I want to know how can I replace the NaN in my dataset with the last average of 5 last values.
Column A
Column B
1
2
2
5
3
5
4
2
5
2
NaN
2
NaN
2
1
2
1
2
1
2
1
NaN
1
2
1
2
For example, in this case the first NaN will be the average of (1,2,3,4,5) and second NaN will be the average of (2,3,4,5, The value of the other NaN).
I have tried
df.fillna(df.mean())
As mentioned, it has been answered here, but the updated version for the latest pandas version is as follow:
data={'col1':[1,2,3,4,5,np.nan,np.nan,1,1,1,1,1,1],
'col2':[2,5,5,2,2,2,2,2,2,2,np.nan,2,2]}
df=pd.DataFrame(data)
window_size = 5
df=df.fillna(df.rolling(window_size+1, min_periods=1).mean())
outputs:
col1 col2
0 1.0 2.0
1 2.0 5.0
2 3.0 5.0
3 4.0 2.0
4 5.0 2.0
5 3.0 2.0
6 3.5 2.0
7 1.0 2.0
8 1.0 2.0
9 1.0 2.0
10 1.0 2.0
11 1.0 2.0
12 1.0 2.0
Related
I have two dataframes shown below:
df_1 =
Lon Lat N
0 2 1 1
1 2 2 3
2 2 3 1
3 3 2 2
and
df_2 =
Lon Lat N
0 1.0 1.0 NaN
1 2.0 1.0 NaN
2 3.0 1.0 NaN
3 4.0 1.0 NaN
4 1.0 2.0 NaN
5 2.0 2.0 NaN
6 3.0 2.0 NaN
7 4.0 2.0 NaN
8 1.0 3.0 NaN
9 2.0 3.0 NaN
10 3.0 3.0 NaN
11 4.0 3.0 NaN
What I want to do is to compare these two dfs and merge them according to Lon and Lat. That is to say NaN in df_2 will be covered with values in df_1 if the corresponding Lon and Lat are identical. The ideal output should be as:
Lon Lat N
0 1.0 1.0 NaN
1 2.0 1.0 1
2 3.0 1.0 NaN
3 4.0 1.0 NaN
4 1.0 2.0 NaN
5 2.0 2.0 3
6 3.0 2.0 2
7 4.0 2.0 NaN
8 1.0 3.0 NaN
9 2.0 3.0 1
10 3.0 3.0 NaN
11 4.0 3.0 NaN
The reason I want to do this is df_1's coordinates Lat and Lon are non-rectangular or unstructured grid, and I need to fill some NaN values so as to get a rectangular meshgrid and make contourf applicable. It would be highly appreciated if you can provide better ways to make the contour plot.
I have tried df_2.combine_first(df_1), but it doesn't work.
Thanks!
df_2.drop(columns = 'N').merge(df_1, on = ['Lon', 'Lat'], how = 'left')
Lon Lat N
0 1.0 1.0 NaN
1 2.0 1.0 1.0
2 3.0 1.0 NaN
3 4.0 1.0 NaN
4 1.0 2.0 NaN
5 2.0 2.0 3.0
6 3.0 2.0 2.0
7 4.0 2.0 NaN
8 1.0 3.0 NaN
9 2.0 3.0 1.0
10 3.0 3.0 NaN
11 4.0 3.0 NaN
If you first create the df_2 with all needed values you can update it with the second DataFrame by using pandas.DataFrame.update.
For this you need to first set the the correct index by using pandas.DataFrame.set_index.
Have a look at this Post for more information.
This is my dataframe:
df = pd.DataFrame(np.array([ [1,5],[1,6],[1,np.nan],[2,np.nan],[2,8],[2,4],[2,np.nan],[2,10],[3,np.nan]]),columns=['id','value'])
id value
0 1 5
1 1 6
2 1 NaN
3 2 NaN
4 2 8
5 2 4
6 2 NaN
7 2 10
8 3 NaN
This is my expected output:
id value
0 1 5
1 1 6
2 1 7
3 2 NaN
4 2 8
5 2 4
6 2 2
7 2 10
8 3 NaN
This is my current output using this code:
df.value.interpolate(method="krogh")
0 5.000000
1 6.000000
2 9.071429
3 10.171429
4 8.000000
5 4.000000
6 2.357143
7 10.000000
8 36.600000
Basically, I want to do two important things here:
Groupby ID then Interpolate using only above values not below row values
This should do the trick:
df["value_interp"]=df.value.combine_first(df.groupby("id")["value"].apply(lambda y: y.expanding().apply(lambda x: x.interpolate(method="krogh").to_numpy()[-1], raw=False)))
Outputs:
id value value_interp
0 1.0 5.0 5.0
1 1.0 6.0 6.0
2 1.0 NaN 7.0
3 2.0 NaN NaN
4 2.0 8.0 8.0
5 2.0 4.0 4.0
6 2.0 NaN 0.0
7 2.0 10.0 10.0
8 3.0 NaN NaN
(It interpolates based only on the previous values within the group - hence index 6 will return 0 not 2)
You can group by id and then loop over groups to make interpolations. For id = 2 interpolation will not give you value 2
import pandas as pd
import numpy as np
df = pd.DataFrame(np.array([ [1,5],[1,6],[1,np.nan],[2,np.nan],[2,8],[2,4],[2,np.nan],[2,10],[3,np.nan]]),columns=['id','value'])
data = []
for name, group in df.groupby('id'):
group_interpolation = group.interpolate(method='krogh', limit_direction='forward', axis=0)
data.append(group_interpolation)
df = (pd.concat(data)).round(1)
Output:
id value
0 1.0 5.0
1 1.0 6.0
2 1.0 7.0
3 2.0 NaN
4 2.0 8.0
5 2.0 4.0
6 2.0 4.7
7 2.0 10.0
8 3.0 NaN
Current pandas.Series.interpolate does not support what you want so to achieve your goal you need to do 2 grouby's that will account for your desire to use only previous rows. The idea is as follows: to combine into one group only missing value (!!!) and previous rows (it might have limitations if you have several missing values in a row, but it serves well for your toy example)
Suppose we have a df:
print(df)
ID Value
0 1 5.0
1 1 6.0
2 1 NaN
3 2 NaN
4 2 8.0
5 2 4.0
6 2 NaN
7 2 10.0
8 3 NaN
Then we will combine any missing values within a group with previous rows:
df["extrapolate"] = df.groupby("ID")["Value"].apply(lambda grp: grp.isnull().cumsum().shift().bfill())
print(df)
ID Value extrapolate
0 1 5.0 0.0
1 1 6.0 0.0
2 1 NaN 0.0
3 2 NaN 1.0
4 2 8.0 1.0
5 2 4.0 1.0
6 2 NaN 1.0
7 2 10.0 2.0
8 3 NaN NaN
You may see, that when grouped by ["ID","extrapolate"] the missing value will fall into the same group as nonnull values of previous rows.
Now we are ready to do extrapolation (with spline of order=1):
df.groupby(["ID","extrapolate"], as_index=False).apply(lambda grp:grp.interpolate(method="spline",order=1)).drop("extrapolate", axis=1)
ID Value
0 1.0 5.0
1 1.0 6.0
2 1.0 7.0
3 2.0 NaN
4 2.0 8.0
5 2.0 4.0
6 2.0 0.0
7 2.0 10.0
8 NaN NaN
Hope this helps.
I have a pandas dataframe with 11 columns. I want to add the sum of all values of columns 9 and column 10 to the end of table. So far I tried 2 methods:
Assigning the data to the cell with dataframe.iloc[rownumber, 8]. This results in an out of bound error.
Creating a vector with some blank: ' ' by using the following code:
total = ['', '', '', '', '', '', '', '', dataframe['Column 9'].sum(), dataframe['Column 10'].sum(), '']
dataframe = dataframe.append(total)
The result was not nice as it added the total vector as a vertical vector at the end rather than a horizontal one. What can I do to solve the issue?
You need use pandas.DataFrame.append with ignore_index=True
so use:
dataframe=dataframe.append(dataframe[['Column 9','Column 10']].sum(),ignore_index=True).fillna('')
Example:
import pandas as pd
import numpy as np
df=pd.DataFrame()
df['col1']=[1,2,3,4]
df['col2']=[2,3,4,5]
df['col3']=[5,6,7,8]
df['col4']=[5,6,7,8]
Using Append:
df=df.append(df[['col2','col3']].sum(),ignore_index=True)
print(df)
col1 col2 col3 col4
0 1.0 2.0 5.0 5.0
1 2.0 3.0 6.0 6.0
2 3.0 4.0 7.0 7.0
3 4.0 5.0 8.0 8.0
4 NaN 14.0 26.0 NaN
Whitout NaN values:
df=df.append(df[['col2','col3']].sum(),ignore_index=True).fillna('')
print(df)
col1 col2 col3 col4
0 1 2.0 5.0 5
1 2 3.0 6.0 6
2 3 4.0 7.0 7
3 4 5.0 8.0 8
4 14.0 26.0
Create new DataFrame with sums. This example DataFrame has columns 'a' and 'b'. df1 is the DataFrame what need to be summed up and df3 is one line DataFrame only with sums:
data = [[df1.a.sum(),df1.b.sum()]]
df3 = pd.DataFrame(data,columns=['a','b'])
Then append it to end:
df1.append(df3)
simply try this:(replace test with your dataframe name)
row wise sum(which you have asked for):
test['Total'] = test[['col9','col10']].sum(axis=1)
print(test)
column wise sum:
test.loc['Total'] = test[['col9','col10']].sum()
test.fillna('',inplace=True)
print(test)
IICU , this is what you need (change numbers 8 & 9 to suit your needs)
df['total']=df.iloc[ : ,[8,9]].sum(axis=1) #horizontal sum
df['total1']=df.iloc[ : ,[8,9]].sum().sum() #Vertical sum
df.loc['total2']=df.iloc[ : ,[8,9]].sum() # vertical sum in rows for only columns 8 & 9
Example
a=np.arange(0, 11, 1)
b=np.random.randint(10, size=(5,11))
df=pd.DataFrame(columns=a, data=b)
0 1 2 3 4 5 6 7 8 9 10
0 0 5 1 3 4 8 6 6 8 1 0
1 9 9 8 9 9 2 3 8 9 3 6
2 5 7 9 0 8 7 8 8 7 1 8
3 0 7 2 8 8 3 3 0 4 8 2
4 9 9 2 5 2 2 5 0 3 4 1
**output**
0 1 2 3 4 5 6 7 8 9 10 total total1
0 0.0 5.0 1.0 3.0 4.0 8.0 6.0 6.0 8.0 1.0 0.0 9.0 48.0
1 9.0 9.0 8.0 9.0 9.0 2.0 3.0 8.0 9.0 3.0 6.0 12.0 48.0
2 5.0 7.0 9.0 0.0 8.0 7.0 8.0 8.0 7.0 1.0 8.0 8.0 48.0
3 0.0 7.0 2.0 8.0 8.0 3.0 3.0 0.0 4.0 8.0 2.0 12.0 48.0
4 9.0 9.0 2.0 5.0 2.0 2.0 5.0 0.0 3.0 4.0 1.0 7.0 48.0
total2 NaN NaN NaN NaN NaN NaN NaN NaN 31.0 17.0 NaN NaN NaN
given the dataframe df
df = pd.DataFrame(data=[[np.nan,1],
[np.nan,np.nan],
[1,2],
[2,3],
[np.nan,np.nan],
[np.nan,np.nan],
[3,4],
[4,5],
[np.nan,np.nan],
[np.nan,np.nan]],columns=['A','B'])
df
Out[16]:
A B
0 NaN 1.0
1 NaN NaN
2 1.0 2.0
3 2.0 3.0
4 NaN NaN
5 NaN NaN
6 3.0 4.0
7 4.0 5.0
8 NaN NaN
9 NaN NaN
I would need to replace the nan using the following rules:
1) if nan is at the beginning replace with the first values after the nan
2) if nan is in the middle of 2 or more values replace the nan with the average of these values
3) if nan is at the end replace with the last value
df
Out[16]:
A B
0 1.0 1.0
1 1.0 1.5
2 1.0 2.0
3 2.0 3.0
4 2.5 3.5
5 2.5 3.5
6 3.0 4.0
7 4.0 5.0
8 4.0 5.0
9 4.0 5.0
Use add between forward filling and backfilling values, then divide by 2 and last replace last and first NaNs:
df = df.bfill().add(df.ffill()).div(2).ffill().bfill()
print (df)
A B
0 1.0 1.0
1 1.0 1.5
2 1.0 2.0
3 2.0 3.0
4 2.5 3.5
5 2.5 3.5
6 3.0 4.0
7 4.0 5.0
8 4.0 5.0
9 4.0 5.0
Detail:
print (df.bfill().add(df.ffill()))
A B
0 NaN 2.0
1 NaN 3.0
2 2.0 4.0
3 4.0 6.0
4 5.0 7.0
5 5.0 7.0
6 6.0 8.0
7 8.0 10.0
8 NaN NaN
9 NaN NaN
I am using the Pandas rolling window tool on a one-column dataframe whose index is in datetime form.
I would like to compute, for each window, the difference between the first value and the last value of said window. How do I refer to the relative index when giving a lambda function? (in the brackets below)
df2 = df.rolling('3s').apply(...)
IIUC:
In [93]: df = pd.DataFrame(np.random.randint(10,size=(9, 3)))
In [94]: df
Out[94]:
0 1 2
0 7 4 5
1 9 9 3
2 1 7 6
3 0 9 2
4 2 3 7
5 6 7 1
6 1 0 1
7 8 4 7
8 0 0 9
In [95]: df.rolling(window=3).apply(lambda x: x[0]-x[-1])
Out[95]:
0 1 2
0 NaN NaN NaN
1 NaN NaN NaN
2 6.0 -3.0 -1.0
3 9.0 0.0 1.0
4 -1.0 4.0 -1.0
5 -6.0 2.0 1.0
6 1.0 3.0 6.0
7 -2.0 3.0 -6.0
8 1.0 0.0 -8.0