pandas filling nans by mean of before and after non-nan values
12
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 1 hour ago
ChrisChris
1,211214
add a comment |
12
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 1 hour ago
ChrisChris
1,211214
add a comment |
12
12
12
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 1 hour ago
ChrisChris
1,211214
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 1 hour ago
ChrisChris
1,211214
asked 1 hour ago
ChrisChris
1,211214
asked 1 hour ago
ChrisChris
1,211214
asked 1 hour ago
ChrisChris
1,211214
asked 1 hour ago
ChrisChris
1,211214
1,211214
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
16
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 1 hour ago
Sandeep KadapaSandeep Kadapa
6,908630
3
That is just brilliant. Thanks a ton :)
– Chris
1 hour ago
@Chris Glad to help.
– Sandeep Kadapa
1 hour ago
3
If first and last elements arenan. Then usedf.bfill().ffill()after using the above solution.
– Dark
1 hour ago
@anon01 Good point
– Chris
1 hour ago
@Dark Great suggestion :) Thanks for the insight
– Chris
1 hour 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
16
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 1 hour ago
Sandeep KadapaSandeep Kadapa
6,908630
3
That is just brilliant. Thanks a ton :)
– Chris
1 hour ago
@Chris Glad to help.
– Sandeep Kadapa
1 hour ago
3
If first and last elements arenan. Then usedf.bfill().ffill()after using the above solution.
– Dark
1 hour ago
@anon01 Good point
– Chris
1 hour ago
@Dark Great suggestion :) Thanks for the insight
– Chris
1 hour ago
add a comment |
16
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 1 hour ago
Sandeep KadapaSandeep Kadapa
6,908630
3
That is just brilliant. Thanks a ton :)
– Chris
1 hour ago
@Chris Glad to help.
– Sandeep Kadapa
1 hour ago
3
If first and last elements arenan. Then usedf.bfill().ffill()after using the above solution.
– Dark
1 hour ago
@anon01 Good point
– Chris
1 hour ago
@Dark Great suggestion :) Thanks for the insight
– Chris
1 hour ago
add a comment |
16
16
16
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 1 hour ago
Sandeep KadapaSandeep Kadapa
6,908630
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 1 hour ago
Sandeep KadapaSandeep Kadapa
6,908630
edited 1 hour ago
answered 1 hour ago
Sandeep KadapaSandeep Kadapa
6,908630
answered 1 hour ago
Sandeep KadapaSandeep Kadapa
6,908630
answered 1 hour ago
Sandeep KadapaSandeep Kadapa
6,908630
6,908630
3
That is just brilliant. Thanks a ton :)
– Chris
1 hour ago
@Chris Glad to help.
– Sandeep Kadapa
1 hour ago
3
If first and last elements arenan. Then usedf.bfill().ffill()after using the above solution.
– Dark
1 hour ago
@anon01 Good point
– Chris
1 hour ago
@Dark Great suggestion :) Thanks for the insight
– Chris
1 hour ago
add a comment |
3
That is just brilliant. Thanks a ton :)
– Chris
1 hour ago
@Chris Glad to help.
– Sandeep Kadapa
1 hour ago
3
If first and last elements arenan. Then usedf.bfill().ffill()after using the above solution.
– Dark
1 hour ago
@anon01 Good point
– Chris
1 hour ago
@Dark Great suggestion :) Thanks for the insight
– Chris
1 hour ago
3
3
That is just brilliant. Thanks a ton :)
– Chris
1 hour ago
That is just brilliant. Thanks a ton :)
– Chris
1 hour ago
@Chris Glad to help.
– Sandeep Kadapa
1 hour ago
@Chris Glad to help.
– Sandeep Kadapa
1 hour ago
3
3
If first and last elements are
nan. Then use df.bfill().ffill() after using the above solution.– Dark
1 hour ago
If first and last elements are
nan. Then use df.bfill().ffill() after using the above solution.– Dark
1 hour ago
@anon01 Good point
– Chris
1 hour ago
@anon01 Good point
– Chris
1 hour ago
@Dark Great suggestion :) Thanks for the insight
– Chris
1 hour ago
@Dark Great suggestion :) Thanks for the insight
– Chris
1 hour ago
add a comment |
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