This question already has an answer here:
How to reset cumsum after change in sign of values?
(1 answer)
Closed 4 years ago.
I have a dataframe with 2 columns, the objective here is simple ; reset the df.cumsum() if a row column is set to False;
df
value condition
0 1 1
1 2 1
2 3 1
3 4 0
4 5 1
the wanted result is as follows :
df
value condition
0 1 1
1 3 1
2 6 1
3 4 0
4 9 1
If i loop over the dataframe as described in this post Python pandas cumsum() reset after hitting max
i can achieve the wanted results, but i was looking for a more vectorized way using pandas standard functions
How about:
df['cSum'] = df.groupby((df.condition == 0).cumsum()).value.cumsum()
Output:
value condition cSum
0 1 1 1
1 2 1 3
2 3 1 6
3 4 0 4
4 5 1 9
You'll group consecutive rows together until you encounter a 0 in the condition column, and then you apply the cumsum within each group separately.
Related
This question already has answers here:
Pandas - Interleave / Zip two DataFrames by row
(5 answers)
Closed 20 days ago.
This post was edited and submitted for review 20 days ago.
Suppose we have:
>>> df1
A B
0 1 a
1 2 a
2 3 a
3 4 a
>>> df2
A B
0 1 b
1 2 b
2 3 b
3 5 b
I would like to merge them on "A" and then list them by interleaving rows like:
A B
0 1 a
0 1 b
1 2 a
1 2 b
2 3 a
2 3 b
I tried merge but it list them column by column. For example if I have 3 or more data frames, merge can merge them on some columns, but my problem would be then to interleave them
If need match by A filter rows by Series.isin in boolean indexing, pass to concat with DataFrame.sort_index:
df = pd.concat([df1[df1.A.isin(df2.A)],
df2[df2.A.isin(df1.A)]]).sort_index(kind='stable')
print (df)
A B
0 1 a
0 1 b
1 2 a
1 2 b
2 3 a
2 3 b
EDIT:
For general data is possible sorting by A and create default index for correct interleaving:
df = (pd.concat([df1[df1.A.isin(df2.A)].sort_values('A', kind='stable').reset_index(drop=True),
df2[df2.A.isin(df1.A)].sort_values('A', kind='stable').reset_index(drop=True)])
.sort_index(kind='stable'))
I have the following dataframe:
d = {'value': [1,1,1,1,1,1,1,1,1,1], 'flag_1': [0,1,0,1,1,1,0,1,1,1],'flag_2':[1,0,1,1,1,1,1,0,1,1],'index':[1,2,3,4,5,6,7,8,9,10]}
df = pd.DataFrame(data=d)
I need to perform the following filter on it:
If flag 1 and flag 2 are equal keep the row with the maximum index from the consecutive indices. Below for rows 4,5,6 and rows 9,10 flag 1 and flag 2 are equal. From the group of consecutive indices 4,5,6 therefore I wish to keep only row 6 and drop rows 4 and 5. For the next group of rows 9 and 10 I wish to keep only row 10. The rows where flag 1 and 2 are not equal should all be retained. I want my final output to look as shown below:
I am really not sure how to achieve what is required so I would be grateful for any advice on how to do it.
IIUC, you can compare consecutive rows with shift. This solution requires a sorted index.
In [5]: df[~df[['flag_1', 'flag_2']].eq(df[['flag_1', 'flag_2']].shift(-1)).all(axis=1)]
Out[5]:
value flag_1 flag_2 index
0 1 0 1 1
1 1 1 0 2
2 1 0 1 3
5 1 1 1 6
6 1 0 1 7
7 1 1 0 8
9 1 1 1 10
Suppose we have the following df
Id PlaceCod Val
1 1 0
1 2 3
2 2 4
2 1 5
3 1 6
How can I convert this DF to this one:
Id Store Warehouse
1 0 3
2 5 4
3 6 null
I've tried to use df.pivot(f.col("PlaceCod")) but got error message 'DataFrame has no pivot attribute'
As posted by #Emma on the comments:
df.groupby('Id').pivot('PlaceCod').agg(F.first('Val'))
Using the above solution my problem was solved!
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')
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])