I have a dataframe in python pandas with several columns taken from a CSV file.
For instance, data =:
Day P1S1 P1S2 P1S3 P2S1 P2S2 P2S3
1 1 2 2 3 1 2
2 2 2 3 5 4 2
And what I need is to get the sum of all columns which name starts with P1... something like P1* with a wildcard.
Something like the following which gives an error:
P1Sum = data["P1*"]
Is there any why to do this with pandas?
I found the answer.
Using the data, dataframe from the question:
from pandas import *
P1Channels = data.filter(regex="P1")
P1Sum = P1Channels.sum(axis=1)
List comprehensions on columns allow more filters in the if condition:
In [1]: df = pd.DataFrame(np.arange(15).reshape(5, 3), columns=['P1S1', 'P1S2', 'P2S1'])
In [2]: df
Out[2]:
P1S1 P1S2 P2S1
0 0 1 2
1 3 4 5
2 6 7 8
3 9 10 11
4 12 13 14
In [3]: df.loc[:, [x for x in df.columns if x.startswith('P1')]].sum(axis=1)
Out[3]:
0 1
1 7
2 13
3 19
4 25
dtype: int64
Thanks for the tip jbssm, for anyone else looking for a sum total, I ended up adding .sum() at the end, so:
P1Sum= P1Channels.sum(axis=1).sum()
Related
I have a dataset that looks like his:
ID date
1 O1-01-2012
1 05-02-2012
1 25-06-2013
1 14-12-2013
1 10-04-2014
2 19-05-2012
2 07-08-2014
2 10-09-2014
2 27-11-2015
2 01-12-2015
3 15-04-2013
3 17-05-2015
3 22-05-2015
3 30-10-2016
3 02-11-2016
I am working with Python and I would like to select the 3 last dates for each ID. Here is the dataset I would like to have:
ID date
1 25-06-2013
1 14-12-2013
1 10-04-2014
2 10-09-2014
2 27-11-2015
2 01-12-2015
3 22-05-2015
3 30-10-2016
3 02-11-2016
I used this code to select the very last date for each ID:
df_2=df.sort_values(by=['date']).drop_duplicates(subset='ID',keep='last')
But how can I select more than one date (for example the 3 last dates, or 4 last dates, etc)?
You might use groupby and tail following way to get 2 last items from each group:
import pandas as pd
df = pd.DataFrame({'ID':[1,1,1,2,2,2,3,3,3],'value':['A','B','C','D','E','F','G','H','I']})
df2 = df.groupby('ID').tail(2)
print(df2)
Output:
ID value
1 1 B
2 1 C
4 2 E
5 2 F
7 3 H
8 3 I
Note that for simplicity sake I used other (already sorted) data for building df.
can try this:
df.sort_values(by=['date']).groupby('ID').tail(3).sort_values(['ID', 'date'])
I tried this but with a non-datetime data type
a = [1,1,1,1,1,2,2,2,2,2,3,3,3,3,3]
b = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o']
import pandas as pd
import numpy as np
a = np.array([a,b])
df=pd.DataFrame(a.T,columns=['ID','Date'])
# the tail would give you the last n number of elements you are interested in
df_ = df.groupby('ID').tail(3)
df_
output:
ID Date
2 1 c
3 1 d
4 1 e
7 2 h
8 2 i
9 2 j
12 3 m
13 3 n
14 3 o
How can I concatenate two Series and create one DataFrame ?
For example, I have series like:
a=pd.Series([1,2,3])
b=pd.Series([4,5,6])
And, I want to get a data frame like:
pd.DataFrame([[1,4], [2,5], [3,6]])
Shortest would be:
pd.DataFrame([a,b]).T
Or:
pd.DataFrame(zip(a,b))
0 1
0 1 4
1 2 5
2 3 6
Or use concat:
>>> pd.concat([a,b],axis=1)
0 1
0 1 4
1 2 5
2 3 6
>>>
Or join:
>>> a.to_frame().join(b.to_frame(name=1))
0 1
0 1 4
1 2 5
2 3 6
>>>
Another possible faster solution could be,
pd.DataFrame(np.vstack((a,b)).T)
This sounds a bit weird, but I think that's exactly what I needed now:
I got several pandas dataframes that contains columns with float numbers, for example:
a b c
0 0 1 2
1 3 4 5
2 6 7 8
Now I want to add a column, with only one row, and the value is equal to the average of column 'a', in this case, is 3.0. So the new dataframe will looks like this:
a b c average
0 0 1 2 3.0
1 3 4 5
2 6 7 8
And all the rows below are empty.
I've tried things like df['average'] = np.mean(df['a']) but that give me a whole column of 3.0. Any help will be appreciated.
Assign a series, this is cleaner.
df['average'] = pd.Series(df['a'].mean(), index=df.index[[0]])
Or, even better, assign with loc:
df.loc[df.index[0], 'average'] = df['a'].mean().item()
Filling NaNs is straightforward, you can do
df['average'] = df['average'].fillna('')
df
a b c average
0 0 1 2 3
1 3 4 5
2 6 7 8
Can do something like:
df['average'] = [np.mean(df['a'])]+['']*(len(df)-1)
Here is a full example:
import pandas as pd
import numpy as np
df = pd.DataFrame(
[(0,1,2), (3,4,5), (6,7,8)],
columns=['a', 'b', 'c'])
print(df)
a b c
0 0 1 2
1 3 4 5
2 6 7 8
df['average'] = ''
df['average'][0] = df['a'].mean()
print(df)
a b c average
0 0 1 2 3
1 3 4 5
2 6 7 8
Using Pandas
I'm trying to determine whether a value in a certain row is greater than the values in all the other columns in the same row.
To do this I'm looping through the rows of a dataframe and using the 'all' function to compare the values in other columns; but it seems this is throwing an error "string indices must be integers"
It seems like this should work: What's wrong with this approach?
for row in dataframe:
if all (i < row['col1'] for i in [row['col2'], row['col3'], row['col4'], row['col5']]):
row['newcol'] = 'value'
Build a mask and pass it to loc:
df.loc[df['col1'] > df.loc[:, 'col2':'col5'].max(axis=1), 'newcol'] = 'newvalue'
The main problem, in my opinion, is using a loop for vectorisable logic.
Below is an example of how your logic can be implemented using numpy.where.
import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randint(0, 9, (5, 10)))
df['new_col'] = np.where(df[1] > df.max(axis=1),
'col1_is_max',
'col1_not_max')
Result:
0 1 2 3 4 5 6 7 8 9 new_col
0 4 1 3 8 3 2 5 1 1 2 col1_not_max
1 2 7 1 2 5 3 5 1 8 5 col1_is_max
2 1 8 2 5 7 4 0 3 6 3 col1_is_max
3 6 4 2 1 7 2 0 8 3 2 col1_not_max
4 0 1 3 3 0 3 7 4 4 1 col1_not_max
I have a pandas dataframe that I groupby, and then perform an aggregate calculation to get the mean for:
grouped = df.groupby(['year_month', 'company'])
means = grouped.agg({'size':['mean']})
Which gives me a dataframe back, but I can't seem to filter it to the specific company and year_month that I want:
means[(means['year_month']=='201412')]
gives me a KeyError
The issue is that you are grouping based on 'year_month' and 'company' . Hence in the means DataFrame, year_month and company would be part of the index (MutliIndex). You cannot access them as you access other columns.
One method to do this would be to get the values of the level 'year_month' of index . Example -
means.loc[means.index.get_level_values('year_month') == '201412']
Demo -
In [38]: df
Out[38]:
A B C
0 1 2 10
1 3 4 11
2 5 6 12
3 1 7 13
4 2 8 14
5 1 9 15
In [39]: means = df.groupby(['A','B']).mean()
In [40]: means
Out[40]:
C
A B
1 2 10
7 13
9 15
2 8 14
3 4 11
5 6 12
In [41]: means.loc[means.index.get_level_values('A') == 1]
Out[41]:
C
A B
1 2 10
7 13
9 15
As already pointed out, you will end up with a 2 level index. You could try to unstack the aggregated dataframe:
means = df.groupby(['year_month', 'company']).agg({'size':['mean']}).unstack(level=1)
This should give you a single 'year_month' index, 'company' as columns and your aggregate size as values. You can then slice by the index:
means.loc['201412']