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.
Related
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
Given df
df = pd.DataFrame({'distance': [0,1,2,np.nan,3,4,5,np.nan,np.nan,6]})
distance
0 0.0
1 1.0
2 2.0
3 NaN
4 3.0
5 4.0
6 5.0
7 NaN
8 NaN
9 6.0
I want to replace the nans with the inbetween mean
Expected output:
distance
0 0.0
1 1.0
2 2.0
3 2.5
4 3.0
5 4.0
6 5.0
7 5.5
8 5.5
9 6.0
I have seen this_answer but it's for a grouping which isn't my case and I couldn't find anything else.
If you don't want df.interpolate you can compute the mean of the surrounding values manually with df.bfill and df.ffill
(df.ffill() + df.bfill()) / 2
Out:
distance
0 0.0
1 1.0
2 2.0
3 2.5
4 3.0
5 4.0
6 5.0
7 5.5
8 5.5
9 6.0
How about using linear interpolation?
print(df.distance.interpolate())
0 0.000000
1 1.000000
2 2.000000
3 2.500000
4 3.000000
5 4.000000
6 5.000000
7 5.333333
8 5.666667
9 6.000000
Name: distance, dtype: float64
I've imported a .csv into pandas and want to extract specific values and put them into a new column whilst maintaining the existing shape.
So df[::3] extracts the data-
1 1
2 4
3 7
4
5
6
7
I want it to look like
1 1
2
3
4 4
5
6
7 7
Here is a solution:
df = pd.read_csv(r"C:/users/k_sego/colsplit.csv",sep=";")
df1 = df[['col1']]
df2 = df[['col2']]
DF = pd.merge(df1,df2, how='outer',left_on=['col1'],right_on=['col2'])
and the result is
col1 col2
0 1.0 1.0
1 2.0 NaN
2 3.0 NaN
3 4.0 4.0
4 5.0 NaN
5 6.0 NaN
6 7.0 7.0
7 NaN NaN
8 NaN NaN
9 NaN NaN
10 NaN NaN
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
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