In the following dataset what's the best way to duplicate row with groupby(['Type']) count < 3 to 3. df is the input, and df1 is my desired outcome. You see row 3 from df was duplicated by 2 times at the end. This is only an example deck. the real data has approximately 20mil lines and 400K unique Types, thus a method that does this efficiently is desired.
>>> df
Type Val
0 a 1
1 a 2
2 a 3
3 b 1
4 c 3
5 c 2
6 c 1
>>> df1
Type Val
0 a 1
1 a 2
2 a 3
3 b 1
4 c 3
5 c 2
6 c 1
7 b 1
8 b 1
Thought about using something like the following but do not know the best way to write the func.
df.groupby('Type').apply(func)
Thank you in advance.
Use value_counts with map and repeat:
counts = df.Type.value_counts()
repeat_map = 3 - counts[counts < 3]
df['repeat_num'] = df.Type.map(repeat_map).fillna(0,downcast='infer')
df = df.append(df.set_index('Type')['Val'].repeat(df['repeat_num']).reset_index(),
sort=False, ignore_index=True)[['Type','Val']]
print(df)
Type Val
0 a 1
1 a 2
2 a 3
3 b 1
4 c 3
5 c 2
6 c 1
7 b 1
8 b 1
Note : sort=False for append is present in pandas>=0.23.0, remove if using lower version.
EDIT : If data contains multiple val columns then make all columns columns as index expcept one column and repeat and then reset_index as:
df = df.append(df.set_index(['Type','Val_1','Val_2'])['Val'].repeat(df['repeat_num']).reset_index(),
sort=False, ignore_index=True)
Related
Let's assume, I have the following data frame.
Id Combinations
1 (A,B)
2 (C,)
3 (A,D)
4 (D,E,F)
5 (F)
I would like to filter out Combination column values with more than value in a set. Something like below. AND I would like count the number of occurrence as whole in Combination column. For example, ID number 2 and 5 should be removed since their value in a set is only 1.
The result I am looking for is:
ID Combination Frequency
1 A 2
1 B 1
3 A 2
3 D 2
4 D 2
4 E 1
4 F 2
Can anyone help to get the above result in Python pandas?
First if necessary convert values to lists:
df['Combinations'] = df['Combinations'].str.strip('(,)').str.split(',')
If need count after filtering only one values by Series.str.len in boolean indexing, then use DataFrame.explode and count values by Series.map with Series.value_counts:
df1 = df[df['Combinations'].str.len().gt(1)].explode('Combinations')
df1['Frequency'] = df1['Combinations'].map(df1['Combinations'].value_counts())
print (df1)
Id Combinations Frequency
0 1 A 2
0 1 B 1
2 3 A 2
2 3 D 2
3 4 D 2
3 4 E 1
3 4 F 1
Or if need count before removing them filter them by Series.duplicated in last step:
df2 = df.explode('Combinations')
df2['Frequency'] = df2['Combinations'].map(df2['Combinations'].value_counts())
df2 = df2[df2['Id'].duplicated(keep=False)]
Alternative:
df2 = df2[df2.groupby('Id').Id.transform('size') > 1]
Or:
df2 = df2[df2['Id'].map(df2['Id'].value_counts() > 1]
print (df2)
Id Combinations Frequency
0 1 A 2
0 1 B 1
2 3 A 2
2 3 D 2
3 4 D 2
3 4 E 1
3 4 F 2
This seems to be easy but couldn't find a working solution for it:
I have a dataframe with 3 columns:
df = pd.DataFrame({'A': [0,0,2,2,2],
'B': [1,1,2,2,3],
'C': [1,1,2,3,4]})
A B C
0 0 1 1
1 0 1 1
2 2 2 2
3 2 2 3
4 2 3 4
I want to select rows based on values of column A, then groupby based on values of column B, and finally transform values of column C into sum. something along the line of this (obviously not working) code:
df[df['A'].isin(['2']), 'C'] = df[df['A'].isin(['2']), 'C'].groupby('B').transform('sum')
desired output for above example is:
A B C
0 0 1 1
1 0 1 1
2 2 2 5
3 2 3 4
I also know how to split dataframe and do it. I am looking more for a solution that does it without the need of split+concat/merge. Thank you.
Is it just
s = df['A'].isin([2])
pd.concat((df[s].groupby(['A','B'])['C'].sum().reset_index(),
df[~s])
)
Output:
A B C
0 2 2 5
1 2 3 4
0 0 1 1
Update: Without splitting, you can assign a new column indicating special values of A:
(df.sort_values('A')
.assign(D=(~df['A'].isin([2])).cumsum())
.groupby(['D','A','B'])['C'].sum()
.reset_index('D',drop=True)
.reset_index()
)
Output:
A B C
0 0 1 1
1 0 1 1
2 2 2 5
3 2 3 4
I am using apply to leverage one dataframe to manipulate a second dataframe and return results. Here is a simplified example that I realize could be more easily answered with "in" logic, but for now let's keep the use of .apply() as a constraint:
import pandas as pd
df1 = pd.DataFrame({'Name':['A','B'],'Value':range(1,3)})
df2 = pd.DataFrame({'Name':['A']*3+['B']*4+['C'],'Value':range(1,9)})
def filter_df(x, df):
return df[df['Name']==x['Name']]
df1.apply(filter_df, axis=1, args=(df2, ))
Which is returning:
0 Name Value
0 A 1
1 A 2
2 ...
1 Name Value
3 B 4
4 B 5
5 ...
dtype: object
What I would like to see instead is one formated DataFrame with Name and Value headers. All advice appreciated!
Name Value
0 A 1
1 A 2
2 A 3
3 B 4
4 B 5
5 B 6
6 B 7
In my opinion, this cannot be done solely based on apply, you need pandas.concat:
result = pd.concat(df1.apply(filter_df, axis=1, args=(df2,)).to_list())
print(result)
Output
Name Value
0 A 1
1 A 2
2 A 3
3 B 4
4 B 5
5 B 6
6 B 7
This sounds a bit weird, but I think that's exactly what I needed now:
I got several pandas dataframes that contains columns with float numbers, for example:
a b c
0 0 1 2
1 3 4 5
2 6 7 8
Now I want to add a column, with only one row, and the value is equal to the average of column 'a', in this case, is 3.0. So the new dataframe will looks like this:
a b c average
0 0 1 2 3.0
1 3 4 5
2 6 7 8
And all the rows below are empty.
I've tried things like df['average'] = np.mean(df['a']) but that give me a whole column of 3.0. Any help will be appreciated.
Assign a series, this is cleaner.
df['average'] = pd.Series(df['a'].mean(), index=df.index[[0]])
Or, even better, assign with loc:
df.loc[df.index[0], 'average'] = df['a'].mean().item()
Filling NaNs is straightforward, you can do
df['average'] = df['average'].fillna('')
df
a b c average
0 0 1 2 3
1 3 4 5
2 6 7 8
Can do something like:
df['average'] = [np.mean(df['a'])]+['']*(len(df)-1)
Here is a full example:
import pandas as pd
import numpy as np
df = pd.DataFrame(
[(0,1,2), (3,4,5), (6,7,8)],
columns=['a', 'b', 'c'])
print(df)
a b c
0 0 1 2
1 3 4 5
2 6 7 8
df['average'] = ''
df['average'][0] = df['a'].mean()
print(df)
a b c average
0 0 1 2 3
1 3 4 5
2 6 7 8
Consider the following dataframe:
>>> import pandas as pd
>>> df = pd.DataFrame({'group': list('aaabbabc')})
>>> df
group
0 a
1 a
2 a
3 b
4 b
5 a
6 b
7 c
I want to count the cumulative number of times each group has occurred. My desired output looks like this:
>>> df
group n
0 a 0
1 a 1
2 a 2
3 b 0
4 b 1
5 a 3
6 b 2
7 c 0
My initial approach was to do something like this:
df['n'] = df.groupby('group').apply(lambda x: list(range(x.shape[0])))
Basically assigning a length n array, zero-indexed, to each group. But that has proven difficult to transpose and join.
You can use groupby + cumcount, and horizontally concat the new column:
>>> pd.concat([df, df.group.groupby(df.group).cumcount()], axis=1).rename(columns={0: 'n'})
group n
0 a 0
1 a 1
2 a 2
3 b 0
4 b 1
5 a 3
6 b 2
7 c 0
Simply use groupby on column name, in this case group and then apply cumcount and finally add a column in dataframe with the result.
df['n']=df.groupby('group').cumcount()
group n
0 a 0
1 a 1
2 a 2
3 b 0
4 b 1
5 a 3
6 b 2
7 c 0
You can use apply method by passing a lambda expression as parameter.
The idea is that you need to find out the count for a group as number of appearances for that group from the previous rows.
df['n'] = df.apply(lambda x: list(df['group'])[:int(x.name)].count(x['group']), axis=1)
Output
group n
0 a 0
1 a 1
2 a 2
3 b 0
4 b 1
5 a 3
6 b 2
7 c 0
Note: cumcount method is build with the help of the apply function.
You can read this in pandas documentation.