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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])

Group by multi-index without losing an index with pandas in python

I currently have a dataset with two indexes, year and zip code, but multiple observations (prices) per zip code. How can I get the average price per zip code, so that I only have distinct observations per zip code and year.
Screenshot of current table

Use DataFrame.mean with level parameter:
df = s.mean(level=[0,1])
Sample:
s = pd.DataFrame({
'B':[5,5,4,5,5,4],
'E':[5,3,6,9,2,4],
'F':list('aaabbb')
}).set_index(['F','B'])['E']
print (s)
F B
a 5 5
5 3
4 6
b 5 9
5 2
4 4
Name: E, dtype: int64
df = s.mean(level=[0,1]).reset_index()
print (df)
F B E
0 a 5 4.0
1 a 4 6.0
2 b 5 5.5
3 b 4 4.0

Dataframe selecting Max for a column but output values of another

I have a dataframe with values similar to below
A10d B10d C10d A B C Strategy
20 10 5 3 5 1 3
The Strategy selects the max of A10d, B10d, C10d and return the value of A,B,C
In this case A10d is the largest and Strategy returns A, value of 3
I am not sure how to create this Strategy column properly, can anyone advise please? Thank you very much for your help

I think you need iloc for select first columns by positions and then get columns names by max with idxmax and replace 10d by whitespace for match columns. Last create new column by lookup:
print (df)
A10d B10d C10d A B C
0 20 10 5 3 5 1
1 20 100 5 3 5 1
df1 = df.iloc[:,:3]
print (df1)
A10d B10d C10d
0 20 10 5
1 20 100 5
s = df1.idxmax(axis=1).str.replace('10d','')
print (s)
0 A
1 B
dtype: object
df['Strategy'] = df.lookup(df.index, s)
print (df)
A10d B10d C10d A B C Strategy
0 20 10 5 3 5 1 3
1 20 100 5 3 5 1 5

Python pandas constructing dataframe by looping over columns

I am trying to develop a new panda dataframe based on data I got from an existing dataframe and then taking into account the previously calculated value in the new dataframe.
As an example, here are two dataframes with the same size.
df1 = pd.DataFrame(np.random.randint(0,10, size = (5, 4)), columns=['1', '2', '3', '4'])
df2 = pd.DataFrame(np.zeros(df1.shape), index=df1.index, columns=df1.columns)
Then I created a list which starts as a starting basis for my second dataframe df2
L = [2,5,6,7]
df2.loc[0] = L
Then for the remaining rows of df2 I want to take the value from the previous time step (df2) and add the value of df1.
for i in df2.loc[1:]:
df2.ix[i] = df2.ix[i-1] + df1
As an example my dataframes should look like this:
>>> df1
1 2 3 4
0 4 6 0 6
1 7 0 7 9
2 9 1 9 9
3 5 2 3 6
4 0 3 2 9
>>> df2
1 2 3 4
0 2 5 6 7
1 9 5 13 16
2 18 6 22 25
3 23 8 25 31
4 23 11 27 40
I know there is something wrong with the indication of indexes in the for loop but I cannot figure out how the argument must be formulated. I would be very thankful for any help on this.

this is a simple cumsum.
df2 = df1.copy()
df2.loc[0] = [2,5,6,7]
desired_df = df2.cumsum()

How to find the top 10 pairs that appear the most in a pandas dataframe in python

I have a pandas dataframe in python with columns 'a', 'b', 'c'. The 'a','b' pairs are unique and repeat multiple times. 'c' is changing all the time. I want to find the 10 pairs 'a','b' that appear the most and put them in a dataframe but don't know how. Any help is appreciated.

I'm not entirely sure I follow you, but assuming you mean you have a DataFrame looking something like
>>> N = 1000
>>> df = pd.DataFrame(np.random.randint(0, 10, (N, 3)), columns="A B C".split())
>>> df.head()
A B C
0 7 4 5
1 5 1 3
2 8 9 8
3 2 3 0
4 2 3 0
and you simply want to count (A, B) pairs, that's easy enough:
>>> df.groupby(["A", "B"]).size().order().iloc[-10:]
A B
6 1 13
1 0 14
4 0 14
7 2 14
1 6 15
8 2 15
1 8 16
2 6 16
6 4 16
7 4 16
dtype: int64
That can be broken down into four parts:
groupby, which groups the data by (A, B) tuples
size, which computes the size of each group
order, which returns the Series sorted by value
iloc, which lets us select the last 10 entries in the Series by position
That results in a Series, but you could make a DataFrame out of it simply by passing it to pd.DataFrame.