I have two dataframes df1 and df2' :- df1` :-
Date
ID
total calls
24-02-2021
1
15
22-02-2021
1
25
20-02-2021
3
100
21-02-2021
4
30
df2:
Date
ID
total calls
match_flag
24-02-2021
1
16
1
22-02-2021
1
25
1
20-02-2021
3
99
1
24-02-2021
2
80
not_found
21-02-2021
4
25
0
I want to first match based on Id and Date if both matches I want to check for an addional condition of total calls and if the difference between total calls in df1 and df2 is +-1 then I want to consider that row as match and update the flag and if it does not satisfy the +-1 condition want to update the flag to 0 and if that date for the ID is not found in df1 then update to not_found
Updating the df1 and df2 matched on ID and DateId
df1:
ID
Call_Date
TId
StartTime
EndTime
total calls
Type
Indicator
DateId
562124
18-10-2021
480271
18-10-2021
18-10-2021
1
Regular Call
SA
20211018
df2 :
ID
total calls
DateId
Start_Time
End_Time
Indicator
Type
match_flag
562124
0
20211018
2021-10-18T13:06:00.000+0000
2021-10-18T13:07:00.000+0000
AD
R
not_found
You can use a merge:
s = df2.merge(df1, on=['Date', 'ID'], how='left')['total calls_y']
df2['match_flag'] = (df2['total calls']
.sub(s).abs().le(1) # is absolute diff ≤ 1?
.astype(int) # convert to int
.mask(s.isna(), 'not_found') # mask missing
)
output:
Date ID total calls match_flag
0 24-02-2021 1 16 1
1 22-02-2021 1 25 1
2 20-02-2021 3 99 1
3 24-02-2021 2 80 not_found
4 21-02-2021 4 25 0
Related
df = pd.DataFrame({'id': ['id1', 'id1','id1', 'id2','id1','id1','id1'],
'activity':['swimming','running','jogging','walking','walking','walking','walking'],
'month':[2,3,4,3,4,4,3]})
pd.crosstab(df['id'], df['activity'])
I'd like to add another column for month in the output to get counts per user within each month for the respective activity.
df.set_index(['id','month'])['activity'].unstack().reset_index()
I get error.
edit: Expected output in the image. I do not know how to create a table.
You can pass a list of columns to pd.crosstab:
x = pd.crosstab([df["id"], df["month"]], df["activity"]).reset_index()
x.columns.name = None
print(x)
Prints:
id month jogging running swimming walking
0 id1 2 0 0 1 0
1 id1 3 0 1 0 1
2 id1 4 1 0 0 2
3 id2 3 0 0 0 1
I have a dataframe that records users, a label, and the start and end date of them being labelled as such
e.g.
user
label
start_date
end_date
1
x
2018-01-01
2018-10-01
2
x
2019-05-10
2020-01-01
3
y
2019-04-01
2022-04-20
1
b
2018-10-01
2020-05-08
etc
where each row is for a given user and a label; a user appears multiple times for different labels
I want to get a count of users for every month for each label, such as this:
date
count_label_x
count_label_y
count_label_b
count_label_
2018-01
10
0
20
5
2018-02
2
5
15
3
2018-03
20
6
8
3
etc
where for instance for the first entry of the previous table, that user should be counted once for every month between his start and end date. The problem boils down to this and since I only have a few labels I can filter labels one by one and produce one output for each label. But how do I check and count users given an interval?
Thanks
You can use date_range combined with to_period to generate the active months, then pivot_table with aggfunc='nunique' to aggregate the unique user (if you want to count the duplicated users use aggfunc='count'):
out = (df
.assign(period=[pd.date_range(a, b, freq='M').to_period('M')
for a,b in zip(df['start_date'], df['end_date'])])
.explode('period')
.pivot_table(index='period', columns='label', values='user',
aggfunc='nunique', fill_value=0)
)
output:
label b x y
period
2018-01 0 1 0
2018-02 0 1 0
2018-03 0 1 0
2018-04 0 1 0
2018-05 0 1 0
...
2021-12 0 0 1
2022-01 0 0 1
2022-02 0 0 1
2022-03 0 0 1
handling NaT
if you have the same start/end and want to count the value:
out = (df
.assign(period=[pd.date_range(a, b, freq='M').to_period('M')
for a,b in zip(df['start_date'], df['end_date'])])
.explode('period')
.assign(period=lambda d: d['period'].fillna(d['start_date'].dt.to_period('M')))
.pivot_table(index='period', columns='label', values='user',
aggfunc='nunique', fill_value=0)
)
i have a df like this:
id
month
1
1
1
3
1
4
1
6
i want to transform it become like this:
id
1
2
3
4
5
6
1
1
0
1
1
0
1
ive tried using this code:
ndf = df[['id']].join(pd.get_dummies(
df['month'])).groupby('id').max()
but it shows like this:
id
1
3
4
6
1
1
1
1
1
how can i insert the middle column (2 and 5) even if it's not in the data?
You can use pd.crosstab
instead, then create new columns using pd.RangeIndex based on the min and max month, and finally use DataFrame.reindex (and optionally DataFrame.reset_index afterwards):
import pandas as pd
new_cols = pd.RangeIndex(df['month'].min(), df['month'].max())
res = (
pd.crosstab(df['id'], df['month'])
.reindex(columns=new_cols, fill_value=0)
.reset_index()
)
Output:
>>> res
id 1 2 3 4 5
0 1 1 0 1 1 0
I have a dataframe of id number and dates:
import pandas as pd
df = pd.DataFrame([['1','01/01/2000'], ['1','01/07/2002'],['1', '04/05/2003'],
['2','01/05/2010'], ['2','08/08/2009'],
['3','12/11/2008']], columns=['id','start_date'])
df
id start_date
0 1 01/01/2000
1 1 01/07/2002
2 1 04/05/2003
3 2 01/05/2010
4 2 08/08/2009
5 3 12/11/2008
I am looking for a way to leave for each id the first TWO dates (i.e. the two earliest dates).
for the example above the output would be:
id start_date
0 1 01/01/2000
1 1 01/07/2002
2 2 08/08/2009
3 2 01/05/2010
4 3 12/11/2008
Thanks!
ensure timestamp
df['start_date']=pd.to_datetime(df['start_date'])
sort values
df=df.sort_values(by=['id','start_date'])
group and select first 2 only
df_=df.groupby('id')['id','start_date'].head(2)
Just group by id and then you can call head. Be sure to sort your values first.
df = df.sort_values(['id', 'start_date'])
df.groupby('id').head(2)
full code:
df = pd.DataFrame([['1','01/01/2000'], ['1','01/07/2002'],['1', '04/05/2003'],
['2','01/05/2010'], ['2','08/08/2009'],
['3','12/11/2008']], columns=['id','start_date'])
# 1. convert 'start_time' column to datetime
df['start_date'] = pd.to_datetime(df['start_date'])
# 2. sort the dataframe ascending by 'start_time'
df.sort_values(by='start_date', ascending=True, inplace=True)
# 3. select only the first two occurances of each id
df.groupby('id').head(2)
output:
id start_date
0 1 2000-01-01
1 1 2002-01-07
5 3 2008-12-11
4 2 2009-08-08
3 2 2010-01-05
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])