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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

Add the values of several columns when the number of columns exceeds 3 - Pandas

I have a pandas dataframe with several columns of dates, numbers and bill amounts. I would like to add the amounts of the other invoices with the 3rd one and change the invoice number by "1111".
Here is an example:
ID customer
Bill1
Date 1
ID Bill 1
Bill2
Date 2
ID Bill 2
Bill3
Date3
ID Bill 3
Bill4
Date 4
ID Bill 4
Bill5
Date 5
ID Bill 5
4
6
2000-10-04
1
45
2000-11-05
2
51
1999-12-05
3
23
2001-11-23
6
76
2011-08-19
12
6
8
2016-05-03
7
39
2017-08-09
8
38
2018-07-14
17
21
2009-05-04
9
Nan
Nan
Nan
12
14
2016-11-16
10
73
2017-05-04
15
Nan
Nan
Nan
Nan
Nan
Nan
Nan
Nan
Nan
And I would like to get this :
ID customer
Bill1
Date 1
ID Bill 1
Bill2
Date 2
ID Bill 2
Bill3
Date3
ID Bill 3
4
6
2000-10-04
1
45
2000-11-05
2
150
1999-12-05
1111
6
8
2016-05-03
7
39
2017-08-09
8
59
2018-07-14
1111
12
14
2016-11-16
10
73
2017-05-04
15
Nan
Nan
Nan
This example is a sample of my data, I may have many more than 5 columns.
Thanks for your help

with a little of data manipulation, you should be able to do it as:
df = df.replace('Nan', np.nan)
idx_col_bill3 = 7
step = 3
idx_col_bill3_id = 10
cols = df.columns
bills = df[cols[range(idx_col_bill3,len(cols), step)]].sum(axis=1)
bills.replace(0, nan, inplace=True)
df = df[cols[range(idx_col_bill3_id)]]
df['Bill3'] = bills
df['ID Bill 3'].iloc._setitem_with_indexer(df['ID Bill 3'].notna(),1111)

Alternative to Excel SUM in Pandas

I have a dataframe (i.e df1) with the below values. I wanted to SUM Row 4 to 9 and put the resulting value in Row3. How can we achieve it? In excel it has been simple SUM formula like this =SUM(B9:B14) but what is the alternative in pandas?
Detail Value
0 Day 23
1 Month Aug
2 Year 2020
3 Total Tickets NaN
4 Pune 2
5 Mumbai 3
6 Thane 33
7 Kolkatta NaN
8 Hyderabad NaN
9 Kerala 283

How fill unstinting numeric values in df column

so I am trying to add rows to data frame that should follow a numeric order 1 to 52
but my data is missing numbers, so I need to add these rows and fill these spots with NaN values or null.
df = pd.DataFrame("Weeks": [1,2,3,15,16,20,21,52],
"Values": [10,10,10,10,50,60,70,40])
Desired output:
Weeks Values
1 10
2 10
3 10
4 NaN
5 NaN
6 NaN
7 NaN
8 NaN
...
52 40
and so on until it reach Weeks = 52
My solution:
new_df = pd.DataFrame("Weeks": "" , "Values":"")
for x in range(1,53):
for i in df.Weeks:
if x == i:
new_df["Weeks"] = x
new_df["Values"] = df.Values[i]
The problem it is super inefficient, anyone know a way to do it in much efficient way?

You could use set_index to set the Weeks as index an reindex with a range up to the maximum week:
df.set_index('Weeks').reindex(range(1,df.Weeks.max()))
Or accounting for the minimum week too:
df.set_index('Weeks').reindex(range(*df.Weeks.agg(('min', 'max'))))
Values
Weeks
1 10.0
2 10.0
3 10.0
4 NaN
5 NaN
6 NaN
7 NaN
8 NaN
9 NaN
10 NaN
11 NaN
12 NaN
13 NaN
14 NaN
15 10.0
16 50.0
17 NaN
...

How to change consecutive repeating values in pandas dataframe series to nan or 0?

I have a pandas dataframe created from measured numbers. When something goes wrong with the measurement, the last value is repeated. I would like to do two things:
1. Change all repeating values either to nan or 0.
2. Keep the first repeating value and change all other values nan or 0.
I have found solutions using "shift" but they drop repeating values. I do not want to drop repeating values.My data frame looks like this:
df = pd.DataFrame(np.random.randn(15, 3))
df.iloc[4:8,0]=40
df.iloc[12:15,1]=22
df.iloc[10:12,2]=0.23
giving a dataframe like this:
0 1 2
0 1.239916 1.109434 0.305490
1 0.248682 1.472628 0.630074
2 -0.028584 -1.116208 0.074299
3 -0.784692 -0.774261 -1.117499
4 40.000000 0.283084 -1.495734
5 40.000000 -0.074763 -0.840403
6 40.000000 0.709794 -1.000048
7 40.000000 0.920943 0.681230
8 -0.701831 0.547689 -0.128996
9 -0.455691 0.610016 0.420240
10 -0.856768 -1.039719 0.230000
11 1.187208 0.964340 0.230000
12 0.116258 22.000000 1.119744
13 -0.501180 22.000000 0.558941
14 0.551586 22.000000 -0.993749
what I would like to be able to do is write some code that would filter the data and give me a data frame like this:
0 1 2
0 1.239916 1.109434 0.305490
1 0.248682 1.472628 0.630074
2 -0.028584 -1.116208 0.074299
3 -0.784692 -0.774261 -1.117499
4 NaN 0.283084 -1.495734
5 NaN -0.074763 -0.840403
6 NaN 0.709794 -1.000048
7 NaN 0.920943 0.681230
8 -0.701831 0.547689 -0.128996
9 -0.455691 0.610016 0.420240
10 -0.856768 -1.039719 NaN
11 1.187208 0.964340 NaN
12 0.116258 NaN 1.119744
13 -0.501180 NaN 0.558941
14 0.551586 NaN -0.993749
or even better keep the first value and change the rest to NaN. Like this:
0 1 2
0 1.239916 1.109434 0.305490
1 0.248682 1.472628 0.630074
2 -0.028584 -1.116208 0.074299
3 -0.784692 -0.774261 -1.117499
4 40.000000 0.283084 -1.495734
5 NaN -0.074763 -0.840403
6 NaN 0.709794 -1.000048
7 NaN 0.920943 0.681230
8 -0.701831 0.547689 -0.128996
9 -0.455691 0.610016 0.420240
10 -0.856768 -1.039719 0.230000
11 1.187208 0.964340 NaN
12 0.116258 22.000000 1.119744
13 -0.501180 NaN 0.558941
14 0.551586 NaN -0.993749

using shift & mask:
df.shift(1) == df compares the next row to the current for consecutive duplicates.
df.mask(df.shift(1) == df)
# outputs
0 1 2
0 0.365329 0.153527 0.143244
1 0.688364 0.495755 1.065965
2 0.354180 -0.023518 3.338483
3 -0.106851 0.296802 -0.594785
4 40.000000 0.149378 1.507316
5 NaN -1.312952 0.225137
6 NaN -0.242527 -1.731890
7 NaN 0.798908 0.654434
8 2.226980 -1.117809 -1.172430
9 -1.228234 -3.129854 -1.101965
10 0.393293 1.682098 0.230000
11 -0.029907 -0.502333 NaN
12 0.107994 22.000000 0.354902
13 -0.478481 NaN 0.531017
14 -1.517769 NaN 1.552974
if you want to remove all the consecutive duplicates, test that the previous row is also the same as the current row
df.mask((df.shift(1) == df) | (df.shift(-1) == df))

Option 1
Specialized solution using diff. Get's at the final desired output.
df.mask(df.diff().eq(0))
0 1 2
0 1.239916 1.109434 0.305490
1 0.248682 1.472628 0.630074
2 -0.028584 -1.116208 0.074299
3 -0.784692 -0.774261 -1.117499
4 40.000000 0.283084 -1.495734
5 NaN -0.074763 -0.840403
6 NaN 0.709794 -1.000048
7 NaN 0.920943 0.681230
8 -0.701831 0.547689 -0.128996
9 -0.455691 0.610016 0.420240
10 -0.856768 -1.039719 0.230000
11 1.187208 0.964340 NaN
12 0.116258 22.000000 1.119744
13 -0.501180 NaN 0.558941
14 0.551586 NaN -0.993749