How to reduce columns in dataframe pandas
7
Here how the datalooks like in df dataframe:
A B C D
0.js 2 1 1 -1
1.js 3 -5 1 -4
total 5 -4 2 -5
And I would get new dataframe df1:
A C
0.js 2 1
1.js 3 1
total 5 2
So basically it should look like this:
df1 = df[df["total"] > 0]
but it should filter on row instead of column and I can't figure it out..
python pandas dataframe
edited 5 hours ago
Scott Boston
54.1k73056
asked 6 hours ago
J OderbergJ Oderberg
404
|
show 2 more comments
7
Here how the datalooks like in df dataframe:
A B C D
0.js 2 1 1 -1
1.js 3 -5 1 -4
total 5 -4 2 -5
And I would get new dataframe df1:
A C
0.js 2 1
1.js 3 1
total 5 2
So basically it should look like this:
df1 = df[df["total"] > 0]
but it should filter on row instead of column and I can't figure it out..
python pandas dataframe
edited 5 hours ago
Scott Boston
54.1k73056
asked 6 hours ago
J OderbergJ Oderberg
404
2
Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.
– Edeki Okoh
6 hours ago
2
Can't you transpose and do the filter?
– Andrew Naguib
6 hours ago
YourDataFrameis just a single row?
– ALollz
6 hours ago
2
Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.
– ALollz
5 hours ago
2
I've updated the question, thanks for the guidance.
– J Oderberg
5 hours ago
|
show 2 more comments
7
7
7
Here how the datalooks like in df dataframe:
A B C D
0.js 2 1 1 -1
1.js 3 -5 1 -4
total 5 -4 2 -5
And I would get new dataframe df1:
A C
0.js 2 1
1.js 3 1
total 5 2
So basically it should look like this:
df1 = df[df["total"] > 0]
but it should filter on row instead of column and I can't figure it out..
python pandas dataframe
edited 5 hours ago
Scott Boston
54.1k73056
asked 6 hours ago
J OderbergJ Oderberg
404
Here how the datalooks like in df dataframe:
A B C D
0.js 2 1 1 -1
1.js 3 -5 1 -4
total 5 -4 2 -5
And I would get new dataframe df1:
A C
0.js 2 1
1.js 3 1
total 5 2
So basically it should look like this:
df1 = df[df["total"] > 0]
but it should filter on row instead of column and I can't figure it out..
python pandas dataframe
python pandas dataframe
edited 5 hours ago
Scott Boston
54.1k73056
asked 6 hours ago
J OderbergJ Oderberg
404
edited 5 hours ago
Scott Boston
54.1k73056
asked 6 hours ago
J OderbergJ Oderberg
404
edited 5 hours ago
Scott Boston
54.1k73056
edited 5 hours ago
Scott Boston
54.1k73056
edited 5 hours ago
Scott Boston
54.1k73056
54.1k73056
asked 6 hours ago
J OderbergJ Oderberg
404
asked 6 hours ago
J OderbergJ Oderberg
404
asked 6 hours ago
J OderbergJ Oderberg
404
404
2
Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.
– Edeki Okoh
6 hours ago
2
Can't you transpose and do the filter?
– Andrew Naguib
6 hours ago
YourDataFrameis just a single row?
– ALollz
6 hours ago
2
Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.
– ALollz
5 hours ago
2
I've updated the question, thanks for the guidance.
– J Oderberg
5 hours ago
|
show 2 more comments
2
Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.
– Edeki Okoh
6 hours ago
2
Can't you transpose and do the filter?
– Andrew Naguib
6 hours ago
YourDataFrameis just a single row?
– ALollz
6 hours ago
2
Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.
– ALollz
5 hours ago
2
I've updated the question, thanks for the guidance.
– J Oderberg
5 hours ago
2
2
Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.
– Edeki Okoh
6 hours ago
Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.
– Edeki Okoh
6 hours ago
2
2
Can't you transpose and do the filter?
– Andrew Naguib
6 hours ago
Can't you transpose and do the filter?
– Andrew Naguib
6 hours ago
Your
DataFrame is just a single row?– ALollz
6 hours ago
Your
DataFrame is just a single row?– ALollz
6 hours ago
2
2
Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.
– ALollz
5 hours ago
Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.
– ALollz
5 hours ago
2
2
I've updated the question, thanks for the guidance.
– J Oderberg
5 hours ago
I've updated the question, thanks for the guidance.
– J Oderberg
5 hours ago
|
show 2 more comments
3 Answers
3
active
oldest
votes
2
You can use, loc with boolean indexing or reindex:
df.loc[:, df.columns[(df.loc['total'] > 0)]]
OR
df.reindex(df.columns[(df.loc['total'] > 0)], axis=1)
Output:
A C
0.js 2 1
1.js 3 1
total 5 2
answered 5 hours ago
Scott BostonScott Boston
54.1k73056
add a comment |
4
You want to use .loc[:, column_mask] i.e.
In [11]: df.loc[:, df.sum() > 0]
Out[11]:
A C
total 5 2
# or
In [12]: df.loc[:, df.iloc[0] > 0]
Out[12]:
A C
total 5 2
answered 6 hours ago
Andy HaydenAndy Hayden
182k52429417
add a comment |
4
Use .where to set negative values to NaN and then dropna setting axis = 1:
df.where(df.gt(0)).dropna(axis=1)
A C
total 5 2
answered 6 hours ago
yatuyatu
8,2011926
1
Okay, that was clever +1
– harvpan
6 hours ago
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
2
You can use, loc with boolean indexing or reindex:
df.loc[:, df.columns[(df.loc['total'] > 0)]]
OR
df.reindex(df.columns[(df.loc['total'] > 0)], axis=1)
Output:
A C
0.js 2 1
1.js 3 1
total 5 2
answered 5 hours ago
Scott BostonScott Boston
54.1k73056
add a comment |
2
You can use, loc with boolean indexing or reindex:
df.loc[:, df.columns[(df.loc['total'] > 0)]]
OR
df.reindex(df.columns[(df.loc['total'] > 0)], axis=1)
Output:
A C
0.js 2 1
1.js 3 1
total 5 2
answered 5 hours ago
Scott BostonScott Boston
54.1k73056
add a comment |
2
2
2
You can use, loc with boolean indexing or reindex:
df.loc[:, df.columns[(df.loc['total'] > 0)]]
OR
df.reindex(df.columns[(df.loc['total'] > 0)], axis=1)
Output:
A C
0.js 2 1
1.js 3 1
total 5 2
answered 5 hours ago
Scott BostonScott Boston
54.1k73056
You can use, loc with boolean indexing or reindex:
df.loc[:, df.columns[(df.loc['total'] > 0)]]
OR
df.reindex(df.columns[(df.loc['total'] > 0)], axis=1)
Output:
A C
0.js 2 1
1.js 3 1
total 5 2
answered 5 hours ago
Scott BostonScott Boston
54.1k73056
answered 5 hours ago
Scott BostonScott Boston
54.1k73056
answered 5 hours ago
Scott BostonScott Boston
54.1k73056
answered 5 hours ago
Scott BostonScott Boston
54.1k73056
54.1k73056
add a comment |
add a comment |
4
You want to use .loc[:, column_mask] i.e.
In [11]: df.loc[:, df.sum() > 0]
Out[11]:
A C
total 5 2
# or
In [12]: df.loc[:, df.iloc[0] > 0]
Out[12]:
A C
total 5 2
answered 6 hours ago
Andy HaydenAndy Hayden
182k52429417
add a comment |
4
You want to use .loc[:, column_mask] i.e.
In [11]: df.loc[:, df.sum() > 0]
Out[11]:
A C
total 5 2
# or
In [12]: df.loc[:, df.iloc[0] > 0]
Out[12]:
A C
total 5 2
answered 6 hours ago
Andy HaydenAndy Hayden
182k52429417
add a comment |
4
4
4
You want to use .loc[:, column_mask] i.e.
In [11]: df.loc[:, df.sum() > 0]
Out[11]:
A C
total 5 2
# or
In [12]: df.loc[:, df.iloc[0] > 0]
Out[12]:
A C
total 5 2
answered 6 hours ago
Andy HaydenAndy Hayden
182k52429417
You want to use .loc[:, column_mask] i.e.
In [11]: df.loc[:, df.sum() > 0]
Out[11]:
A C
total 5 2
# or
In [12]: df.loc[:, df.iloc[0] > 0]
Out[12]:
A C
total 5 2
answered 6 hours ago
Andy HaydenAndy Hayden
182k52429417
answered 6 hours ago
Andy HaydenAndy Hayden
182k52429417
answered 6 hours ago
Andy HaydenAndy Hayden
182k52429417
answered 6 hours ago
Andy HaydenAndy Hayden
182k52429417
182k52429417
add a comment |
add a comment |
4
Use .where to set negative values to NaN and then dropna setting axis = 1:
df.where(df.gt(0)).dropna(axis=1)
A C
total 5 2
answered 6 hours ago
yatuyatu
8,2011926
1
Okay, that was clever +1
– harvpan
6 hours ago
add a comment |
4
Use .where to set negative values to NaN and then dropna setting axis = 1:
df.where(df.gt(0)).dropna(axis=1)
A C
total 5 2
answered 6 hours ago
yatuyatu
8,2011926
1
Okay, that was clever +1
– harvpan
6 hours ago
add a comment |
4
4
4
Use .where to set negative values to NaN and then dropna setting axis = 1:
df.where(df.gt(0)).dropna(axis=1)
A C
total 5 2
answered 6 hours ago
yatuyatu
8,2011926
Use .where to set negative values to NaN and then dropna setting axis = 1:
df.where(df.gt(0)).dropna(axis=1)
A C
total 5 2
answered 6 hours ago
yatuyatu
8,2011926
edited 6 hours ago
answered 6 hours ago
yatuyatu
8,2011926
answered 6 hours ago
yatuyatu
8,2011926
answered 6 hours ago
yatuyatu
8,2011926
8,2011926
1
Okay, that was clever +1
– harvpan
6 hours ago
add a comment |
1
Okay, that was clever +1
– harvpan
6 hours ago
1
1
Okay, that was clever +1
– harvpan
6 hours ago
Okay, that was clever +1
– harvpan
6 hours ago
add a comment |
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2
Do you want to only keep the columns where the condition is met ALWAYS, even if the condition is not met once in another column.
– Edeki Okoh
6 hours ago
2
Can't you transpose and do the filter?
– Andrew Naguib
6 hours ago
Your
DataFrameis just a single row?– ALollz
6 hours ago
2
Then please provide a Minimal, Complete, and Verifiable example with multiple lines and your expected output, because I don't think the current answers are going to do what you are looking for.
– ALollz
5 hours ago
2
I've updated the question, thanks for the guidance.
– J Oderberg
5 hours ago