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How to use each vector entry to fill NAN's of a separate groups in a dataframe

Say I have a vector ValsHR which looks like this:
valsHR=[78.8, 82.3, 91.0]
And I have a dataframe MainData
Age Patient HR
21 1 NaN
21 1 NaN
21 1 NaN
30 2 NaN
30 2 NaN
24 3 NaN
24 3 NaN
24 3 NaN
I want to fill the NaNs so that the first value in valsHR will only fill in the NaNs for patient 1, the second will fill the NaNs for patient 2 and the third will fill in for patient 3.
So far I've tried using this:
mainData['HR'] = mainData['HR'].fillna(ValsHR) but it fills all the NaNs with the first value in the vector.
I've also tried to use this:
mainData['HR'] = mainData.groupby('Patient').fillna(ValsHR) fills the NaNs with values that aren't in the valsHR vector at all.
I was wondering if anyone knew a way to do this?

Create dictionary by Patient values with missing values, map to original column and replace missing values only:
print (df)
Age Patient HR
0 21 1 NaN
1 21 1 NaN
2 21 1 NaN
3 30 2 100.0 <- value is not replaced
4 30 2 NaN
5 24 3 NaN
6 24 3 NaN
7 24 3 NaN
p = df.loc[df.HR.isna(), 'Patient'].unique()
valsHR = [78.8, 82.3, 91.0]
df['HR'] = df['HR'].fillna(df['Patient'].map(dict(zip(p, valsHR))))
print (df)
Age Patient HR
0 21 1 78.8
1 21 1 78.8
2 21 1 78.8
3 30 2 100.0
4 30 2 82.3
5 24 3 91.0
6 24 3 91.0
7 24 3 91.0
If some groups has no NaNs:
print (df)
Age Patient HR
0 21 1 NaN
1 21 1 NaN
2 21 1 NaN
3 30 2 100.0 <- group 2 is not replaced
4 30 2 100.0 <- group 2 is not replaced
5 24 3 NaN
6 24 3 NaN
7 24 3 NaN
p = df.loc[df.HR.isna(), 'Patient'].unique()
valsHR = [78.8, 82.3, 91.0]
df['HR'] = df['HR'].fillna(df['Patient'].map(dict(zip(p, valsHR))))
print (df)
Age Patient HR
0 21 1 78.8
1 21 1 78.8
2 21 1 78.8
3 30 2 100.0
4 30 2 100.0
5 24 3 82.3
6 24 3 82.3
7 24 3 82.3

It is simply mapping, if all of NaN should be replaced
import pandas as pd
from io import StringIO
valsHR=[78.8, 82.3, 91.0]
vals = {i:k for i,k in enumerate(valsHR, 1)}
df = pd.read_csv(StringIO("""Age Patient
21 1
21 1
21 1
30 2
30 2
24 3
24 3
24 3"""), sep="\s+")
df["HR"] = df["Patient"].map(vals)
>>> df
Age Patient HR
0 21 1 78.8
1 21 1 78.8
2 21 1 78.8
3 30 2 82.3
4 30 2 82.3
5 24 3 91.0
6 24 3 91.0
7 24 3 91.0

How to calculate a groupby mean and variance in a pandas DataFrame?

I have a DataFrame and I want to calculate the mean and the variance for each row for each person. Moreover, there is a column date and the chronological order must be respect when calculating the mean and the variance; the dataframe is already sorted by date. The date are just the number of day after the earliest date. The mean for the earliest date of a person row is simply the value in the column Points and the variance should be NAN or 0. Then, for the second date, the mean should be the means of the value in the column Points for this date and the previous one. Here is my code to generate the dataframe:
import pandas as pd
import numpy as np
data=[["Al",0, 12],["Bob",2, 10],["Carl",5, 12],["Al",5, 5],["Bob",9, 2]
,["Al",22, 4],["Bob",22, 16],["Carl",33, 2],["Al",45, 7],["Bob",68, 4]
,["Al",72, 11],["Bob",79, 5]]
df= pd.DataFrame(data, columns=["Name", "Date", "Points"])
print(df)
Name Date Points
0 Al 0 12
1 Bob 2 10
2 Carl 5 12
3 Al 5 5
4 Bob 9 2
5 Al 22 4
6 Bob 22 16
7 Carl 33 2
8 Al 45 7
9 Bob 68 4
10 Al 72 11
11 Bob 79 5
Here is my code to obtain the mean and the variance:
df['Mean'] = df.apply(
lambda x: df[(df.Name == x.Name) & (df.Date < x.Date)].Points.mean(),
axis=1)
df['Variance'] = df.apply(
lambda x: df[(df.Name == x.Name)& (df.Date < x.Date)].Points.var(),
axis=1)
However, the mean is shifted by one row and the variance by two rows. The dataframe obtained when sort by Nameand Dateis:
Name Date Points Mean Variance
0 Al 0 12 NaN NaN
3 Al 5 5 12.000000 NaN
5 Al 22 4 8.50000 24.500000
8 Al 45 7 7.000000 19.000000
10 Al 72 11 7.000000 12.666667
1 Bob 2 10 NaN NaN
4 Bob 9 2 10.000000 NaN
6 Bob 22 16 6.000000 32.000000
9 Bob 68 4 9.333333 49.333333
11 Bob 79 5 8.000000 40.000000
2 Carl 5 12 NaN NaN
7 Carl 33 2 12.000000 NaN
Instead, the dataframe should be as below:
Name Date Points Mean Variance
0 Al 0 12 12 NaN
3 Al 5 5 8.5 24.5
5 Al 22 4 7 19
8 Al 45 7 7 12.67
10 Al 72 11 7.8 ...
1 Bob 2 10 10 NaN
4 Bob 9 2 6 32
6 Bob 22 16 9.33 49.33
9 Bob 68 4 8 40
11 Bob 79 5 7.4 ...
2 Carl 5 12 12 NaN
7 Carl 33 2 7 50
What should I change ?

Transpose the columns in aggregate function in Pandas

I am having an aggregate function with group by to get the summarized values.
My data set:
df=pd.DataFrame({"A":['a','a','a','a','a','a','b','b','b','b'],
"Sales":[2,3,7,1,4,3,5,6,9,10],
"Units":[12,2,2,33,6,2,4,8,3,5],
"Week":[1,2,2,1,2,1,1,2,2,1]})
Upon this, I am applying the function:
def my_agg(x):
names = {
'Sales': x['Sales'].sum(),
'Units': x['Sales'].sum()
}
return pd.Series(names, index=['Sales','Units'])
dfA= df.groupby(['A','Week']).apply(my_agg)
which gives me output:
Sales Units
A Week
a 1 6 6
2 14 14
b 1 15 15
2 15 15
I want to transpose week into columns. Like this:
REQUIRED OUTPUT:
Week 1 2
A Sales Units Sales Units
a 6 6 14 14
b 15 15 15 15
ALSO, please suggest for OUTPUT 2:
Sales Units
A Week 1 2
a 6 14 6 14
b 15 15 15 15

unstack with swaplevel
s=dfA.unstack()
s
Out[127]:
Sales Units
Week 1 2 1 2
A
a 6 14 6 14
b 15 15 15 15
s.swaplevel(0,1,axis=1).sort_index(level=0,axis=1)
Out[128]:
Week 1 2
Sales Units Sales Units
A
a 6 6 14 14
b 15 15 15 15

Output 1
df.pivot_table(index='A', columns='Week', aggfunc='sum').swaplevel(1, 0, 1)
Week 1 2 1 2
Sales Sales Units Units
A
a 6 14 47 10
b 15 15 9 11
Output 2
df.pivot_table(index='A', columns='Week', aggfunc='sum')
Sales Units
Week 1 2 1 2
A
a 6 14 47 10
b 15 15 9 11

Compare two pandas dataframe with different size

I have one massive pandas dataframe with this structure:
df1:
A B
0 0 12
1 0 15
2 0 17
3 0 18
4 1 45
5 1 78
6 1 96
7 1 32
8 2 45
9 2 78
10 2 44
11 2 10
And a second one, smaller like this:
df2
G H
0 0 15
1 1 45
2 2 31
I want to add a column to my first dataframe following this rule: column df1.C = df2.H when df1.A == df2.G
I manage to do it with for loops, but the database is massive and the code run really slowly so I am looking for a Pandas-way or numpy to do it.
Many thanks,
Boris

If you only want to match mutual rows in both dataframes:
import pandas as pd
df1 = pd.DataFrame({'Name':['Sara'],'Special ability':['Walk on water']})
df1
Name Special ability
0 Sara Walk on water
df2 = pd.DataFrame({'Name':['Sara', 'Gustaf', 'Patrik'],'Age':[4,12,11]})
df2
Name Age
0 Sara 4
1 Gustaf 12
2 Patrik 11
df = df2.merge(df1, left_on='Name', right_on='Name', how='left')
df
Name Age Special ability
0 Sara 4 NaN
1 Gustaf 12 Walk on water
2 Patrik 11 NaN
This Can allso be done with more than one matching argument: (In this example Patrik from df1 does not exist in df2 becuse they have different ages and therfore will not merge)
df1 = pd.DataFrame({'Name':['Sara','Patrik'],'Special ability':['Walk on water','FireBalls'],'Age':[12,83]})
df1
Name Special ability Age
0 Sara Walk on water 12
1 Patrik FireBalls 83
df2 = pd.DataFrame({'Name':['Sara', 'Gustaf', 'Patrik'],'Age':[4,12,11]})
df2
Name Age
0 Sara 4
1 Gustaf 12
2 Patrik 11
df = df2.merge(df1,left_on=['Name','Age'],right_on=['Name','Age'],how='left')
df
Name Age Special ability
0 Sara 12 Walk on water
1 Gustaf 12 NaN
2 Patrik 11 NaN

You probably want to use a merge:
df=df1.merge(df2,left_on="A",right_on="G")
will give you a dataframe with 3 columns, but the third one's name will be H
df.columns=["A","B","C"]
will then give you the column names you want

You can use map by Series created by set_index:
df1['C'] = df1['A'].map(df2.set_index('G')['H'])
print (df1)
A B C
0 0 12 15
1 0 15 15
2 0 17 15
3 0 18 15
4 1 45 45
5 1 78 45
6 1 96 45
7 1 32 45
8 2 45 31
9 2 78 31
10 2 44 31
11 2 10 31
Or merge with drop and rename:
df = df1.merge(df2,left_on="A",right_on="G", how='left')
.drop('G', axis=1)
.rename(columns={'H':'C'})
print (df)
A B C
0 0 12 15
1 0 15 15
2 0 17 15
3 0 18 15
4 1 45 45
5 1 78 45
6 1 96 45
7 1 32 45
8 2 45 31
9 2 78 31
10 2 44 31
11 2 10 31

Here's one vectorized NumPy approach -
idx = np.searchsorted(df2.G.values, df1.A.values)
df1['C'] = df2.H.values[idx]
idx could be computed in a simpler way with : df2.G.searchsorted(df1.A), but don't think that would be anymore efficient, because we want to use the underlying array with .values for performance as done earlier.

Multiindex on DataFrames and sum in Pandas

I am currently trying to make use of Pandas MultiIndex attribute. I am trying to group an existing DataFrame-object df_original based on its columns in a smart way, and was therefore thinking of MultiIndex.
print df_original =
by_currency by_portfolio A B C
1 AUD a 1 2 3
2 AUD b 4 5 6
3 AUD c 7 8 9
4 AUD d 10 11 12
5 CHF a 13 14 15
6 CHF b 16 17 18
7 CHF c 19 20 21
8 CHF d 22 23 24
Now, what I would like to have is a MultiIndex DataFrame-object, with A, B and C, and by_portfolio as indices. Looking like
CHF AUD
A a 13 1
b 16 4
c 19 7
d 22 10
B a 14 2
b 17 5
c 20 8
d 23 11
C a 15 3
b 18 6
c 21 9
d 24 12
I have tried making all columns in df_original and the sought after indices into list-objects, and from there create a new DataFrame. This seems a bit cumbersome, and I can't figure out how to add the actual values after.
Perhaps some sort of groupby is better for this purpose? Thing is I will need to be able to add this data to another, similar, DataFrame, so I will need the resulting DataFrame to be able to be added to another one later on.
Thanks

You can use a combination of stack and unstack:
In [50]: df.set_index(['by_currency', 'by_portfolio']).stack().unstack(0)
Out[50]:
by_currency AUD CHF
by_portfolio
a A 1 13
B 2 14
C 3 15
b A 4 16
B 5 17
C 6 18
c A 7 19
B 8 20
C 9 21
d A 10 22
B 11 23
C 12 24
To obtain your desired result, we only need to swap the levels of the index:
In [51]: df2 = df.set_index(['by_currency', 'by_portfolio']).stack().unstack(0)
In [52]: df2.columns.name = None
In [53]: df2.index = df2.index.swaplevel(0,1)
In [55]: df2 = df2.sort_index()
In [56]: df2
Out[56]:
AUD CHF
by_portfolio
A a 1 13
b 4 16
c 7 19
d 10 22
B a 2 14
b 5 17
c 8 20
d 11 23
C a 3 15
b 6 18
c 9 21
d 12 24