, then generate new Sub-DataFrame from 'sales' DataFrame. The expected results are
dataSamsung = [{'name': 'Samsung', 'model': 'S1'},{'name': 'Samsung', 'model': 'J1'}]
dataLG = [{'name': 'LG', 'model': 'L1'}]
I tried:
customer={}
for i in companyList:
customer[i] = df[df.name.str.contains('i')]
but this gives me a wrong answer. Could you help me to fix this case?
python pandas dataframe
edited Nov 25 '18 at 1:32
U9-Forward
18.6k51744
asked Nov 25 '18 at 0:22
HTBHTB
41
add a comment |
0
I have a set:
CompanyList={'Apple','LG','Samsung'}
and a pandas DataFrame:
sales=[{'name':'Samsung Korea','model':'S1'},
{'name':'Samsung Vienam','model':'J1'},
{'name':'LG America','model':'L1'}
]
df=pd.DataFrame(sales)
I'd like to go through the CompanyList, then generate new Sub-DataFrame from 'sales' DataFrame. The expected results are
dataSamsung = [{'name': 'Samsung', 'model': 'S1'},{'name': 'Samsung', 'model': 'J1'}]
dataLG = [{'name': 'LG', 'model': 'L1'}]
I tried:
customer={}
for i in companyList:
customer[i] = df[df.name.str.contains('i')]
but this gives me a wrong answer. Could you help me to fix this case?
python pandas dataframe
edited Nov 25 '18 at 1:32
U9-Forward
18.6k51744
asked Nov 25 '18 at 0:22
HTBHTB
41
Thats asetnot alist!
– U9-Forward
Nov 25 '18 at 0:22
This is a typo. Useinot'i', the latter is just a string, the former references an element incompanyList.
– jpp
Nov 25 '18 at 1:37
it works after fixing, thanks @jpp
– HTB
Nov 26 '18 at 18:26
add a comment |
0
0
0
I have a set:
CompanyList={'Apple','LG','Samsung'}
and a pandas DataFrame:
sales=[{'name':'Samsung Korea','model':'S1'},
{'name':'Samsung Vienam','model':'J1'},
{'name':'LG America','model':'L1'}
]
df=pd.DataFrame(sales)
I'd like to go through the CompanyList, then generate new Sub-DataFrame from 'sales' DataFrame. The expected results are
dataSamsung = [{'name': 'Samsung', 'model': 'S1'},{'name': 'Samsung', 'model': 'J1'}]
dataLG = [{'name': 'LG', 'model': 'L1'}]
I tried:
customer={}
for i in companyList:
customer[i] = df[df.name.str.contains('i')]
but this gives me a wrong answer. Could you help me to fix this case?
python pandas dataframe
edited Nov 25 '18 at 1:32
U9-Forward
18.6k51744
asked Nov 25 '18 at 0:22
HTBHTB
41
I have a set:
CompanyList={'Apple','LG','Samsung'}
and a pandas DataFrame:
sales=[{'name':'Samsung Korea','model':'S1'},
{'name':'Samsung Vienam','model':'J1'},
{'name':'LG America','model':'L1'}
]
df=pd.DataFrame(sales)
I'd like to go through the CompanyList, then generate new Sub-DataFrame from 'sales' DataFrame. The expected results are
dataSamsung = [{'name': 'Samsung', 'model': 'S1'},{'name': 'Samsung', 'model': 'J1'}]
dataLG = [{'name': 'LG', 'model': 'L1'}]
I tried:
customer={}
for i in companyList:
customer[i] = df[df.name.str.contains('i')]
but this gives me a wrong answer. Could you help me to fix this case?
python pandas dataframe
python pandas dataframe
edited Nov 25 '18 at 1:32
U9-Forward
18.6k51744
asked Nov 25 '18 at 0:22
HTBHTB
41
edited Nov 25 '18 at 1:32
U9-Forward
18.6k51744
asked Nov 25 '18 at 0:22
HTBHTB
41
edited Nov 25 '18 at 1:32
U9-Forward
18.6k51744
edited Nov 25 '18 at 1:32
U9-Forward
18.6k51744
edited Nov 25 '18 at 1:32
U9-Forward
18.6k51744
18.6k51744
asked Nov 25 '18 at 0:22
HTBHTB
41
asked Nov 25 '18 at 0:22
HTBHTB
41
asked Nov 25 '18 at 0:22
HTBHTB
41
41
Thats asetnot alist!
– U9-Forward
Nov 25 '18 at 0:22
This is a typo. Useinot'i', the latter is just a string, the former references an element incompanyList.
– jpp
Nov 25 '18 at 1:37
it works after fixing, thanks @jpp
– HTB
Nov 26 '18 at 18:26
add a comment |
Thats asetnot alist!
– U9-Forward
Nov 25 '18 at 0:22
This is a typo. Useinot'i', the latter is just a string, the former references an element incompanyList.
– jpp
Nov 25 '18 at 1:37
it works after fixing, thanks @jpp
– HTB
Nov 26 '18 at 18:26
Thats a
set not a list!– U9-Forward
Nov 25 '18 at 0:22
Thats a
set not a list!– U9-Forward
Nov 25 '18 at 0:22
This is a typo. Use
i not 'i', the latter is just a string, the former references an element in companyList.– jpp
Nov 25 '18 at 1:37
This is a typo. Use
i not 'i', the latter is just a string, the former references an element in companyList.– jpp
Nov 25 '18 at 1:37
it works after fixing, thanks @jpp
– HTB
Nov 26 '18 at 18:26
it works after fixing, thanks @jpp
– HTB
Nov 26 '18 at 18:26
add a comment |
1 Answer
1
active
oldest
votes
0
Try apply:
df['name']=df['name'].apply(lambda x: [i for i in CompanyList if i in x][0])
apply with list comprehension.
answered Nov 25 '18 at 0:26
U9-ForwardU9-Forward
18.6k51744
Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?
– HTB
Nov 26 '18 at 18:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
Try apply:
df['name']=df['name'].apply(lambda x: [i for i in CompanyList if i in x][0])
apply with list comprehension.
answered Nov 25 '18 at 0:26
U9-ForwardU9-Forward
18.6k51744
Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?
– HTB
Nov 26 '18 at 18:03
add a comment |
0
Try apply:
df['name']=df['name'].apply(lambda x: [i for i in CompanyList if i in x][0])
apply with list comprehension.
answered Nov 25 '18 at 0:26
U9-ForwardU9-Forward
18.6k51744
Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?
– HTB
Nov 26 '18 at 18:03
add a comment |
0
0
0
Try apply:
df['name']=df['name'].apply(lambda x: [i for i in CompanyList if i in x][0])
apply with list comprehension.
answered Nov 25 '18 at 0:26
U9-ForwardU9-Forward
18.6k51744
Try apply:
df['name']=df['name'].apply(lambda x: [i for i in CompanyList if i in x][0])
apply with list comprehension.
answered Nov 25 '18 at 0:26
U9-ForwardU9-Forward
18.6k51744
answered Nov 25 '18 at 0:26
U9-ForwardU9-Forward
18.6k51744
answered Nov 25 '18 at 0:26
U9-ForwardU9-Forward
18.6k51744
answered Nov 25 '18 at 0:26
U9-ForwardU9-Forward
18.6k51744
18.6k51744
Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?
– HTB
Nov 26 '18 at 18:03
add a comment |
Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?
– HTB
Nov 26 '18 at 18:03
Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?
– HTB
Nov 26 '18 at 18:03
Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?
– HTB
Nov 26 '18 at 18:03
add a comment |
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Thats a
setnot alist!– U9-Forward
Nov 25 '18 at 0:22
This is a typo. Use
inot'i', the latter is just a string, the former references an element incompanyList.– jpp
Nov 25 '18 at 1:37
it works after fixing, thanks @jpp
– HTB
Nov 26 '18 at 18:26