pandas filling nans by mean of before and after non-nan values
11
I would like to fill df's nan with an average of adjacent elements.
Consider a dataframe:
df = pd.DataFrame({'val': [1,np.nan, 4, 5, np.nan, 10, 1,2,5, np.nan, np.nan, 9]})
val
0 1.0
1 NaN
2 4.0
3 5.0
4 NaN
5 10.0
6 1.0
7 2.0
8 5.0
9 NaN
10 NaN
11 9.0
My desired output is:
val
0 1.0
1 2.5
2 4.0
3 5.0
4 7.5
5 10.0
6 1.0
7 2.0
8 5.0
9 7.0 <<< deadend
10 7.0 <<< deadend
11 9.0
I've looked into other solutions such as Fill cell containing NaN with average of value before and after, but this won't work in case of two or more consecutive np.nans.
Any help is greatly appreciated!
python pandas
asked 40 mins ago
ChrisChris
1,206214
add a comment |
11
I would like to fill df's nan with an average of adjacent elements.
Consider a dataframe:
df = pd.DataFrame({'val': [1,np.nan, 4, 5, np.nan, 10, 1,2,5, np.nan, np.nan, 9]})
val
0 1.0
1 NaN
2 4.0
3 5.0
4 NaN
5 10.0
6 1.0
7 2.0
8 5.0
9 NaN
10 NaN
11 9.0
My desired output is:
val
0 1.0
1 2.5
2 4.0
3 5.0
4 7.5
5 10.0
6 1.0
7 2.0
8 5.0
9 7.0 <<< deadend
10 7.0 <<< deadend
11 9.0
I've looked into other solutions such as Fill cell containing NaN with average of value before and after, but this won't work in case of two or more consecutive np.nans.
Any help is greatly appreciated!
python pandas
asked 40 mins ago
ChrisChris
1,206214
add a comment |
11
11
11
1
I would like to fill df's nan with an average of adjacent elements.
Consider a dataframe:
df = pd.DataFrame({'val': [1,np.nan, 4, 5, np.nan, 10, 1,2,5, np.nan, np.nan, 9]})
val
0 1.0
1 NaN
2 4.0
3 5.0
4 NaN
5 10.0
6 1.0
7 2.0
8 5.0
9 NaN
10 NaN
11 9.0
My desired output is:
val
0 1.0
1 2.5
2 4.0
3 5.0
4 7.5
5 10.0
6 1.0
7 2.0
8 5.0
9 7.0 <<< deadend
10 7.0 <<< deadend
11 9.0
I've looked into other solutions such as Fill cell containing NaN with average of value before and after, but this won't work in case of two or more consecutive np.nans.
Any help is greatly appreciated!
python pandas
asked 40 mins ago
ChrisChris
1,206214
I would like to fill df's nan with an average of adjacent elements.
Consider a dataframe:
df = pd.DataFrame({'val': [1,np.nan, 4, 5, np.nan, 10, 1,2,5, np.nan, np.nan, 9]})
val
0 1.0
1 NaN
2 4.0
3 5.0
4 NaN
5 10.0
6 1.0
7 2.0
8 5.0
9 NaN
10 NaN
11 9.0
My desired output is:
val
0 1.0
1 2.5
2 4.0
3 5.0
4 7.5
5 10.0
6 1.0
7 2.0
8 5.0
9 7.0 <<< deadend
10 7.0 <<< deadend
11 9.0
I've looked into other solutions such as Fill cell containing NaN with average of value before and after, but this won't work in case of two or more consecutive np.nans.
Any help is greatly appreciated!
python pandas
python pandas
asked 40 mins ago
ChrisChris
1,206214
asked 40 mins ago
ChrisChris
1,206214
asked 40 mins ago
ChrisChris
1,206214
asked 40 mins ago
ChrisChris
1,206214
asked 40 mins ago
ChrisChris
1,206214
1,206214
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
14
Use ffill + bfill and divide by 2:
df = (df.ffill()+df.bfill())/2
print(df)
val
0 1.0
1 2.5
2 4.0
3 5.0
4 7.5
5 10.0
6 1.0
7 2.0
8 5.0
9 7.0
10 7.0
11 9.0
EDIT : If 1st and last element contains NaN then use (Dark
suggestion):
df = pd.DataFrame({'val':[np.nan,1,np.nan, 4, 5, np.nan,
10, 1,2,5, np.nan, np.nan, 9,np.nan,]})
df = (df.ffill()+df.bfill())/2
df = df.bfill().ffill()
print(df)
val
0 1.0
1 1.0
2 2.5
3 4.0
4 5.0
5 7.5
6 10.0
7 1.0
8 2.0
9 5.0
10 7.0
11 7.0
12 9.0
13 9.0
answered 34 mins ago
Sandeep KadapaSandeep Kadapa
6,873630
3
That is just brilliant. Thanks a ton :)
– Chris
33 mins ago
@Chris Glad to help.
– Sandeep Kadapa
27 mins ago
3
If first and last elements arenan. Then usedf.bfill().ffill()after using the above solution.
– Dark
12 mins ago
@anon01 Good point
– Chris
10 mins ago
@Dark Great suggestion :) Thanks for the insight
– Chris
9 mins ago
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
14
Use ffill + bfill and divide by 2:
df = (df.ffill()+df.bfill())/2
print(df)
val
0 1.0
1 2.5
2 4.0
3 5.0
4 7.5
5 10.0
6 1.0
7 2.0
8 5.0
9 7.0
10 7.0
11 9.0
EDIT : If 1st and last element contains NaN then use (Dark
suggestion):
df = pd.DataFrame({'val':[np.nan,1,np.nan, 4, 5, np.nan,
10, 1,2,5, np.nan, np.nan, 9,np.nan,]})
df = (df.ffill()+df.bfill())/2
df = df.bfill().ffill()
print(df)
val
0 1.0
1 1.0
2 2.5
3 4.0
4 5.0
5 7.5
6 10.0
7 1.0
8 2.0
9 5.0
10 7.0
11 7.0
12 9.0
13 9.0
answered 34 mins ago
Sandeep KadapaSandeep Kadapa
6,873630
3
That is just brilliant. Thanks a ton :)
– Chris
33 mins ago
@Chris Glad to help.
– Sandeep Kadapa
27 mins ago
3
If first and last elements arenan. Then usedf.bfill().ffill()after using the above solution.
– Dark
12 mins ago
@anon01 Good point
– Chris
10 mins ago
@Dark Great suggestion :) Thanks for the insight
– Chris
9 mins ago
add a comment |
14
Use ffill + bfill and divide by 2:
df = (df.ffill()+df.bfill())/2
print(df)
val
0 1.0
1 2.5
2 4.0
3 5.0
4 7.5
5 10.0
6 1.0
7 2.0
8 5.0
9 7.0
10 7.0
11 9.0
EDIT : If 1st and last element contains NaN then use (Dark
suggestion):
df = pd.DataFrame({'val':[np.nan,1,np.nan, 4, 5, np.nan,
10, 1,2,5, np.nan, np.nan, 9,np.nan,]})
df = (df.ffill()+df.bfill())/2
df = df.bfill().ffill()
print(df)
val
0 1.0
1 1.0
2 2.5
3 4.0
4 5.0
5 7.5
6 10.0
7 1.0
8 2.0
9 5.0
10 7.0
11 7.0
12 9.0
13 9.0
answered 34 mins ago
Sandeep KadapaSandeep Kadapa
6,873630
3
That is just brilliant. Thanks a ton :)
– Chris
33 mins ago
@Chris Glad to help.
– Sandeep Kadapa
27 mins ago
3
If first and last elements arenan. Then usedf.bfill().ffill()after using the above solution.
– Dark
12 mins ago
@anon01 Good point
– Chris
10 mins ago
@Dark Great suggestion :) Thanks for the insight
– Chris
9 mins ago
add a comment |
14
14
14
Use ffill + bfill and divide by 2:
df = (df.ffill()+df.bfill())/2
print(df)
val
0 1.0
1 2.5
2 4.0
3 5.0
4 7.5
5 10.0
6 1.0
7 2.0
8 5.0
9 7.0
10 7.0
11 9.0
EDIT : If 1st and last element contains NaN then use (Dark
suggestion):
df = pd.DataFrame({'val':[np.nan,1,np.nan, 4, 5, np.nan,
10, 1,2,5, np.nan, np.nan, 9,np.nan,]})
df = (df.ffill()+df.bfill())/2
df = df.bfill().ffill()
print(df)
val
0 1.0
1 1.0
2 2.5
3 4.0
4 5.0
5 7.5
6 10.0
7 1.0
8 2.0
9 5.0
10 7.0
11 7.0
12 9.0
13 9.0
answered 34 mins ago
Sandeep KadapaSandeep Kadapa
6,873630
Use ffill + bfill and divide by 2:
df = (df.ffill()+df.bfill())/2
print(df)
val
0 1.0
1 2.5
2 4.0
3 5.0
4 7.5
5 10.0
6 1.0
7 2.0
8 5.0
9 7.0
10 7.0
11 9.0
EDIT : If 1st and last element contains NaN then use (Dark
suggestion):
df = pd.DataFrame({'val':[np.nan,1,np.nan, 4, 5, np.nan,
10, 1,2,5, np.nan, np.nan, 9,np.nan,]})
df = (df.ffill()+df.bfill())/2
df = df.bfill().ffill()
print(df)
val
0 1.0
1 1.0
2 2.5
3 4.0
4 5.0
5 7.5
6 10.0
7 1.0
8 2.0
9 5.0
10 7.0
11 7.0
12 9.0
13 9.0
answered 34 mins ago
Sandeep KadapaSandeep Kadapa
6,873630
edited 8 mins ago
answered 34 mins ago
Sandeep KadapaSandeep Kadapa
6,873630
answered 34 mins ago
Sandeep KadapaSandeep Kadapa
6,873630
answered 34 mins ago
Sandeep KadapaSandeep Kadapa
6,873630
6,873630
3
That is just brilliant. Thanks a ton :)
– Chris
33 mins ago
@Chris Glad to help.
– Sandeep Kadapa
27 mins ago
3
If first and last elements arenan. Then usedf.bfill().ffill()after using the above solution.
– Dark
12 mins ago
@anon01 Good point
– Chris
10 mins ago
@Dark Great suggestion :) Thanks for the insight
– Chris
9 mins ago
add a comment |
3
That is just brilliant. Thanks a ton :)
– Chris
33 mins ago
@Chris Glad to help.
– Sandeep Kadapa
27 mins ago
3
If first and last elements arenan. Then usedf.bfill().ffill()after using the above solution.
– Dark
12 mins ago
@anon01 Good point
– Chris
10 mins ago
@Dark Great suggestion :) Thanks for the insight
– Chris
9 mins ago
3
3
That is just brilliant. Thanks a ton :)
– Chris
33 mins ago
That is just brilliant. Thanks a ton :)
– Chris
33 mins ago
@Chris Glad to help.
– Sandeep Kadapa
27 mins ago
@Chris Glad to help.
– Sandeep Kadapa
27 mins ago
3
3
If first and last elements are
nan. Then use df.bfill().ffill() after using the above solution.– Dark
12 mins ago
If first and last elements are
nan. Then use df.bfill().ffill() after using the above solution.– Dark
12 mins ago
@anon01 Good point
– Chris
10 mins ago
@anon01 Good point
– Chris
10 mins ago
@Dark Great suggestion :) Thanks for the insight
– Chris
9 mins ago
@Dark Great suggestion :) Thanks for the insight
– Chris
9 mins ago
add a comment |
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