use group by to get n smallest values but with duplicates - python

Suppose I have pandas DataFrame like this:
>>> df = pd.DataFrame({'id':[1,1,1,1,1,2,2,2,2,2,2,3,4],'value':[1,1,1,1,3,1,2,2,3,3,4,1,1]})
>>> df
id value
1 1
1 1
1 1
1 1
1 3
2 1
2 2
2 2
2 3
2 3
2 4
3 1
4 1
I want to get a new DataFrame with top 2 (well really n values) values for each id including duplicates, like this:
id value
0 1 1
1 1 1
3 1 1
4 1 1
5 1 3
6 2 1
7 2 2
8 2 2
9 3 1
10 4 1
I've tried using head() and nsmallest() but I think those will not include duplicates. Is there a better way to do this?
Edited to make it clear I want more than 2 records per group if there are more than 2 duplictes


Use DataFrame.drop_duplicates in first step, then get top values and last use DataFrame.merge:
df1 = df.drop_duplicates(['id','value']).sort_values(['id','value']).groupby('id').head(2)
df = df.merge(df1)
print (df)
id value
0 1 1
1 1 1
2 1 2
3 1 2
4 2 1
5 2 2
6 2 2
7 3 1
8 4 1
df = pd.DataFrame({'id':[1,1,1,1,1,2,2,2,2,2,2,3,4],'value':[1,1,1,1,3,1,2,2,3,3,4,1,1]})
df1 = df.drop_duplicates(['id','value']).sort_values(['id','value']).groupby('id').head(2)
df = df.merge(df1)
print (df)
id value
0 1 1
1 1 1
2 1 1
3 1 1
4 1 3
5 2 1
6 2 2
7 2 2
8 3 1
9 4 1
Or use custom lambda function with GroupBy.transform and filter in boolean indexing:
df = df[df.groupby('id')['value'].transform(lambda x: x.isin(sorted(set(x))[:2]))]
print (df)
id value
0 1 1
1 1 1
2 1 2
3 1 2
5 2 1
6 2 2
7 2 2
11 3 1
12 4 1
df = df[df.groupby('id')['value'].transform(lambda x: x.isin(sorted(set(x))[:2]))]
print (df)
id value
0 1 1
1 1 1
2 1 1
3 1 1
4 1 3
5 2 1
6 2 2
7 2 2
11 3 1
12 4 1

Related

Create a new column that starts counting from 0 until the value in another column changes

I have a dataframe df that looks like this:
ID
Months
Borrow_Rank
1
0
1
1
1
1
1
2
1
2
0
1
2
1
1
2
2
1
3
0
2
3
1
2
4
0
1
4
1
1
I want to create a new variable Months_Adjusted that starts counting from 0 for as long as Borrow_Rank remains the same.
ID
Months
Borrow_Rank
Months_Adjusted
1
0
1
0
1
1
1
1
1
2
1
2
2
0
1
3
2
1
1
4
2
2
1
5
3
0
2
0
3
1
2
1
4
0
1
0
4
1
1
1
Thank you all and I apologise if I could have written the question better. This is my first post.

import pandas as pd
df = pd.DataFrame({'Borrow_Rank':[1,1,1,1,1,1,1,2,2,2,2,2,3,3,3,1,1,1]})
selector = (df['Borrow_Rank'] != df['Borrow_Rank'].shift()).cumsum()
df['Months_Adjusted'] = df.groupby(selector).cumcount()

Here is a solution using itertools.groupby -
from itertools import groupby
df['Months_Adjusted'] = pd.concat(
[pd.Series(range(len(list(g))))
for k, g in groupby(df['Borrow_Rank'])],
ignore_index=True)
Output
ID Months Borrow_Rank Months_Adjusted
0 1 0 1 0
1 1 1 1 1
2 1 2 1 2
3 2 0 1 3
4 2 1 1 4
5 2 2 1 5
6 3 0 2 0
7 3 1 2 1
8 4 0 1 0
9 4 1 1 1

How to count consecutive same values in a pythonic way that looks iterative

So I am trying to count the number of consecutive same values in a dataframe and put that information into a new column in the dataframe, but I want the count to look iterative.
Here is what I have so far:
df = pd.DataFrame(np.random.randint(0,3, size=(15,4)), columns=list('ABCD'))
df['subgroupA'] = (df.A != df.A.shift(1)).cumsum()
dfg = df.groupby(by='subgroupA', as_index=False).apply(lambda grp: len(grp))
dfg.rename(columns={None: 'numConsec'}, inplace=True)
df = df.merge(dfg, how='left', on='subgroupA')
df
Here is the result:
A B C D subgroupA numConsec
0 2 1 1 1 1 1
1 1 2 1 0 2 2
2 1 0 2 1 2 2
3 0 1 2 0 3 1
4 1 0 0 1 4 1
5 0 2 2 1 5 2
6 0 2 1 1 5 2
7 1 0 0 1 6 1
8 0 2 0 0 7 4
9 0 0 0 2 7 4
10 0 2 1 1 7 4
11 0 2 2 0 7 4
12 1 2 0 1 8 1
13 0 1 1 0 9 1
14 1 1 1 0 10 1
The problem is, in the numConsec column, I don't want the full count for every row. I want it to reflect how it looks as you iteratively look at the dataframe. The problem is, my dataframe is too large to iteratively loop through and make the counts, as that would be too slow. I need to do it in a pythonic way and make it look like this:
A B C D subgroupA numConsec
0 2 1 1 1 1 1
1 1 2 1 0 2 1
2 1 0 2 1 2 2
3 0 1 2 0 3 1
4 1 0 0 1 4 1
5 0 2 2 1 5 1
6 0 2 1 1 5 2
7 1 0 0 1 6 1
8 0 2 0 0 7 1
9 0 0 0 2 7 2
10 0 2 1 1 7 3
11 0 2 2 0 7 4
12 1 2 0 1 8 1
13 0 1 1 0 9 1
14 1 1 1 0 10 1
Any ideas?

Keep only rows if the count of an object is maximum

I have a pandas data frame as follows
A B
1 2
1 2
1 0
1 2
2 3
2 3
2 1
3 0
3 0
3 1
3 2
I would like to get the following output
A B
1 2
1 2
1 2
2 3
2 3
3 0
3 0
This means that I need only rows where the count of A is maximum. Is there any solution to this?
Many thanks!

You can combine groupby() with Series.mode():
df_out = df[df.groupby("A")["B"].transform(lambda x: x == x.mode()[0])]
print(df_out)
Prints:
A B
0 1 2
1 1 2
3 1 2
4 2 3
5 2 3
7 3 0
8 3 0

Is this what you're looking for?
df.set_index(['A','B']).loc[df.groupby(['A','B']).size().groupby(level=0).idxmax()].reset_index()

Padding and reshaping pandas dataframe

I have a dataframe with the following form:
data = pd.DataFrame({'ID':[1,1,1,2,2,2,2,3,3],'Time':[0,1,2,0,1,2,3,0,1],
'sig':[2,3,1,4,2,0,2,3,5],'sig2':[9,2,8,0,4,5,1,1,0],
'group':['A','A','A','B','B','B','B','A','A']})
print(data)
ID Time sig sig2 group
0 1 0 2 9 A
1 1 1 3 2 A
2 1 2 1 8 A
3 2 0 4 0 B
4 2 1 2 4 B
5 2 2 0 5 B
6 2 3 2 1 B
7 3 0 3 1 A
8 3 1 5 0 A
I want to reshape and pad such that each 'ID' has the same number of Time values, the sig1,sig2 are padded with zeros (or mean value within ID) and the group carries the same letter value. The output after repadding would be :
data_pad = pd.DataFrame({'ID':[1,1,1,1,2,2,2,2,3,3,3,3],'Time':[0,1,2,3,0,1,2,3,0,1,2,3],
'sig1':[2,3,1,0,4,2,0,2,3,5,0,0],'sig2':[9,2,8,0,0,4,5,1,1,0,0,0],
'group':['A','A','A','A','B','B','B','B','A','A','A','A']})
print(data_pad)
ID Time sig1 sig2 group
0 1 0 2 9 A
1 1 1 3 2 A
2 1 2 1 8 A
3 1 3 0 0 A
4 2 0 4 0 B
5 2 1 2 4 B
6 2 2 0 5 B
7 2 3 2 1 B
8 3 0 3 1 A
9 3 1 5 0 A
10 3 2 0 0 A
11 3 3 0 0 A
My end goal is to ultimately reshape this into something with shape (number of ID, number of time points, number of sequences {2 here}).
It seems that if I pivot data, it fills in with nan values, which is fine for the signal values, but not the groups. I am also hoping to avoid looping through data.groupby('ID'), since my actual data has a large number of groups and the looping would likely be very slow.

Here's one approach creating the new index with pd.MultiIndex.from_product and using it to reindex on the Time column:
df = data.set_index(['ID', 'Time'])
# define a the new index
ix = pd.MultiIndex.from_product([df.index.levels[0],
df.index.levels[1]],
names=['ID', 'Time'])
# reindex using the above multiindex
df = df.reindex(ix, fill_value=0)
# forward fill the missing values in group
df['group'] = df.group.mask(df.group.eq(0)).ffill()
print(df.reset_index())
ID Time sig sig2 group
0 1 0 2 9 A
1 1 1 3 2 A
2 1 2 1 8 A
3 1 3 0 0 A
4 2 0 4 0 B
5 2 1 2 4 B
6 2 2 0 5 B
7 2 3 2 1 B
8 3 0 3 1 A
9 3 1 5 0 A
10 3 2 0 0 A
11 3 3 0 0 A

IIUC:
(data.pivot_table(columns='Time', index=['ID','group'], fill_value=0)
.stack('Time')
.sort_index(level=['ID','Time'])
.reset_index()
)
Output:
ID group Time sig sig2
0 1 A 0 2 9
1 1 A 1 3 2
2 1 A 2 1 8
3 1 A 3 0 0
4 2 B 0 4 0
5 2 B 1 2 4
6 2 B 2 0 5
7 2 B 3 2 1
8 3 A 0 3 1
9 3 A 1 5 0
10 3 A 2 0 0
11 3 A 3 0 0

splitting/grouping pandas dataframe column

I have a dataframe with a column populated with groups of 1s and 0s. How can I assign each group a consecutive number beginning from 1?
I have tried a for loop across rows, but I need a column operation for fast performance.
d = {'col1': [1,1,1,0,0,1,1,0,0,0,1,1]}
df1 = pd.DataFrame(data=d)
df1
col1
0 1
1 1
2 1
3 0
4 0
5 1
6 1
7 0
8 0
9 0
10 1
11 1
I need the following output:
col1 col2
0 1 1
1 1 1
2 1 1
3 0 2
4 0 2
5 1 3
6 1 3
7 0 4
8 0 4
9 0 4
10 1 5
11 1 5

You can compare shifted values for not equal and add cumulative sum by Series.cumsum:
df1['col2'] = df1['col1'].ne(df1['col1'].shift()).cumsum()
print (df1)
col1 col2
0 1 1
1 1 1
2 1 1
3 0 2
4 0 2
5 1 3
6 1 3
7 0 4
8 0 4
9 0 4
10 1 5
11 1 5

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