We Keep Coding

how to extract 'day' of each week from dataframe according to specific condition [duplicate]

I'm using groupby on a pandas dataframe to drop all rows that don't have the minimum of a specific column. Something like this:
df1 = df.groupby("item", as_index=False)["diff"].min()
However, if I have more than those two columns, the other columns (e.g. otherstuff in my example) get dropped. Can I keep those columns using groupby, or am I going to have to find a different way to drop the rows?
My data looks like:
item diff otherstuff
0 1 2 1
1 1 1 2
2 1 3 7
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
and should end up like:
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
but what I'm getting is:
item diff
0 1 1
1 2 -6
2 3 0
I've been looking through the documentation and can't find anything. I tried:
df1 = df.groupby(["item", "otherstuff"], as_index=false)["diff"].min()
df1 = df.groupby("item", as_index=false)["diff"].min()["otherstuff"]
df1 = df.groupby("item", as_index=false)["otherstuff", "diff"].min()
But none of those work (I realized with the last one that the syntax is meant for aggregating after a group is created).

Method #1: use idxmin() to get the indices of the elements of minimum diff, and then select those:
>>> df.loc[df.groupby("item")["diff"].idxmin()]
item diff otherstuff
1 1 1 2
6 2 -6 2
7 3 0 0
[3 rows x 3 columns]
Method #2: sort by diff, and then take the first element in each item group:
>>> df.sort_values("diff").groupby("item", as_index=False).first()
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
[3 rows x 3 columns]
Note that the resulting indices are different even though the row content is the same.

You can use DataFrame.sort_values with DataFrame.drop_duplicates:
df = df.sort_values(by='diff').drop_duplicates(subset='item')
print (df)
item diff otherstuff
6 2 -6 2
7 3 0 0
1 1 1 2
If possible multiple minimal values per groups and want all min rows use boolean indexing with transform for minimal values per groups:
print (df)
item diff otherstuff
0 1 2 1
1 1 1 2 <-multiple min
2 1 1 7 <-multiple min
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
print (df.groupby("item")["diff"].transform('min'))
0 1
1 1
2 1
3 -6
4 -6
5 -6
6 -6
7 0
8 0
Name: diff, dtype: int64
df = df[df.groupby("item")["diff"].transform('min') == df['diff']]
print (df)
item diff otherstuff
1 1 1 2
2 1 1 7
6 2 -6 2
7 3 0 0

The above answer worked great if there is / you want one min. In my case there could be multiple mins and I wanted all rows equal to min which .idxmin() doesn't give you. This worked
def filter_group(dfg, col):
return dfg[dfg[col] == dfg[col].min()]
df = pd.DataFrame({'g': ['a'] * 6 + ['b'] * 6, 'v1': (list(range(3)) + list(range(3))) * 2, 'v2': range(12)})
df.groupby('g',group_keys=False).apply(lambda x: filter_group(x,'v1'))
As an aside, .filter() is also relevant to this question but didn't work for me.

I tried everyone's method and I couldn't get it to work properly. Instead I did the process step-by-step and ended up with the correct result.
df.sort_values(by='item', inplace=True, ignore_index=True)
df.drop_duplicates(subset='diff', inplace=True, ignore_index=True)
df.sort_values(by=['diff'], inplace=True, ignore_index=True)
For a little more explanation:
Sort items by the minimum value you want
Drop the duplicates of the column you want to sort with
Resort the data because the data is still sorted by the minimum values

If you know that all of your "items" have more than one record you can sort, then use duplicated:
df.sort_values(by='diff').duplicated(subset='item', keep='first')

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])

Count the number of events in Python

In continuation to my previous Question I need some more help.
The dataframe is like
time eve_id sub_id flag
0 5 2 0
1 5 2 0
2 5 2 1
3 5 2 1
4 5 2 0
5 4 25 0
6 4 30 0
7 5 2 1
I need to count the eve_id in the time flag goes 0 to 1,
and count the eve_id for the time flag is 1 to 1
the output will look like this
time flag count
0 0 2
2 1 2
4 0 3
Can someone help me here ?

First we make a grouper indicator which checks if the difference between two rows is not equal to 0, which indicates a difference.
Then we groupby on this indicator and use agg. Since pandas 0.25.0 we have named aggregations:
s = df['flag'].diff().ne(0).cumsum()
grpd = df.groupby(s).agg(time=('time', 'first'),
flag=('flag', 'first'),
count=('flag', 'size')).reset_index(drop=True)
Output
time flag count
0 0 0 2
1 2 1 2
2 4 0 3
3 7 1 1
If time is your index, use:
grpd = df.assign(time=df.index).groupby(s).agg(time=('time', 'first'),
flag=('flag', 'first'),
count=('flag', 'size')).reset_index(drop=True)
notice: the row extra is because there's a difference between the last row and the row before as well

Change aggregate function sum to GroupBy.size:
df1 = (df.groupby([df['flag'].ne(df['flag'].shift()).cumsum(), 'flag'])
.size()
.reset_index(level=0, drop=True)
.reset_index(name='count'))
print (df1)
flag count
0 0 2
1 1 2
2 0 3
3 1 1

How to change the string in data frame to integer ID with pandas fast?

I have a data set including the user ID, item ID(both string) and rating like that:
A12VH45Q3H5R5I B000NWJTKW 5.0
A3J8AQWNNI3WSN B000NWJTKW 4.0
A1XOBWIL4MILVM B000NWJTKW 1.0
I'd like to change the IDs to integer like:
1 1 5.0
2 1 4.0
3 1 1.0
I have tried a traditional way, creating a big dictionary and mark each string ID with an integer one. But it took extremely long time. So could you please tell me how to finish it in a more fast way? Thanks in advance.

You can apply factorize:
In [244]:
df[[0,1]] = df[[0,1]].apply(lambda x: pd.factorize(x)[0] + 1)
df
Out[244]:
0 1 2
0 1 1 5
1 2 1 4
2 3 1 1

You could also encode the column as a categorical and then get the codes.
df['User_ID_code'] = df.User_ID.astype('category').cat.codes
>>> df
User_ID Item_ID Rating User_ID_code
0 A12VH45Q3H5R5I B000NWJTKW 5 0
1 A3J8AQWNNI3WSN B000NWJTKW 4 2
2 A1XOBWIL4MILVM B000NWJTKW 1 1

Python pandas join on with overwrite

I realize this question is similar to join or merge with overwrite in pandas, but the accepted answer does not work for me since I want to use the on='keys' from df.join().
I have a DataFrame df which looks like this:
keys values
0 0 0.088344
1 0 0.088344
2 0 0.088344
3 0 0.088344
4 0 0.088344
5 1 0.560857
6 1 0.560857
7 1 0.560857
8 2 0.978736
9 2 0.978736
10 2 0.978736
11 2 0.978736
12 2 0.978736
13 2 0.978736
14 2 0.978736
Then I have a Series s (which is a result from some df.groupy.apply()) with the same keys:
keys
0 0.183328
1 0.239322
2 0.574962
Name: new_values, dtype: float64
Basically I want to replace the 'values' in the df with the values in the Series, by keys so every keys block gets the same new value. Currently, I do it as follows:
df = df.join(s, on='keys')
df['values'] = df['new_values']
df = df.drop('new_values', axis=1)
The obtained (and desired) result is then:
keys values
0 0 0.183328
1 0 0.183328
2 0 0.183328
3 0 0.183328
4 0 0.183328
5 1 0.239322
6 1 0.239322
7 1 0.239322
8 2 0.574962
9 2 0.574962
10 2 0.574962
11 2 0.574962
12 2 0.574962
13 2 0.574962
14 2 0.574962
That is, I add it as a new column and by using on='keys' it gets the corrects shape. Then I assign values to be new_values and remove the new_values column. This of course works perfectly, the only problem being that I find it extremely ugly.
Is there a better way to do this?

You could try something like:
df = df[df.columns[df.columns!='values']].join(s, on='keys')
Make sure s is named 'values' instead of 'new_values'.
To my knowledge, pandas doesn't have the ability to join with "force overwrite" or "overwrite with warning".