After merging of two data frames:
output = pd.merge(df1, df2, on='ID', how='outer')
I have data frame like this:
index x y z
0 2 NaN 3
0 NaN 3 3
1 2 NaN 4
1 NaN 3 4
...
How to merge rows with the same index?
Expected output:
index x y z
0 2 3 3
1 2 3 4
Perhaps, you could take mean on them.
In [418]: output.groupby('index', as_index=False).mean()
Out[418]:
index x y z
0 0 2.0 3.0 3
1 1 2.0 3.0 4
We can group the DataFrame by the 'index' and then... we can just get the first values with .first() or minimum with .min() etc. depending on the case of course. What do you want to get if the values in z differ?
In [28]: gr = df.groupby('index', as_index=False)
In [29]: gr.first()
Out[29]:
index x y z
0 0 2.0 3.0 3
1 1 2.0 3.0 4
In [30]: gr.max()
Out[30]:
index x y z
0 0 2.0 3.0 3
1 1 2.0 3.0 4
In [31]: gr.min()
Out[31]:
index x y z
0 0 2.0 3.0 3
1 1 2.0 3.0 4
In [32]: gr.mean()
Out[32]:
index x y z
0 0 2.0 3.0 3
1 1 2.0 3.0 4
Related
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
I have this dataframe.
from pandas import DataFrame
import pandas as pd
df = pd.DataFrame({'name': ['A','D','M','T','B','C','D','E','A','L'],
'id': [1,1,1,2,2,3,3,3,3,5],
'rate': [3.5,4.5,2.0,5.0,4.0,1.5,2.0,2.0,1.0,5.0]})
>> df
name id rate
0 A 1 3.5
1 D 1 4.5
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
5 C 3 1.5
6 D 3 2.0
7 E 3 2.0
8 A 3 1.0
9 L 5 5.0
df = df.groupby('id')['rate'].mean()
what i want is this:
1) find mean of every 'id'.
2) give the number of ids (length) which has mean >= 3.
3) give back all rows of dataframe (where mean of any id >= 3.
Expected output:
Number of ids (length) where mean >= 3: 3
>> dataframe where (mean(id) >=3)
>>df
name id rate
0 A 1 3.0
1 D 1 4.0
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
5 L 5 5.0
Use GroupBy.transform for means by all groups with same size like original DataFrame, so possible filter by boolean indexing:
df = df[df.groupby('id')['rate'].transform('mean') >=3]
print (df)
name id rate
0 A 1 3.5
1 D 1 4.5
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
9 L 5 5.0
Detail:
print (df.groupby('id')['rate'].transform('mean'))
0 3.333333
1 3.333333
2 3.333333
3 4.500000
4 4.500000
5 1.625000
6 1.625000
7 1.625000
8 1.625000
9 5.000000
Name: rate, dtype: float64
Alternative solution with DataFrameGroupBy.filter:
df = df.groupby('id').filter(lambda x: x['rate'].mean() >=3)
I tried to solve this problem on my own, but I unfortunately haven't made much progress and would really appreciate anyone who can help me out.
My current dataframe contains 3 columns: 2 healthy columns and one column with some missing values, denoted as NaN.
df
Out[18]:
x1 x2 x3
0 A 1 2.0
1 B 0 NaN
2 A 0 1.0
3 A 1 2.0
4 A 0 NaN
5 B 1 1.0
6 A 1 1.0
7 B 0 2.0
8 B 0 2.0
I would like to fill the missing values in 'x3' by taking the median value of groupby of 'x1' and 'x2'.
groupby_df = df.groupby(['x1', 'x2'])['x3'].median()
groupby_df
Out[22]:
x1 x2
A 0 1.0
1 2.0
B 0 2.0
1 1.0
So, for instance, the NaN value corresponding to (B, 0) would be replaced by 2 and (A,0) by 1. I unfortunately can't figure out this part. Is there an elegant "DataFrame way" of filling in the NaN values with the computed median using groupby?
Thank You
using fillna inside groupby
df['x3']=df.groupby(['x1','x2'])['x3'].apply(lambda x : x.fillna(x.median()))
df
Out[928]:
x1 x2 x3
0 A 1 2.0
1 B 0 2.0
2 A 0 1.0
3 A 1 2.0
4 A 0 1.0
5 B 1 1.0
6 A 1 1.0
7 B 0 2.0
8 B 0 2.0
I have a dataframe having 4 columns(A,B,C,D). D has some NaN entries. I want to fill the NaN values by the average value of D having same value of A,B,C.
For example,if the value of A,B,C,D are x,y,z and Nan respectively,then I want the NaN value to be replaced by the average of D for the rows where the value of A,B,C are x,y,z respectively.
df['D'].fillna(df.groupby(['A','B','C'])['D'].transform('mean')) would be faster than apply
In [2400]: df
Out[2400]:
A B C D
0 1 1 1 1.0
1 1 1 1 NaN
2 1 1 1 3.0
3 3 3 3 5.0
In [2401]: df['D'].fillna(df.groupby(['A','B','C'])['D'].transform('mean'))
Out[2401]:
0 1.0
1 2.0
2 3.0
3 5.0
Name: D, dtype: float64
In [2402]: df['D'] = df['D'].fillna(df.groupby(['A','B','C'])['D'].transform('mean'))
In [2403]: df
Out[2403]:
A B C D
0 1 1 1 1.0
1 1 1 1 2.0
2 1 1 1 3.0
3 3 3 3 5.0
Details
In [2396]: df.shape
Out[2396]: (10000, 4)
In [2398]: %timeit df['D'].fillna(df.groupby(['A','B','C'])['D'].transform('mean'))
100 loops, best of 3: 3.44 ms per loop
In [2397]: %timeit df.groupby(['A','B','C'])['D'].apply(lambda x: x.fillna(x.mean()))
100 loops, best of 3: 5.34 ms per loop
I think you need:
df.D = df.groupby(['A','B','C'])['D'].apply(lambda x: x.fillna(x.mean()))
Sample:
df = pd.DataFrame({'A':[1,1,1,3],
'B':[1,1,1,3],
'C':[1,1,1,3],
'D':[1,np.nan,3,5]})
print (df)
A B C D
0 1 1 1 1.0
1 1 1 1 NaN
2 1 1 1 3.0
3 3 3 3 5.0
df.D = df.groupby(['A','B','C'])['D'].apply(lambda x: x.fillna(x.mean()))
print (df)
A B C D
0 1 1 1 1.0
1 1 1 1 2.0
2 1 1 1 3.0
3 3 3 3 5.0
Link to duplicate of this question for further information:
Pandas Dataframe: Replacing NaN with row average
Another suggested way of doing it mentioned in the link is using a simple fillna on the transpose:
df.T.fillna(df.mean(axis=1)).T
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]]))