I have a DataFrame with (several) grouping variables and (several) value variables. My goal is to set the last n non nan values to nan. So let's take a simple example:
df = pd.DataFrame({'id':[1,1,1,2,2,],
'value':[1,2,np.nan, 9,8]})
df
Out[1]:
id value
0 1 1.0
1 1 2.0
2 1 NaN
3 2 9.0
4 2 8.0
The desired result for n=1 would look like the following:
Out[53]:
id value
0 1 1.0
1 1 NaN
2 1 NaN
3 2 9.0
4 2 NaN
Use with groupby().cumcount():
N=1
groups = df.loc[df['value'].notna()].groupby('id')
enum = groups.cumcount()
sizes = groups['value'].transform('size')
df['value'] = df['value'].where(enum < sizes - N)
Output:
id value
0 1 1.0
1 1 NaN
2 1 NaN
3 2 9.0
4 2 NaN
You can check cumsum after groupby get how many notna value per-row
df['value'].where(df['value'].notna().iloc[::-1].groupby(df['id']).cumsum()>1,inplace=True)
df
Out[86]:
id value
0 1 1.0
1 1 NaN
2 1 NaN
3 2 9.0
4 2 NaN
One option: create a reversed cumcount on the non-NA values:
N = 1
m = (df
.loc[df['value'].notna()]
.groupby('id')
.cumcount(ascending=False)
.lt(N)
)
df.loc[m[m].index, 'value'] = np.nan
Similar approach with boolean masking:
m = df['value'].notna()
df['value'] = df['value'].mask(m[::-1].groupby(df['id']).cumsum().le(N))
output:
id value
0 1 1.0
1 1 NaN
2 1 NaN
3 2 9.0
4 2 NaN
Related
I am working with data like the following. The dataframe is sorted by the date:
category value Date
0 1 24/5/2019
1 NaN 24/5/2019
1 1 26/5/2019
2 2 1/6/2019
1 2 23/7/2019
2 NaN 18/8/2019
2 3 20/8/2019
7 3 1/9/2019
1 NaN 12/9/2019
2 NaN 13/9/2019
I would like to replace the "NaN" values with the previous mean for that specific category.
What is the best way to do this in pandas?
Some approaches I considered:
1) This litte riff:
df['mean' = df.groupby('category')['time'].apply(lambda x: x.shift().expanding().mean()))
source
This gets me the the correct means in but in another column, and it does not replace the NaNs.
2) This riff replaces the NaNs with the average of the columns:
df = df.groupby(df.columns, axis = 1).transform(lambda x: x.fillna(x.mean()))
Source 2
Both of these do not exactly give what I want. If someone could guide me on this it would be much appreciated!
You can replace value by new Series from shift + expanding + mean, first value of 1 group is not replaced, because no previous NaN values exits:
df['Date'] = pd.to_datetime(df['Date'])
s = df.groupby('category')['value'].apply(lambda x: x.shift().expanding().mean())
df['value'] = df['value'].fillna(s)
print (df)
category value Date
0 0 1.0 2019-05-24
1 1 NaN 2019-05-24
2 1 1.0 2019-05-26
3 2 2.0 2019-01-06
4 1 2.0 2019-07-23
5 2 2.0 2019-08-18
6 2 3.0 2019-08-20
7 7 3.0 2019-01-09
8 1 1.5 2019-12-09
9 2 2.5 2019-09-13
You can use pandas.Series.fillna to replace NaN values:
df['value']=df['value'].fillna(df.groupby('category')['value'].transform(lambda x: x.shift().expanding().mean()))
print(df)
category value Date
0 0 1.0 24/5/2019
1 1 NaN 24/5/2019
2 1 1.0 26/5/2019
3 2 2.0 1/6/2019
4 1 2.0 23/7/2019
5 2 2.0 18/8/2019
6 2 3.0 20/8/2019
7 7 3.0 1/9/2019
8 1 1.5 12/9/2019
9 2 2.5 13/9/2019
I am trying to impute/fill values using rows with similar columns' values.
For example, I have this dataframe:
one | two | three
1 1 10
1 1 nan
1 1 nan
1 2 nan
1 2 20
1 2 nan
1 3 nan
1 3 nan
I wanted to using the keys of column one and two which is similar and if column three is not entirely nan then impute the existing value from a row of similar keys with value in column '3'.
Here is my desired result:
one | two | three
1 1 10
1 1 10
1 1 10
1 2 20
1 2 20
1 2 20
1 3 nan
1 3 nan
You can see that keys 1 and 3 do not contain any value because the existing value does not exists.
I have tried using groupby+fillna():
df['three'] = df.groupby(['one','two'])['three'].fillna()
which gave me an error.
I have tried forward fill which give me rather strange result where it forward fill the column 2 instead. I am using this code for forward fill.
df['three'] = df.groupby(['one','two'], sort=False)['three'].ffill()
If only one non NaN value per group use ffill (forward filling) and bfill (backward filling) per group, so need apply with lambda:
df['three'] = df.groupby(['one','two'], sort=False)['three']
.apply(lambda x: x.ffill().bfill())
print (df)
one two three
0 1 1 10.0
1 1 1 10.0
2 1 1 10.0
3 1 2 20.0
4 1 2 20.0
5 1 2 20.0
6 1 3 NaN
7 1 3 NaN
But if multiple value per group and need replace NaN by some constant - e.g. mean by group:
print (df)
one two three
0 1 1 10.0
1 1 1 40.0
2 1 1 NaN
3 1 2 NaN
4 1 2 20.0
5 1 2 NaN
6 1 3 NaN
7 1 3 NaN
df['three'] = df.groupby(['one','two'], sort=False)['three']
.apply(lambda x: x.fillna(x.mean()))
print (df)
one two three
0 1 1 10.0
1 1 1 40.0
2 1 1 25.0
3 1 2 20.0
4 1 2 20.0
5 1 2 20.0
6 1 3 NaN
7 1 3 NaN
You can sort data by the column with missing values then groupby and forwardfill:
df.sort_values('three', inplace=True)
df['three'] = df.groupby(['one','two'])['three'].ffill()
If I create a new dataframe based on a user parameter, say a = 2. Therefore my dataframe df shrinks to 4 (ax2) columns into df_new. For example:
df_new = pd.DataFrame(data = {'col_01_01': [float('nan'),float('nan'),1,2,float('nan')], 'col_02_01': [float('nan'),float('nan'),1,2,float('nan')],'col_01_02': [0,0,0,0,1],'col_02_02': [1,0,0,1,1],'output':[1,0,1,1,1]})
To be more precise on the output column, let's look at the first row. [(nan,nan,0,1)] -> apply notna()-function to the first two entries and a comparison '==1' to the third and fourth row. -> This gives [(false, false, false, true)] -> compare these with an OR-expression and receive the desired result True -> 1
In the second row we find [(nan,nan,0,0)] therefore we find the output to be 0, since there is no valid value in the first two cols and 0 in the last two.
For a parameter a=3 we find 6 columns.
The result loos like this:
col_01_01 col_02_01 col_01_02 col_02_02 output
0 NaN NaN 0 1 1
1 NaN NaN 0 0 0
2 1.0 1.0 0 0 1
3 2.0 2.0 0 1 1
4 NaN NaN 1 1 1
You can use vectorised operations with notnull and eq:
null_cols = ['col_01_01', 'col_02_01']
int_cols = ['col_01_02', 'col_02_02']
df['output'] = (df[null_cols].notnull().any(1) | df[int_cols].eq(1).any(1)).astype(int)
print(df)
col_01_01 col_02_01 col_01_02 col_02_02 output
0 NaN NaN 0 1 1
1 NaN NaN 0 0 0
2 1.0 1.0 0 0 1
3 2.0 2.0 0 1 1
4 NaN NaN 1 1 1
I want to fill in data between each row in a dataframe with an average of current and next row (where columns are numeric)
starting data:
time value value_1 value-2
0 0 0 4 3
1 2 1 6 6
intermediate df:
time value value_1 value-2
0 0 0 4 3
1 1 0 4 3 #duplicate of row 0
2 2 1 6 6
3 3 1 6 6 #duplicate of row 2
I would like to create df_1:
time value value_1 value-2
0 0 0 4 3
1 1 0.5 5 4.5 #average of row 0 and 2
2 2 1 6 6
3 3 2 8 8 #average of row 2 and 4
To to this I appended a copy of the starting dataframe to create the intermediate dataframe shown above:
df = df_0.append(df_0)
df.sort_values(['time'], ascending=[True], inplace=True)
df = df.reset_index()
df['value_shift'] = df['value'].shift(-1)
df['value_shift_1'] = df['value_1'].shift(-1)
df['value_shift_2'] = df['value_2'].shift(-1)
then I was thinking of applying a function to each column:
def average_vals(numeric_val):
#average every odd row
if int(row.name) % 2 != 0:
#take average of value and value_shift for each value
#but this way I need to create 3 separate functions
Is there a way to do this without writing a separate function for each column and applying to each column one by one (in real data I have tens of columns)?
How about this method using DataFrame.reindex and DataFrame.interpolate
df.reindex(np.arange(len(df.index) * 2) / 2).interpolate().reset_index(drop=True)
Explanation
Reindex, in half steps reindex(np.arange(len(df.index) * 2) / 2)
This gives a DataFrame like this:
time value value_1 value-2
0.0 0.0 0.0 4.0 3.0
0.5 NaN NaN NaN NaN
1.0 2.0 1.0 6.0 6.0
1.5 NaN NaN NaN NaN
Then use DataFrame.interpolate to fill in the NaN values .... the default will be linear interpolation, so mean in this case.
Finaly, use .reset_index(drop=True) to fix your index.
Should give
time value value_1 value-2
0 0.0 0.0 4.0 3.0
1 1.0 0.5 5.0 4.5
2 2.0 1.0 6.0 6.0
3 2.0 1.0 6.0 6.0
This question already has answers here:
Pandas: filling missing values by mean in each group
(12 answers)
Closed last year.
I Know that the fillna() method can be used to fill NaN in whole dataframe.
df.fillna(df.mean()) # fill with mean of column.
How to limit mean calculation to the group (and the column) where the NaN is.
Exemple:
import pandas as pd
import numpy as np
df = pd.DataFrame({
'a': pd.Series([1,1,1,2,2,2]),
'b': pd.Series([1,2,np.NaN,1,np.NaN,4])
})
print df
Input
a b
0 1 1
1 1 2
2 1 NaN
3 2 1
4 2 NaN
5 2 4
Output (after groupby('a') & replace NaN by mean of group)
a b
0 1 1.0
1 1 2.0
2 1 1.5
3 2 1.0
4 2 2.5
5 2 4.0
IIUC then you can call fillna with the result of groupby on 'a' and transform on 'b':
In [44]:
df['b'] = df['b'].fillna(df.groupby('a')['b'].transform('mean'))
df
Out[44]:
a b
0 1 1.0
1 1 2.0
2 1 1.5
3 2 1.0
4 2 2.5
5 2 4.0
If you have multiple NaN values then I think the following should work:
In [47]:
df.fillna(df.groupby('a').transform('mean'))
Out[47]:
a b
0 1 1.0
1 1 2.0
2 1 1.5
3 2 1.0
4 2 2.5
5 2 4.0
EDIT
In [49]:
df = pd.DataFrame({
'a': pd.Series([1,1,1,2,2,2]),
'b': pd.Series([1,2,np.NaN,1,np.NaN,4]),
'c': pd.Series([1,np.NaN,np.NaN,1,np.NaN,4]),
'd': pd.Series([np.NaN,np.NaN,np.NaN,1,np.NaN,4])
})
df
Out[49]:
a b c d
0 1 1 1 NaN
1 1 2 NaN NaN
2 1 NaN NaN NaN
3 2 1 1 1
4 2 NaN NaN NaN
5 2 4 4 4
In [50]:
df.fillna(df.groupby('a').transform('mean'))
Out[50]:
a b c d
0 1 1.0 1.0 NaN
1 1 2.0 1.0 NaN
2 1 1.5 1.0 NaN
3 2 1.0 1.0 1.0
4 2 2.5 2.5 2.5
5 2 4.0 4.0 4.0
You get all NaN for 'd' as all values are NaN for group 1 for d
We first compute the group means, ignoring the missing values:
group_means = df.groupby('a')['b'].agg(lambda v: np.nanmean(v))
Next, we use groupby again, this time fetching the corresponding values:
df_new = df.groupby('a').apply(lambda t: t.fillna(group_means.loc[t['a'].iloc[0]]))