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Pandas: return the occurrences of the most frequent value for each group (possibly without apply)

Let's assume the input dataset:
test1 = [[0,7,50], [0,3,51], [0,3,45], [1,5,50],[1,0,50],[2,6,50]]
df_test = pd.DataFrame(test1, columns=['A','B','C'])
that corresponds to:
A B C
0 0 7 50
1 0 3 51
2 0 3 45
3 1 5 50
4 1 0 50
5 2 6 50
I would like to obtain the a dataset grouped by 'A', together with the most common value for 'B' in each group, and the occurrences of that value:
A most_freq freq
0 3 2
1 5 1
2 6 1
I can obtain the first 2 columns with:
grouped = df_test.groupby("A")
out_df = pd.DataFrame(index=grouped.groups.keys())
out_df['most_freq'] = df_test.groupby('A')['B'].apply(lambda x: x.value_counts().idxmax())
but I am having problems the last column.
Also: is there a faster way that doesn't involve 'apply'? This solution doesn't scale well with lager inputs (I also tried dask).
Thanks a lot!

Use SeriesGroupBy.value_counts which sorting by default, so then add DataFrame.drop_duplicates for top values after Series.reset_index:
df = (df_test.groupby('A')['B']
.value_counts()
.rename_axis(['A','most_freq'])
.reset_index(name='freq')
.drop_duplicates('A'))
print (df)
A most_freq freq
0 0 3 2
2 1 0 1
4 2 6 1

Trying to group by, then sort a dataframe based on multiple values [duplicate]

Suppose I have pandas DataFrame like this:
df = pd.DataFrame({'id':[1,1,1,2,2,2,2,3,4], 'value':[1,2,3,1,2,3,4,1,1]})
which looks like:
id value
0 1 1
1 1 2
2 1 3
3 2 1
4 2 2
5 2 3
6 2 4
7 3 1
8 4 1
I want to get a new DataFrame with top 2 records for each id, like this:
id value
0 1 1
1 1 2
3 2 1
4 2 2
7 3 1
8 4 1
I can do it with numbering records within group after groupby:
dfN = df.groupby('id').apply(lambda x:x['value'].reset_index()).reset_index()
which looks like:
id level_1 index value
0 1 0 0 1
1 1 1 1 2
2 1 2 2 3
3 2 0 3 1
4 2 1 4 2
5 2 2 5 3
6 2 3 6 4
7 3 0 7 1
8 4 0 8 1
then for the desired output:
dfN[dfN['level_1'] <= 1][['id', 'value']]
Output:
id value
0 1 1
1 1 2
3 2 1
4 2 2
7 3 1
8 4 1
But is there more effective/elegant approach to do this? And also is there more elegant approach to number records within each group (like SQL window function row_number()).

Did you try
df.groupby('id').head(2)
Output generated:
id value
id
1 0 1 1
1 1 2
2 3 2 1
4 2 2
3 7 3 1
4 8 4 1
(Keep in mind that you might need to order/sort before, depending on your data)
EDIT: As mentioned by the questioner, use
df.groupby('id').head(2).reset_index(drop=True)
to remove the MultiIndex and flatten the results:
id value
0 1 1
1 1 2
2 2 1
3 2 2
4 3 1
5 4 1

Since 0.14.1, you can now do nlargest and nsmallest on a groupby object:
In [23]: df.groupby('id')['value'].nlargest(2)
Out[23]:
id
1 2 3
1 2
2 6 4
5 3
3 7 1
4 8 1
dtype: int64
There's a slight weirdness that you get the original index in there as well, but this might be really useful depending on what your original index was.
If you're not interested in it, you can do .reset_index(level=1, drop=True) to get rid of it altogether.
(Note: From 0.17.1 you'll be able to do this on a DataFrameGroupBy too but for now it only works with Series and SeriesGroupBy.)

Sometimes sorting the whole data ahead is very time consuming.
We can groupby first and doing topk for each group:
g = df.groupby(['id']).apply(lambda x: x.nlargest(topk,['value'])).reset_index(drop=True)

df.groupby('id').apply(lambda x : x.sort_values(by = 'value', ascending = False).head(2).reset_index(drop = True))
Here sort values ascending false gives similar to nlargest and True gives similar to nsmallest.
The value inside the head is the same as the value we give inside nlargest to get the number of values to display for each group.
reset_index is optional and not necessary.

This works for duplicated values
If you have duplicated values in top-n values, and want only unique values, you can do like this:
import pandas as pd
ifile = "https://raw.githubusercontent.com/bhishanpdl/Shared/master/data/twitter_employee.tsv"
df = pd.read_csv(ifile,delimiter='\t')
print(df.query("department == 'Audit'")[['id','first_name','last_name','department','salary']])
id first_name last_name department salary
24 12 Shandler Bing Audit 110000
25 14 Jason Tom Audit 100000
26 16 Celine Anston Audit 100000
27 15 Michale Jackson Audit 70000
If we do not remove duplicates, for the audit department we get top 3 salaries as 110k,100k and 100k.
If we want to have not-duplicated salaries per each department, we can do this:
(df.groupby('department')['salary']
.apply(lambda ser: ser.drop_duplicates().nlargest(3))
.droplevel(level=1)
.sort_index()
.reset_index()
)
This gives
department salary
0 Audit 110000
1 Audit 100000
2 Audit 70000
3 Management 250000
4 Management 200000
5 Management 150000
6 Sales 220000
7 Sales 200000
8 Sales 150000

To get the first N rows of each group, another way is via groupby().nth[:N]. The outcome of this call is the same as groupby().head(N). For example, for the top-2 rows for each id, call:
N = 2
df1 = df.groupby('id', as_index=False).nth[:N]
To get the largest N values of each group, I suggest two approaches.
First sort by "id" and "value" (make sure to sort "id" in ascending order and "value" in descending order by using the ascending parameter appropriately) and then call groupby().nth[].
N = 2
df1 = df.sort_values(by=['id', 'value'], ascending=[True, False])
df1 = df1.groupby('id', as_index=False).nth[:N]
Another approach is to rank the values of each group and filter using these ranks.
# for the entire rows
N = 2
msk = df.groupby('id')['value'].rank(method='first', ascending=False) <= N
df1 = df[msk]
# for specific column rows
df1 = df.loc[msk, 'value']
Both of these are much faster than groupby().apply() and groupby().nlargest() calls as suggested in the other answers on here(1, 2, 3). On a sample with 100k rows and 8000 groups, a %timeit test showed that it was 24-150 times faster than those solutions.
Also, instead of slicing, you can also pass a list/tuple/range to a .nth() call:
df.groupby('id', as_index=False).nth([0,1])
# doesn't even have to be consecutive
# the following returns 1st and 3rd row of each id
df.groupby('id', as_index=False).nth([0,2])

How to find the maximum value of a column with pandas?

I have a table with 40 columns and 1500 rows. I want to find the maximum value among the 30-32nd (3 columns). How can it be done? I want to return the maximum value among these 3 columns and the index of dataframe.
print(Max_kVA_df.iloc[30:33].max())

hi you can refer this example
import pandas as pd
df=pd.DataFrame({'col1':[1,2,3,4,5],
'col2':[4,5,6,7,8],
'col3':[2,3,4,5,7]
})
print(df)
#print(df.iloc[:,0:3].max())# Mention range of the columns which you want, In your case change 0:3 to 30:33, here 33 will be excluded
ser=df.iloc[:,0:3].max()
print(ser.max())
Output
8

Select values by positions and use np.max:
Sample: for maximum by first 5 rows:
np.random.seed(123)
df = pd.DataFrame(np.random.randint(10, size=(10, 3)), columns=list('ABC'))
print (df)
A B C
0 2 2 6
1 1 3 9
2 6 1 0
3 1 9 0
4 0 9 3
print (df.iloc[0:5])
A B C
0 2 2 6
1 1 3 9
2 6 1 0
3 1 9 0
4 0 9 3
print (np.max(df.iloc[0:5].max()))
9

Or use iloc this way:
print(df.iloc[[30, 31], 2].max())

Finding min of values across multiple columns in pandas

I am trying to find min of values across columns in a pandas data frame where cols are ranged and split. For example, I have the dataframe in pandas as shown in the image.
I am iterating over the dataframe for more logic and would like to get the min of values in columns between T3:T6 and T11:T14 in separate variables.
Tried print(df.iloc[2,2:,2:4].min(axis=1))
I expect 9 and 13 for Row1 when I iterate.

create a simple dataframe:
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.randint(0,10,size=(10, 4)), columns=list('ABCD'))
A B C D
0 2 0 5 1
1 9 7 5 5
2 5 5 3 0
3 0 6 3 8
4 4 4 4 0
5 8 2 1 4
6 4 1 1 8
7 6 5 2 9
8 2 4 3 0
9 4 7 1 8
use the min() function:
df.min()
result:
A 0
B 0
C 1
D 0
and if you wish to select specific columns, use the loc:
df.loc[:,'B':'C'].min()
B 0
C 1
Bonus: Take pandas to another level - paint the minimum:
df.style.apply(lambda x: ['background-color : red; font-size: 16px' if v==x.min() else 'font-size: 16px' for _,v in enumerate(x) ],axis=0)

print(df[['T'+str(x) for x in range(3,7)]].min(axis=1)]
print(df[['T'+str(x) for x in range(11,15)]].min(axis=1)]
should print the mins for all the rows of t3, t4, t5,16 and t11, t12, t13,14 separately

For test dataframe:
df = pd.DataFrame({'A':[x for x in range(100)], 'B': [x for x in range(10,110)], 'C' : [x for x in range(20,120)] })
Create a function that can be applied to each row to find the minimum:
def test(row):
print(row[['A','B']].min())
Then use apply to run the function on each row:
df.apply(lambda row: test(row), axis=1)
This will print the minimum of whichever columns you put in the "test function"

Python pandas: Append rows of DataFrame and delete the appended rows

import pandas as pd
df = pd.DataFrame({
'id':[1,2,3,4,5,6,7,8,9,10,11],
'text': ['abc','zxc','qwe','asf','efe','ert','poi','wer','eer','poy','wqr']})
I have a DataFrame with columns:
id text
1 abc
2 zxc
3 qwe
4 asf
5 efe
6 ert
7 poi
8 wer
9 eer
10 poy
11 wqr
I have a list L = [1,3,6,10] which contains list of id's.
I am trying to append the text column using a list such that, from my list first taking 1 and 3(first two values in a list) and appending text column in my DataFrame with id = 1 which has id's 2, then deleting rows with id column 2 similarly then taking 3 and 6 and then appending text column where id = 4,5 to id 3 and then delete rows with id = 4 and 5 and iteratively for elements in list (x, x+1)
My final output would look like this:
id text
1 abczxc # joining id 1 and 2
3 qweasfefe # joining id 3,4 and 5
6 ertpoiwereer # joining id 6,7,8,9
10 poywqr # joining id 10 and 11

You can use isin with cumsum for Series, which is use for groupby with apply join function:
s = df.id.where(df.id.isin(L)).ffill().astype(int)
df1 = df.groupby(s)['text'].apply(''.join).reset_index()
print (df1)
id text
0 1 abczxc
1 3 qweasfefe
2 6 ertpoiwereer
3 10 poywqr
It working because:
s = df.id.where(df.id.isin(L)).ffill().astype(int)
print (s)
0 1
1 1
2 3
3 3
4 3
5 6
6 6
7 6
8 6
9 10
10 10
Name: id, dtype: int32

I changed the values not in list to np.nan and then ffill and groupby. Though #Jezrael's approach is much better. I need to remember to use cumsum:)
l = [1,3,6,10]
df.id[~df.id.isin(l)] = np.nan
df = df.ffill().groupby('id').sum()
text
id
1.0 abczxc
3.0 qweasfefe
6.0 ertpoiwereer
10.0 poywqr

Use pd.cut to create you bins then groupby with a lambda function to join your text in that group.
df.groupby(pd.cut(df.id,L+[np.inf],right=False, labels=[i for i in L])).apply(lambda x: ''.join(x.text))
EDIT:
(df.groupby(pd.cut(df.id,L+[np.inf],
right=False,
labels=[i for i in L]))
.apply(lambda x: ''.join(x.text)).reset_index().rename(columns={0:'text'}))
Output:
id text
0 1 abczxc
1 3 qweasfefe
2 6 ertpoiwereer
3 10 poywqr