Naming a column in pandas that is dependent on a function
Suppose I have a pandas dataframe, df.
a = len(series) - 6
df['M20'] = ...
It is irrelevant to show what the dataset series looks like, but len(series) = 26, so a = 20.
Is there a way of writing something like df['Ma'] instead of df['M20']?
I am running a monthly process, meaning next month len(series) would equal 27, a would equal 21, and so I don't want to have to manually change all of the 20s to 21s etc.
python pandas function dataframe
asked Jan 2 at 13:28
MRHarvMRHarv
959
add a comment |
Suppose I have a pandas dataframe, df.
a = len(series) - 6
df['M20'] = ...
It is irrelevant to show what the dataset series looks like, but len(series) = 26, so a = 20.
Is there a way of writing something like df['Ma'] instead of df['M20']?
I am running a monthly process, meaning next month len(series) would equal 27, a would equal 21, and so I don't want to have to manually change all of the 20s to 21s etc.
python pandas function dataframe
asked Jan 2 at 13:28
MRHarvMRHarv
959
add a comment |
Suppose I have a pandas dataframe, df.
a = len(series) - 6
df['M20'] = ...
It is irrelevant to show what the dataset series looks like, but len(series) = 26, so a = 20.
Is there a way of writing something like df['Ma'] instead of df['M20']?
I am running a monthly process, meaning next month len(series) would equal 27, a would equal 21, and so I don't want to have to manually change all of the 20s to 21s etc.
python pandas function dataframe
asked Jan 2 at 13:28
MRHarvMRHarv
959
Suppose I have a pandas dataframe, df.
a = len(series) - 6
df['M20'] = ...
It is irrelevant to show what the dataset series looks like, but len(series) = 26, so a = 20.
Is there a way of writing something like df['Ma'] instead of df['M20']?
I am running a monthly process, meaning next month len(series) would equal 27, a would equal 21, and so I don't want to have to manually change all of the 20s to 21s etc.
python pandas function dataframe
python pandas function dataframe
asked Jan 2 at 13:28
MRHarvMRHarv
959
asked Jan 2 at 13:28
MRHarvMRHarv
959
asked Jan 2 at 13:28
MRHarvMRHarv
959
asked Jan 2 at 13:28
MRHarvMRHarv
959
asked Jan 2 at 13:28
MRHarvMRHarv
959
959
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Use format:
a = len(series) - 6
df['M{}'.format(a)] = ...
df['M{}'.format(len(series) - 6)] = ...
Or f-strings for python 3.6+:
df[f'M{a}'] = ...
df[f'M{len(series) - 6}'] = ...
answered Jan 2 at 13:30
jezraeljezrael
346k25302376
1
Marvellous, thanks.
– MRHarv
Jan 2 at 13:43
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
Use format:
a = len(series) - 6
df['M{}'.format(a)] = ...
df['M{}'.format(len(series) - 6)] = ...
Or f-strings for python 3.6+:
df[f'M{a}'] = ...
df[f'M{len(series) - 6}'] = ...
answered Jan 2 at 13:30
jezraeljezrael
346k25302376
1
Marvellous, thanks.
– MRHarv
Jan 2 at 13:43
add a comment |
Use format:
a = len(series) - 6
df['M{}'.format(a)] = ...
df['M{}'.format(len(series) - 6)] = ...
Or f-strings for python 3.6+:
df[f'M{a}'] = ...
df[f'M{len(series) - 6}'] = ...
answered Jan 2 at 13:30
jezraeljezrael
346k25302376
1
Marvellous, thanks.
– MRHarv
Jan 2 at 13:43
add a comment |
Use format:
a = len(series) - 6
df['M{}'.format(a)] = ...
df['M{}'.format(len(series) - 6)] = ...
Or f-strings for python 3.6+:
df[f'M{a}'] = ...
df[f'M{len(series) - 6}'] = ...
answered Jan 2 at 13:30
jezraeljezrael
346k25302376
Use format:
a = len(series) - 6
df['M{}'.format(a)] = ...
df['M{}'.format(len(series) - 6)] = ...
Or f-strings for python 3.6+:
df[f'M{a}'] = ...
df[f'M{len(series) - 6}'] = ...
answered Jan 2 at 13:30
jezraeljezrael
346k25302376
answered Jan 2 at 13:30
jezraeljezrael
346k25302376
answered Jan 2 at 13:30
jezraeljezrael
346k25302376
answered Jan 2 at 13:30
jezraeljezrael
346k25302376
346k25302376
1
Marvellous, thanks.
– MRHarv
Jan 2 at 13:43
add a comment |
1
Marvellous, thanks.
– MRHarv
Jan 2 at 13:43
1
1
Marvellous, thanks.
– MRHarv
Jan 2 at 13:43
Marvellous, thanks.
– MRHarv
Jan 2 at 13:43
add a comment |
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