Here is the snippet:
test = pd.DataFrame({'uid':[1,1,2,2,3,3],
'start_time':[datetime(2017,7,20),datetime(2017,6,20),datetime(2017,5,20),datetime(2017,4,20),datetime(2017,3,20),datetime(2017,2,20)],
'amount': [10,11,12,13,14,15]})
Output:
amount start_time uid
0 10 2017-07-20 1
1 11 2017-06-20 1
2 12 2017-05-20 2
3 13 2017-04-20 2
4 14 2017-03-20 3
5 15 2017-02-20 3
Desired Output:
amount start_time uid
0 10 2017-07-20 1
2 12 2017-05-20 2
4 14 2017-03-20 3
I want to group by uid and mind the row with the latest start_time. Basically, I want to remove duplicate uid by only selecting the uid with the latest start_time.
I tried test.groupby(['uid'])['start_time'].max() but it doesn't work as it only returns back the uid and start_time column. I need the amount column as well.
Update: Thanks to #jezrael & #EdChum, you guys always help me out on this forum, thank you so much！
I tested both solutions in terms of execution time on a dataset of 1136 rows and 30 columns:
Method A: test.sort_values('start_time', ascending=False).drop_duplicates('uid')
Total execution time: 3.21 ms
Method B: test.loc[test.groupby('uid')['start_time'].idxmax()]
Total execution time: 65.1 ms
I guess groupby requires more time to compute.
Use idxmax to return the index of the latest time and use this to index the original df:
In[35]:
test.loc[test.groupby('uid')['start_time'].idxmax()]
Out[35]:
amount start_time uid
0 10 2017-07-20 1
2 12 2017-05-20 2
4 14 2017-03-20 3
Use sort_values by column start_time with drop_duplicates by uid:
df = test.sort_values('start_time', ascending=False).drop_duplicates('uid')
print (df)
amount start_time uid
0 10 2017-07-20 1
2 12 2017-05-20 2
4 14 2017-03-20 3
If need output with ordered uid:
print (test.sort_values('start_time', ascending=False)
.drop_duplicates('uid')
.sort_values('uid'))
Related
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)
)
We have a dataframe containing an 'ID' and 'DAY' columns, which shows when a specific customer made a complaint. We need to drop duplicates from the 'ID' column, but only if the duplicates happened 30 days apart, tops. Please see the example below:
Current Dataset:
ID DAY
0 1 22.03.2020
1 1 18.04.2020
2 2 10.05.2020
3 2 13.01.2020
4 3 30.03.2020
5 3 31.03.2020
6 3 24.02.2021
Goal:
ID DAY
0 1 22.03.2020
1 2 10.05.2020
2 2 13.01.2020
3 3 30.03.2020
4 3 24.02.2021
Any suggestions? I have tried groupby and then creating a loop to calculate the difference between each combination, but because the dataframe has millions of rows this would take forever...
You can compute the difference between successive dates per group and use it to form a mask to remove days that are less than 30 days apart:
df['DAY'] = pd.to_datetime(df['DAY'], dayfirst=True)
mask = (df
.sort_values(by=['ID', 'DAY'])
.groupby('ID')['DAY']
.diff().lt('30d')
.sort_index()
)
df[~mask]
NB. the potential drawback of this approach is that if the customer makes a new complaint within the 30days, this restarts the threshold for the next complaint
output:
ID DAY
0 1 2020-03-22
2 2 2020-10-05
3 2 2020-01-13
4 3 2020-03-30
6 3 2021-02-24
Thus another approach might be to resample the data per group to 30days:
(df
.groupby('ID')
.resample('30d', on='DAY').first()
.dropna()
.convert_dtypes()
.reset_index(drop=True)
)
output:
ID DAY
0 1 2020-03-22
1 2 2020-01-13
2 2 2020-10-05
3 3 2020-03-30
4 3 2021-02-24
You can try group by ID column and diff the DAY column in each group
df['DAY'] = pd.to_datetime(df['DAY'], dayfirst=True)
from datetime import timedelta
m = timedelta(days=30)
out = df.groupby('ID').apply(lambda group: group[~group['DAY'].diff().abs().le(m)]).reset_index(drop=True)
print(out)
ID DAY
0 1 2020-03-22
1 2 2020-05-10
2 2 2020-01-13
3 3 2020-03-30
4 3 2021-02-24
To convert to original date format, you can use dt.strftime
out['DAY'] = out['DAY'].dt.strftime('%d.%m.%Y')
print(out)
ID DAY
0 1 22.03.2020
1 2 10.05.2020
2 2 13.01.2020
3 3 30.03.2020
4 3 24.02.2021
(using sql or pandas)
I want to delete records if the Date difference between two records is less than 30 days.
But first record of ID must be remained.
#example
ROW ID DATE
1 A 2020-01-01 -- first
2 A 2020-01-03
3 A 2020-01-31
4 A 2020-02-05
5 A 2020-02-28
6 A 2020-03-09
7 B 2020-03-06 -- first
8 B 2020-05-07
9 B 2020-06-02
#expected results
ROW ID DATE
1 A 2020-01-01
4 A 2020-02-05
6 A 2020-03-09
7 B 2020-03-06
8 B 2020-05-07
ROW 2,3 are within 30 days from ROW 1
ROW 5 is within 30 days from ROW 4
ROW 9 is within 30 days from ROW 8
To cope with your task it is not possible to call any
vectorized methods.
The cause is that after a row is recognized as a duplicate, then
this row "does not count" when you check further rows.
E.g. after rows 2020-01-03 and 2020-01-31 were deleted (as
"too close" to the previous row) then 2020-02-05 row should be
left, because now the distance to the previous row (2020-01-01)
is big enough.
So I came up with a solution based on a "function with memory":
def isDupl(elem):
if isDupl.prev is None:
isDupl.prev = elem
return False
dDiff = (elem - isDupl.prev).days
rv = dDiff <= 30
if not rv:
isDupl.prev = elem
return rv
This function should be invoked for each DATE in the
current group (with same ID) but before that isDupl.prev
must be set to None.
So the function to apply to each group of rows is:
def isDuplGrp(grp):
isDupl.prev = None
return grp.DATE.apply(isDupl)
And to get the expected result, run:
df[~(df.groupby('ID').apply(isDuplGrp).reset_index(level=0, drop=True))]
(you may save it back to df).
The result is:
ROW ID DATE
0 1 A 2020-01-01
3 4 A 2020-02-05
5 6 A 2020-03-09
6 7 B 2020-03-06
7 8 B 2020-05-07
And finally, a remark about the other solution:
It contains rows:
3 4 A 2020-02-05
4 5 A 2020-02-28
which are only 23 days apart, so this solution is wrong.
The same pertains to rows:
5 A 2020-02-28
6 A 2020-03-09
which are also too close in time.
You can try this:
Convert date to datetime64
Get the first date from each group df.groupby('ID')['DATE'].transform('first')
Add a filter to keep only dates greater than 30 days
Append the first date of each group to the dataframe
Code:
df['DATE'] = pd.to_datetime(df['DATE'])
df1 = df[(df['DATE'] - df.groupby('ID')['DATE'].transform('first')) >= pd.Timedelta(30, unit='D')]
df1 = df1.append(df.groupby('ID', as_index=False).agg('first')).sort_values(by=['ID', 'DATE'])
print(df1)
ROW ID DATE
0 1 A 2020-01-01
2 3 A 2020-01-31
3 4 A 2020-02-05
4 5 A 2020-02-28
5 6 A 2020-03-09
1 7 B 2020-03-06
7 8 B 2020-05-07
8 9 B 2020-06-02
import pandas as pd
def nearest(items, pivot):
return min(items, key=lambda x: abs(x - pivot))
df = pd.read_csv("C:/Files/input.txt", dtype=str)
duplicatesDf = df[df.duplicated(subset=['CLASS_ID', 'START_TIME', 'TEACHER_ID'], keep=False)]
duplicatesDf['START_TIME'] = pd.to_datetime(duplicatesDf['START_TIME'], format='%Y/%m/%d %H:%M:%S.%f')
print duplicatesDf
print df['START_TIME'].dt.date
df:
ID,CLASS_ID,START_TIME,TEACHER_ID,END_TIME
1,123,2020/06/01 20:47:26.000,o1,2020/06/02 00:00:00.000
2,123,2020/06/01 20:47:26.000,o1,2020/06/04 20:47:26.000
3,789,2020/06/01 20:47:26.000,o3,2020/06/03 14:47:26.000
4,789,2020/06/01 20:47:26.000,o3,2020/06/03 14:40:00.000
5,456,2020/06/01 20:47:26.000,o5,2020/06/08 20:00:26.000
So, I've got a dataframe like mentioned above. As you can see, I have multiple records with the same CLASS_ID,START_DATE and TEACHER_ID. Whenever, multiple records like these are present, I would like to retain only 1 record based on the condition that, the retained record should have its END_DATE nearest to its START_DATE(by minute level precision).
In this case,
for CLASS_ID 123, the record with ID 1 will be retained, as its END_DATE 2020/06/02 00:00:00.000 is nearest to its START_DATE 2020/06/01 20:47:26.000 as compared to record with ID 2 whose END_DATE is 2020/06/04 20:47:26.000. Similarly for CLASS_ID 789, record with ID 4 will be retained.
Hence the expected output will be:
ID,CLASS_ID,START_TIME,TEACHER_ID,END_TIME
1,123,2020/06/01 20:47:26.000,o1,2020/06/02 00:00:00.000
4,789,2020/06/01 20:47:26.000,o3,2020/06/03 14:40:00.000
5,456,2020/06/01 20:47:26.000,o5,2020/06/08 20:00:26.000
I've been going through the following links,
https://stackoverflow.com/a/32237949,
https://stackoverflow.com/a/33043374
to find a solution but have unfortunately reached an impasse.
Hence, would some kind soul mind helping me out a bit. Many thanks.
IIUC, we can use .loc and idxmin() after creating a condtional column to measure the elapsed time between the start and the end, we will apply idxmin() as a groupby operation on your CLASS_ID column.
df.loc[
df.assign(mins=(df["END_TIME"] - df["START_TIME"]))
.groupby("CLASS_ID")["mins"]
.idxmin()
]
ID CLASS_ID START_TIME TEACHER_ID END_TIME
0 1 123 2020-06-01 20:47:26 o1 2020-06-02 00:00:00
4 5 456 2020-06-01 20:47:26 o5 2020-06-08 20:00:26
3 4 789 2020-06-01 20:47:26 o3 2020-06-03 14:40:00
in steps.
Time Delta.
print(df.assign(mins=(df["END_TIME"] - df["START_TIME"]))[['CLASS_ID','mins']])
CLASS_ID mins
0 123 0 days 03:12:34
1 123 3 days 00:00:00
2 789 1 days 18:00:00
3 789 1 days 17:52:34
4 456 6 days 23:13:00
minimum index from time delta column whilst grouping with CLASS_ID
print(df.assign(mins=(df["END_TIME"] - df["START_TIME"]) )
.groupby("CLASS_ID")["mins"]
.idxmin())
CLASS_ID
123 0
456 4
789 3
Name: mins, dtype: int64
ip app device os channel click_time is_attributed
0 83230 3 1 33 888 2017-11-06 14:32:21 0
1 17357 3 1 19 379 2017-11-06 14:33:34 0
2 35810 3 1 13 379 2017-11-06 14:34:12 0
3 45745 14 1 33 888 2017-11-06 14:34:52 0
4 161007 3 1 13 379 2017-11-06 14:35:08 0
Here is the dataframe and I want to add one column which represents the time (seconds) delta value between every specified condition.
For example, let's take os-channel as an identifier and the timedelta in line-3 (os=33&channel=888) should be the time gap that is from the record last seen os=33&channel=88, which can be found in line-0. So the timedelta should be the gap between 2017-11-06 14:34:52 and 2017-11-06 14:32:21. Is there is no os=33&channel=888 before, the outcome should be Nan.
So how can I realize this in pandas ?
Assuming click_time is already datetime
df.groupby([“os”, “channel”]).click_time.diff()
Create a new column
df.assign(click_diff=df.groupby([“os”, “channel”]).click_time.diff())