I have a pandas dataframe with id and date as the 2 columns - the date column has all the way to seconds.
data = {'id':[17,17,17,17,17,18,18,18,18],'date':['2018-01-16','2018-01-26','2018-01-27','2018-02-11',
'2018-03-14','2018-01-28','2018-02-12','2018-02-25','2018-03-04'],
}
df1 = pd.DataFrame(data)
I would like to have a new column - (tslt) - 'time_since_last_transaction'. The first transaction for each unique user_id could be a number say 1. Each subsequent transaction for that user should measure the difference between the 1st time stamp for that user and its current time stamp to generate a time difference in seconds.
I used the datetime and timedelta etc. but did not have too much of luck. Any help would be appreciated.
You can try groupby().transform():
df1['date'] = pd.to_datetime(df1['date'])
df1['diff'] = df1['date'].sub(df1.groupby('id').date.transform('min')).dt.total_seconds()
Output:
id date diff
0 17 2018-01-16 0.0
1 17 2018-01-26 864000.0
2 17 2018-01-27 950400.0
3 17 2018-02-11 2246400.0
4 17 2018-03-14 4924800.0
5 18 2018-01-28 0.0
6 18 2018-02-12 1296000.0
7 18 2018-02-25 2419200.0
8 18 2018-03-04 3024000.0
Related
So I have two data frames. The first data frame contains numerical data that is used to "score" the second data frame which contains simulation data.
df1 = base records
df2 = simulation records
Part 1: What I am trying to accomplish is to query df1 'base records' to find the row that has the most recent timestamp to that in the df2 'simulation records' where the "Name" & "Time" columns match exactly.
Part 2: Then I want to use an if then function to determine whether a value in the simulation record row fall between a range created using two values from the base record row and return a boolean.
low range = df1['Po']-df1['Ref']
high range = df1['Po']+df1['Ref']
if df2['Sim'] falls in between the low range & high range of its most recent df1 base record then I want to return true in the new column "Sim Score"
otherwise return false
Part 3: I want to repeat Part 1 & Part 2 for each row in the simulation records.
helpful information:
df1 (base records) have more or less rows than df2 (simulation records)
df1 has more columns than df2
some columns in df1 have the same name but different values in df2
ideally want to be able to slice both dataframes where the if then function only sees the two rows used in the comparison
only need the most recent df1 base record to compare to the df2 simulation record
previously accomplished this in google sheets with if then & query combination formula dragged down entire sheet (want to replace with python & pandas)
df1 base records example (columns that matter)
Timestamp Name Time Po Ref
7/11/2022 11:30:00 trial 20 mins 5 2
7/10/2022 04:00:00 trial 20 mins 4 4
7/09/2022 02:45:00 trial 20 mins 2 2
6/28/2022 03:45:00 trial 20 mins 3 6
df2 simulation records example (columns that matter)
Timestamp Name Time Sim
7/10/2022 05:15:00 trial 20 mins 7
7/11/2022 12:45:00 trial 20 mins 4
7/12/2022 03:30:00 trial 20 mins 8
desired result of new column added to df2
Timestamp Name Time Sim Sim Score
7/10/2022 05:15:00 trial 20 mins 7 True
7/11/2022 12:45:00 trial 20 mins 4 True
7/12/2022 03:30:00 trial 20 mins 8 False
Use pandas.DataFrame.reindex, its method offer nearest to find the computable index(e.g., string can not calculate distance)
Or use merge_asof, its direction offer nearest.
Method 1:
reindex() with method='nearest'
df1['Timestamp'] = pd.to_datetime(df1['Timestamp'])
df1.set_index('Timestamp', inplace=True)
df1['l_r'] = df1['Po'] - df1['Ref']
df1['h_r'] = df1['Po'] + df1['Ref']
print(df1)
###
Name Time Po Ref l_r h_r
Timestamp
2022-07-11 11:30:00 trial 20 mins 5 2 3 7
2022-07-10 04:00:00 trial 20 mins 4 4 0 8
2022-07-09 02:45:00 trial 20 mins 2 2 0 4
2022-06-28 03:45:00 trial 20 mins 3 6 -3 9
df2['Timestamp'] = pd.to_datetime(df2['Timestamp'])
df2.set_index('Timestamp', inplace=True)
print(df2)
###
Name Time Sim
Timestamp
2022-07-10 05:15:00 trial 20 mins 7
2022-07-11 12:45:00 trial 20 mins 4
2022-07-12 03:30:00 trial 20 mins 8
temp = df2.join(df1.reindex(df2.index, method='nearest'), lsuffix='_left', rsuffix='_right')
print(temp)
As you can see, this is df2.join(df1),
join multiple DataFrame objects by index at once.
with method='nearest', in this case, it would join df2 and df1 by the nearest Timestamp index.
df2['Sim Score'] = temp['Sim'].between(temp['l_r'], temp['h_r']).values
df2.reset_index(inplace=True)
print(df2)
###
Timestamp Name Time Sim Sim Score
0 2022-07-10 05:15:00 trial 20 mins 7 True
1 2022-07-11 12:45:00 trial 20 mins 4 True
2 2022-07-12 03:30:00 trial 20 mins 8 False
Method 2:
merge_asof() with direction='nearest'
this way is not executed with indexed value, therefore we don't have to set column Timestamp as index. But it needs binding objects(in this case we merge on column Timestamp)sorted.
df1['Timestamp'] = pd.to_datetime(df1['Timestamp'])
# df1.set_index('Timestamp', inplace=True)
df1['l_r'] = df1['Po'] - df1['Ref']
df1['h_r'] = df1['Po'] + df1['Ref']
df1.sort_values(by='Timestamp', inplace=True)
print(df1)
###
Timestamp Name Time Po Ref l_r h_r
3 2022-06-28 03:45:00 trial 20 mins 3 6 -3 9
2 2022-07-09 02:45:00 trial 20 mins 2 2 0 4
1 2022-07-10 04:00:00 trial 20 mins 4 4 0 8
0 2022-07-11 11:30:00 trial 20 mins 5 2 3 7
df2['Timestamp'] = pd.to_datetime(df2['Timestamp'])
# df2.set_index('Timestamp', inplace=True)
df2.sort_values(by='Timestamp', inplace=True)
print(df2)
###
Timestamp Name Time Sim
0 2022-07-10 05:15:00 trial 20 mins 7
1 2022-07-11 12:45:00 trial 20 mins 4
2 2022-07-12 03:30:00 trial 20 mins 8
temp = pd.merge_asof(df2 ,df1[['Timestamp', 'l_r', 'h_r']], on='Timestamp', direction='nearest')
print(temp)
As you can see, this is pd.merge_asof(df2, df1),
This is similar to a left-join except that we match on nearest key rather than equal keys. Both DataFrames must be sorted by the key.
For each row in the left DataFrame:
A “nearest” search selects the row in the right DataFrame whose ‘on’ key is closest in absolute distance to the left’s key.
df2['Sim Score'] = temp['Sim'].between(temp['l_r'], temp['h_r']).values
print(df2)
###
Timestamp Name Time Sim Sim Score
0 2022-07-10 05:15:00 trial 20 mins 7 True
1 2022-07-11 12:45:00 trial 20 mins 4 True
2 2022-07-12 03:30:00 trial 20 mins 8 False
Frankly speaking, working on indexed things would be faster if you have a large dataset.
Method 2 (on multiple keys)
I remodified df1 adding different Name and Time
df1 = pd.DataFrame({'Timestamp':['7/11/2022 11:30:00','7/11/2022 11:30:00','7/10/2022 04:00:00','7/10/2022 04:00:00','7/09/2022 02:45:00','6/28/2022 03:45:00'],
'Name':['trial','trial','trial','non-trial','trial','trial'],
'Time':['20 mins','30 mins','20 mins','20 mins','20 mins','20 mins'],
'Po':[5, 6, 4, 1, 2, 3],
'Ref':[2, 2, 4, 3, 2, 6]})
df1['Timestamp'] = pd.to_datetime(df1['Timestamp'])
df1['l_r'] = df1['Po'] - df1['Ref']
df1['h_r'] = df1['Po'] + df1['Ref']
df1.sort_values(by='Timestamp', inplace=True)
print(df1)
###
Timestamp Name Time Po Ref l_r h_r
5 2022-06-28 03:45:00 trial 20 mins 3 6 -3 9
4 2022-07-09 02:45:00 trial 20 mins 2 2 0 4
2 2022-07-10 04:00:00 trial 20 mins 4 4 0 8
3 2022-07-10 04:00:00 non-trial 20 mins 1 3 -2 4
0 2022-07-11 11:30:00 trial 20 mins 5 2 3 7
1 2022-07-11 11:30:00 trial 30 mins 6 2 4 8
print(df2)
###
Timestamp Name Time Sim
0 2022-07-10 05:15:00 trial 20 mins 7
1 2022-07-11 12:45:00 trial 20 mins 4
2 2022-07-12 03:30:00 trial 20 mins 8
Important:
can only merge_asof on a single key, therefore others would utilize by= to deal with.
temp = pd.merge_asof(df2, df1[['Timestamp', 'Name', 'Time', 'l_r', 'h_r']], on='Timestamp', by=['Name','Time'], direction='nearest')
print(temp)
df2['Sim Score'] = temp['Sim'].between(temp['l_r'], temp['h_r']).values
print(df2)
###
Timestamp Name Time Sim Sim Score
0 2022-07-10 05:15:00 trial 20 mins 7 True
1 2022-07-11 12:45:00 trial 20 mins 4 True
2 2022-07-12 03:30:00 trial 20 mins 8 False
Reference:
pandas.DataFrame.join
pandas.merge_asof
merging/join concept
Because you don't provide code to construct the dataframe, I will sketch a potential solution:
First, I will assume that you have only one timestamp per day (which it looks like in your examples). Accordingly, I would truncate or split the timestamp to only have the date in one column. This is done so we can join the dataframes based on the date, i.e. use set_index("date_column") for both dataframes (use an inner-join to only keep the rows where the date was present in both dataframes). Finally, you can use apply() to check your condition:
df_joined['Sim Score'] = df_joined.apply(lambda row: (row['Po']-row['Ref'] <= row['Sim']) and (row['Po']+row['Ref'] >= row['Sim']), axis = 1)
You can do it via pandasql:
But note that you better add a unique constraint to one of the columns (e.g. a number of trial)
from pandasql import sqldf
df3 = sqldf('''
SELECT df2.Timestamp AS Date, df1.Name, df1.Time, df2.Sim,
CASE
WHEN Sim >= (df1.Po - df1.Ref) AND Sim <= (df1.Po + df1.Ref) THEN 'True'
WHEN Sim < (df1.Po - df1.Ref) OR Sim > (df1.Po + df1.Ref) THEN 'False'
END AS 'Sim Score'
FROM df1, df2
WHERE df2.Name == df1.Name AND df2.Time == df1.Time
ORDER BY Date ASC
''')
Also to work with datetime format in sqldf you need to name your Timestamp column as date in the query. If you need to get only let's say first/earliest 5 results add LIMIT 5 in the end of the query.
If you need to get closest date in df2 to df1 try this:
from pandasql import sqldf
df3 = sqldf('''
SELECT df2.Timestamp AS Date1, df2.Timestamp AS Date2,
df1.Name, df1.Time, df2.Sim,
CASE
WHEN Sim >= (df1.Po - df1.Ref) AND Sim <= (df1.Po + df1.Ref) THEN 'True'
WHEN Sim < (df1.Po - df1.Ref) OR Sim > (df1.Po + df1.Ref) THEN 'False'
END AS 'Sim Score'
FROM df1, df2
WHERE df2.Name == df1.Name AND df2.Time == df1.Time
and Date1 <= Date2
group by Date2
ORDER BY Date1 ASC
''')
I'm creating a pandas DataFrame with random dates and random integers values and I want to resample it by month and compute the average value of integers. This can be done with the following code:
def random_dates(start='2018-01-01', end='2019-01-01', n=300):
start_u = start.value//10**9
end_u = end.value//10**9
return pd.to_datetime(np.random.randint(start_u, end_u, n), unit='s')
start = pd.to_datetime('2018-01-01')
end = pd.to_datetime('2019-01-01')
dates = random_dates(start, end)
ints = np.random.randint(100, size=300)
df = pd.DataFrame({'Month': dates, 'Integers': ints})
print(df.resample('M', on='Month').mean())
The thing is that the resampled months always starts from day one and I want all months to start from day 15. I'm using pandas 1.1.4 and I've tried using origin='15/01/2018' or offset='15' and none of them works with 'M' resample rule (they do work when I use 30D but it is of no use). I've also tried to use '2SM'but it also doesn't work.
So my question is if is there a way of changing the resample rule or I will have to add an offset in my data?
Assume that the source DataFrame is:
Month Amount
0 2020-05-05 1
1 2020-05-14 1
2 2020-05-15 10
3 2020-05-20 10
4 2020-05-30 10
5 2020-06-15 20
6 2020-06-20 20
To compute your "shifted" resample, first shift Month column so that
the 15-th day of month becomes the 1-st:
df.Month = df.Month - pd.Timedelta('14D')
and then resample:
res = df.resample('M', on='Month').mean()
The result is:
Amount
Month
2020-04-30 1
2020-05-31 10
2020-06-30 20
If you want, change dates in the index to month periods:
res.index = res.index.to_period('M')
Then the result will be:
Amount
Month
2020-04 1
2020-05 10
2020-06 20
Edit: Not a working solution for OP's request. See short discussion in the comments.
Interesting problem. I suggest to resample using 'SMS' - semi-month start frequency (1st and 15th). Instead of keeping just the mean values, keep the count and sum values and recalculate the weighted mean for each monthly period by its two sub-period (for example: 15/1 to 15/2 is composed of 15/1-31/1 and 1/2-15/2).
The advantages here is that unlike with an (improper use of an) offset, we are certain we always start on the 15th of the month till the 14th of the next month.
df_sm = df.resample('SMS', on='Month').aggregate(['sum', 'count'])
df_sm
Integers
sum count
Month
2018-01-01 876 16
2018-01-15 864 16
2018-02-01 412 10
2018-02-15 626 12
...
2018-12-01 492 10
2018-12-15 638 16
Rolling sum and rolling count; Find the mean out of them:
df_sm['sum_rolling'] = df_sm['Integers']['sum'].rolling(2).sum()
df_sm['count_rolling'] = df_sm['Integers']['count'].rolling(2).sum()
df_sm['mean'] = df_sm['sum_rolling'] / df_sm['count_rolling']
df_sm
Integers count_sum count_rolling mean
sum count
Month
2018-01-01 876 16 NaN NaN NaN
2018-01-15 864 16 1740.0 32.0 54.375000
2018-02-01 412 10 1276.0 26.0 49.076923
2018-02-15 626 12 1038.0 22.0 47.181818
...
2018-12-01 492 10 1556.0 27.0 57.629630
2018-12-15 638 16 1130.0 26.0 43.461538
Now, just filter the odd indices of df_sm:
df_sm.iloc[1::2]['mean']
Month
2018-01-15 54.375000
2018-02-15 47.181818
2018-03-15 51.000000
2018-04-15 44.897436
2018-05-15 52.450000
2018-06-15 33.722222
2018-07-15 41.277778
2018-08-15 46.391304
2018-09-15 45.631579
2018-10-15 54.107143
2018-11-15 58.058824
2018-12-15 43.461538
Freq: 2SMS-15, Name: mean, dtype: float64
The code:
df_sm = df.resample('SMS', on='Month').aggregate(['sum', 'count'])
df_sm['sum_rolling'] = df_sm['Integers']['sum'].rolling(2).sum()
df_sm['count_rolling'] = df_sm['Integers']['count'].rolling(2).sum()
df_sm['mean'] = df_sm['sum_rolling'] / df_sm['count_rolling']
df_out = df_sm[1::2]['mean']
Edit: Changed a name of one of the columns to make it clearer
I have a file, df, that I wish to take the delta of every 7 day period and reflect the timestamp for that particular period
df:
Date Value
10/15/2020 75
10/14/2020 70
10/13/2020 65
10/12/2020 60
10/11/2020 55
10/10/2020 50
10/9/2020 45
10/8/2020 40
10/7/2020 35
10/6/2020 30
10/5/2020 25
10/4/2020 20
10/3/2020 15
10/2/2020 10
10/1/2020 5
Desired Output:
10/15/2020 to 10/9/2020 is 7 days with the delta being: 75 - 45 = 30
10/9/2020 timestamp would be: 30 and so on
Date Value
10/9/2020 30
10/2/2020 30
This is what I am doing:
df= df['Delta']=df.iloc[:,6].sub(df.iloc[:,0]),Date=pd.Series
(pd.date_range(pd.Timestamp('2020-10-
15'),
periods=7, freq='7d')))[['Delta','Date']]
I am also thinking I may be able to do this:
Edit I updated callDate to Date
for row in df.itertuples():
Date = datetime.strptime(row.Date, "%m/%d/%y %I:%M %p")
previousRecord = df['Date'].shift(-6).strptime(row.Date, "%m/%d/%y %I:%M
%p")
Delta = Date - previousRecord
Any suggestion is appreciated
Don't iterate through the dataframe. You can use a merge:
(df.merge(df.assign(Date=df['Date'] - pd.to_timedelta('6D')),
on='Date')
.assign(Value = lambda x: x['Value_y']-x['Value_x'])
[['Date','Value']]
)
Output:
Date Value
0 2020-10-09 30
1 2020-10-08 30
2 2020-10-07 30
3 2020-10-06 30
4 2020-10-05 30
5 2020-10-04 30
6 2020-10-03 30
7 2020-10-02 30
8 2020-10-01 30
The last block of code you wrote is the way I would do it. Only problem is in Delta = Date - previousRecord, there is nothing called Date here. You should instead be accessing the value associated with callDate.
I have a pandas dataframe with a date column
I'm trying to create a function and apply it to the dataframe to create a column that returns the number of days in the month/year specified
so far i have:
from calendar import monthrange
def dom(x):
m = dfs["load_date"].dt.month
y = dfs["load_date"].dt.year
monthrange(y,m)
days = monthrange[1]
return days
This however does not work when I attempt to apply it to the date column.
Additionally, I would like to be able to identify whether or not it is the current month, and if so return the number of days up to the current date in that month as opposed to days in the entire month.
I am not sure of the best way to do this, all I can think of is to check the month/year against datetime's today and then use a delta
thanks in advance
For pt.1 of your question, you can cast to pd.Period and retrieve days_in_month:
import pandas as pd
# create a sample df:
df = pd.DataFrame({'date': pd.date_range('2020-01', '2021-01', freq='M')})
df['daysinmonths'] = df['date'].apply(lambda t: pd.Period(t, freq='S').days_in_month)
# df['daysinmonths']
# 0 31
# 1 29
# 2 31
# ...
For pt.2, you can take the timestamp of 'now' and create a boolean mask for your date column, i.e. where its year/month is less than "now". Then calculate the cumsum of the daysinmonth column for the section where the mask returns True. Invert the order of that series to get the days until now.
now = pd.Timestamp('now')
m = (df['date'].dt.year <= now.year) & (df['date'].dt.month < now.month)
df['daysuntilnow'] = df['daysinmonths'][m].cumsum().iloc[::-1].reset_index(drop=True)
Update after comment: to get the elapsed days per month, you can do
df['dayselapsed'] = df['daysinmonths']
m = (df['date'].dt.year == now.year) & (df['date'].dt.month == now.month)
if m.any():
df.loc[m, 'dayselapsed'] = now.day
df.loc[(df['date'].dt.year >= now.year) & (df['date'].dt.month > now.month), 'dayselapsed'] = 0
output
df
Out[13]:
date daysinmonths daysuntilnow dayselapsed
0 2020-01-31 31 213.0 31
1 2020-02-29 29 182.0 29
2 2020-03-31 31 152.0 31
3 2020-04-30 30 121.0 30
4 2020-05-31 31 91.0 31
5 2020-06-30 30 60.0 30
6 2020-07-31 31 31.0 31
7 2020-08-31 31 NaN 27
8 2020-09-30 30 NaN 0
9 2020-10-31 31 NaN 0
10 2020-11-30 30 NaN 0
11 2020-12-31 31 NaN 0
I'm new to python and I'm facing the following problem. I have a dataframe composed of 2 columns, one of them is date (datetime64[ns]). I want to keep all records within the last 12 months. My code is the following:
today=start_time.date()
last_year = today + relativedelta(months = -12)
new_df = df[pd.to_datetime(df.mydate) >= last_year]
when I run it I get the following message:
TypeError: type object 2017-06-05
Any ideas?
last_year seems to bring me the date that I want in the following format: 2017-06-05
Create a time delta object in pandas to increment the date (12 months). Call pandas.Timstamp('now') to get the current date. And then create a date_range. Here is an example for getting monthly data for 12 months.
import pandas as pd
import datetime
list_1 = [i for i in range(0, 12)]
list_2 = [i for i in range(13, 25)]
list_3 = [i for i in range(26, 38)]
data_frame = pd.DataFrame({'A': list_1, 'B': list_2, 'C':list_3}, pd.date_range(pd.Timestamp('now'), pd.Timestamp('now') + pd.Timedelta (weeks=53), freq='M'))
We create a timestamp for the current date and enter that as our start date. Then we create a timedelta to increment that date by 53 weeks (or 52 if you'd like) which gets us 12 months of data. Below is the output:
A B C
2018-06-30 05:05:21.335625 0 13 26
2018-07-31 05:05:21.335625 1 14 27
2018-08-31 05:05:21.335625 2 15 28
2018-09-30 05:05:21.335625 3 16 29
2018-10-31 05:05:21.335625 4 17 30
2018-11-30 05:05:21.335625 5 18 31
2018-12-31 05:05:21.335625 6 19 32
2019-01-31 05:05:21.335625 7 20 33
2019-02-28 05:05:21.335625 8 21 34
2019-03-31 05:05:21.335625 9 22 35
2019-04-30 05:05:21.335625 10 23 36
2019-05-31 05:05:21.335625 11 24 37
Try
today = datetime.datetime.now()
You can use pandas functionality with datetime objects. The syntax is often more intuitive and obviates the need for additional imports.
last_year = pd.to_datetime('today') + pd.DateOffset(years=-1)
new_df = df[pd.to_datetime(df.mydate) >= last_year]
As such, we would need to see all your code to be sure of the reason behind your error; for example, how is start_time defined?