I've separate columns for start( timestamp ) and end( timestamp) and i need to get the earliest starttime and last endtime for each date.
number start end test time
0 1 2020-02-01 06:27:38 2020-02-01 08:29:42 1 02:02:04
1 1 2020-02-01 08:41:03 2020-02-01 11:05:30 2 02:24:27
2 1 2020-02-01 11:20:22 2020-02-01 13:03:49 1 01:43:27
3 1 2020-02-01 13:38:18 2020-02-01 16:04:31 2 02:26:13
4 1 2020-02-01 16:26:46 2020-02-01 17:42:49 1 01:16:03
5 1 2020-02-02 10:11:00 2020-02-02 12:11:00 1 02:00:00
I want the output for each date as : Date Min Max
I'm fairly new to Pandas and most of the solutions i've across is finding the min and max datetime from column. While what i want to do is min and max datetime for each date, where the timestamps are spread over two columns
expected output (ignore the date and time formats please)
date min max
1/2/2020 6:27 17:42
2/2/2020 10:11 12:11
I believe you need to start by creating a date column and later performing groupby with date.
df['date'] = df['start'].dt.date
df['start_hm'] = df['start'].dt.strftime('%H:%M')
df['end_hm'] = df['end'].dt.strftime('%H:%M')
output = df.groupby('date').agg(min = pd.NamedAgg(column = 'start_hm',aggfunc='min'),
max = pd.NamedAgg(column='end_hm',aggfunc='max'))
Output:
min max
date
2020-02-01 06:27 17:42
2020-02-02 10:11 12:11
Related
I have the following df:
time_series date sales
store_0090_item_85261507 1/2020 1,0
store_0090_item_85261501 2/2020 0,0
store_0090_item_85261500 3/2020 6,0
Being 'date' = Week/Year.
So, I tried use the following code:
df['date'] = df['date'].apply(lambda x: datetime.strptime(x + '/0', "%U/%Y/%w"))
But, return this df:
time_series date sales
store_0090_item_85261507 2020-01-05 1,0
store_0090_item_85261501 2020-01-12 0,0
store_0090_item_85261500 2020-01-19 6,0
But, the first day of the first week of 2020 is 2019-12-29, considering sunday as first day. How can I have the first day 2020-12-29 of the first week of 2020 and not 2020-01-05?
From the datetime module's documentation:
%U: Week number of the year (Sunday as the first day of the week) as a zero padded decimal number. All days in a new year preceding the first Sunday are considered to be in week 0.
Edit: My originals answer doesn't work for input 1/2023 and using ISO 8601 date values doesn't work for 1/2021, so I've edited this answer by adding a custom function
Here is a way with a custom function
import pandas as pd
from datetime import datetime, timedelta
##############################################
# to demonstrate issues with certain dates
print(datetime.strptime('0/2020/0', "%U/%Y/%w")) # 2019-12-29 00:00:00
print(datetime.strptime('1/2020/0', "%U/%Y/%w")) # 2020-01-05 00:00:00
print(datetime.strptime('0/2021/0', "%U/%Y/%w")) # 2020-12-27 00:00:00
print(datetime.strptime('1/2021/0', "%U/%Y/%w")) # 2021-01-03 00:00:00
print(datetime.strptime('0/2023/0', "%U/%Y/%w")) # 2023-01-01 00:00:00
print(datetime.strptime('1/2023/0', "%U/%Y/%w")) # 2023-01-01 00:00:00
#################################################
df = pd.DataFrame({'date':["1/2020", "2/2020", "3/2020", "1/2021", "2/2021", "1/2023", "2/2023"]})
print(df)
def get_first_day(date):
date0 = datetime.strptime('0/' + date.split('/')[1] + '/0', "%U/%Y/%w")
date1 = datetime.strptime('1/' + date.split('/')[1] + '/0', "%U/%Y/%w")
date = datetime.strptime(date + '/0', "%U/%Y/%w")
return date if date0 == date1 else date - timedelta(weeks=1)
df['new_date'] = df['date'].apply(lambda x:get_first_day(x))
print(df)
Input
date
0 1/2020
1 2/2020
2 3/2020
3 1/2021
4 2/2021
5 1/2023
6 2/2023
Output
date new_date
0 1/2020 2019-12-29
1 2/2020 2020-01-05
2 3/2020 2020-01-12
3 1/2021 2020-12-27
4 2/2021 2021-01-03
5 1/2023 2023-01-01
6 2/2023 2023-01-08
You'll want to use ISO week parsing directives, Ex:
import pandas as pd
date = pd.Series(["1/2020", "2/2020", "3/2020"])
pd.to_datetime(date+"/1", format="%V/%G/%u")
0 2019-12-30
1 2020-01-06
2 2020-01-13
dtype: datetime64[ns]
you can also shift by one day if the week should start on Sunday:
pd.to_datetime(date+"/1", format="%V/%G/%u") - pd.Timedelta('1d')
0 2019-12-29
1 2020-01-05
2 2020-01-12
dtype: datetime64[ns]
I need to find the timeframe from the master based on the input time.
cust_id starttime
0 1 2000-01-01 09:00:03
1 2 2000-01-01 18:01:03
output i needed is
cust_id starttime timeframe
0 1 2000-01-01 09:00:03 morning
1 2 2000-01-01 18:01:03 evening
Code for creating master timeframe details
mastdf={'timeframe':['morning','latemorning','midnoon','evening'],'start_time':['8:00:00','11:00:00','13:00:00','17:00:00'],'end_time':['10:59:59','13:59:59','16:59:59','7:59:59']}enter code here
Code for creating input dataframe
inputdf={'cust_id':[1,2],'starttime':['2000-01-01 09:00:03', '2000-01-01 18:01:03']}
Use cut for binning but first convert values to timedeltas by to_timedelta, create bins with add endpoint 24H and for timeframe between 00:00:00 to 8:00:00 is used fillna by last value of column timeframe:
mastdf={'timeframe':['morning','latemorning','midnoon','evening'],
'start_time':['8:00:00','11:00:00','13:00:00','17:00:00'],
'end_time':['10:59:59','13:59:59','16:59:59','7:59:59']}
mastdf = pd.DataFrame(mastdf)
print (mastdf)
timeframe start_time end_time
0 morning 8:00:00 10:59:59
1 latemorning 11:00:00 13:59:59
2 midnoon 13:00:00 16:59:59
3 evening 17:00:00 7:59:59
inputdf={'cust_id':[1,2],'starttime':['2000-01-01 09:00:03', '2000-01-01 18:01:03']}
inputdf = pd.DataFrame(inputdf)
inputdf['starttime'] = pd.to_datetime(inputdf['starttime'])
start = pd.to_timedelta(mastdf['start_time']).tolist() + [pd.Timedelta(24, unit='h')]
s = pd.to_timedelta(inputdf['starttime'].dt.strftime('%H:%M:%S'))
last = mastdf['timeframe'].iat[-1]
inputdf['timeframe'] = pd.cut(s,
bins=start,
labels=mastdf['timeframe'], right=False).fillna(last)
print (inputdf)
cust_id starttime timeframe
0 1 2000-01-01 09:00:03 morning
1 2 2000-01-01 18:01:03 evening
I would like to calculate a mean of a time delta serie excluding 00:00:00 values.
Then this is my time serie:
1 00:28:00
3 01:57:00
5 00:00:00
7 01:27:00
9 00:00:00
11 01:30:00
I try to replace 5 and 9 row per NaN and then apply .mean() to the serie. mean() doesn´t include NaN values and I get the desired value.
How can I do that stuff?
I´am trying:
`df["time_column"].replace('0 days 00:00:00', np.NaN).mean()`
but no values are replaced
One idea is use 0 Timedelta object:
out = df["time_column"].replace(pd.Timedelta(0), np.NaN).mean()
print (out)
0 days 01:20:30
I have a dataframe that contains hourly temperature data from 1990-2019 for 25 different locations. I want to count the amount of hours that a value is above or below a certain threshold and then plot that amount as a sum of the hours for every year. I know I can use a bar chart or histogram to plot, but am unsure how to aggregate the data to perform this task.
Dataframe:
time Antwerp Rotterdam ...
1990-01-01 00:00:00 2 4 ...
1990-01-01 01:00:00 3 4 ...
1990-01-01 02:00:00 2 4 ...
...
Do I need to use the groupby function?
Sample data to demonstrate:
time Antwerp Rotterdam Los Angeles
0 1990-01-01 00:00:00 0 2 15
1 1990-01-01 01:00:00 1 4 14
2 1990-01-01 02:00:00 3 5 15
3 1990-01-01 03:00:00 2 6 16
Now I am looking for the amount of hours that one city is equal to or less than 5 degrees during the year 1990. Expected output:
time Antwerp Rotterdam Los Angeles
1990 4 3 0
Ideally I would want to be able to select whatever temperature value I want.
I think you need DatetimeIndex, compare, e.g. for greater by DataFrame.gt and then count Trues values by aggregate sum:
df['time'] = pd.to_datetime(df['time'])
df = df.set_index('time')
N = 2
df = df.gt(N).groupby(df.index.year).sum()
print (df)
Antwerp Rotterdam
time
1990 0.0 1.0
1991 1.0 2.0
If want low or equal use DataFrame.le:
N = 3
df = df.le(N).groupby(df.index.year).sum()
print (df)
Antwerp Rotterdam
time
1990 1.0 0.0
1991 2.0 0.0
This is without using pandas functions.
# get the time column as a list by timelist = list(df['time'])
def get_hour_ud(df, threshold):
# timelist = list(df['time'])
# df['time'] = ['1990-01-01 00:00:00', '1990-01-01 01:00:00', '1990-01-01 02:00:00'] # remove this line
timelist = list(df['time'])
hour_list = [int(a.split(' ')[1].split(':')[0]) for a in timelist]
up_cnt = sum(a>threshold for a in hour_list)
low_cnt = sum(a<threshold for a in hour_list)
print(up_cnt)
print(low_cnt)
return up_cnt, low_cnt
UsageDate CustID1 CustID2 .... CustIDn
0 2018-01-01 00:00:00 1.095
1 2018-01-01 01:00:00 1.129
2 2018-01-01 02:00:00 1.165
3 2018-01-01 04:00:00 1.697
.
.
m 2018-31-01 23:00:00 1.835 (m,n)
The dataframe (df) has m rows and n columns. m is a Hourly TimeSeries Index which starts from first hour of month to last hour of month.
The columns are the customers which are almost 100,000.
The values at each cell of Dataframe are energy consumption values.
For every customer, I need to calculate:
1) Mean of every hour usage - so basically average of 1st hour of every day in a month, 2nd hour of every day in a month etc.
2) Summation of usage of every customer
3) Top 3 usage hours - for a customer x, it can be "2018-01-01 01:00:00",
"2018-11-01 05:00:00" "2018-21-01 17:00:00"
4) Bottom 3 usage hours - Similar explanation as above
5) Mean of usage for every customer in the month
My main point of trouble is how to aggregate data both for every customer and the hour of day, or day together.
For summation of usage for every customer, I tried:
df_temp = pd.DataFrame(columns=["TotalUsage"])
for col in df.columns:
`df_temp[col,"TotalUsage"] = df[col].apply.sum()`
However, this and many version of this which I tried are not helping me solve the problem.
Please help me with an approach and how to think about such problems.
Also, since the dataframe is large, it would be helpful if we can talk about Computational Complexity and how can we decrease computation time.
This looks like a job for pandas.groupby.
(I didn't test the code because I didn't have a good sample dataset from which to work. If there are errors, let me know.)
For some of your requirements, you'll need to add a column with the hour:
df['hour']=df['UsageDate'].dt.hour
1) Mean by hour.
mean_by_hour=df.groupby('hour').mean()
2) Summation by user.
sum_by_uers=df.sum()
3) Top usage by customer. Bottom 3 usage hours - Similar explanation as above.--I don't quite understand your desired output, you might be asking too many different questions in this question. If you want the hour and not the value, I think you may have to iterate through the columns. Adding an example may help.
4) Same comment.
5) Mean by customer.
mean_by_cust = df.mean()
I am not sure if this is all the information you are looking for but it will point you in the right direction:
import pandas as pd
import numpy as np
# sample data for 3 days
np.random.seed(1)
data = pd.DataFrame(pd.date_range('2018-01-01', periods= 72, freq='H'), columns=['UsageDate'])
data2 = pd.DataFrame(np.random.rand(72,5), columns=[f'ID_{i}' for i in range(5)])
df = data.join([data2])
# print('Sample Data:')
# print(df.head())
# print()
# mean of every month and hour per year
# groupby year month hour then find the mean of every hour in a given year and month
mean_data = df.groupby([df['UsageDate'].dt.year, df['UsageDate'].dt.month, df['UsageDate'].dt.hour]).mean()
mean_data.index.names = ['UsageDate_year', 'UsageDate_month', 'UsageDate_hour']
# print('Mean Data:')
# print(mean_data.head())
# print()
# use set_index with max and head
top_3_Usage_hours = df.set_index('UsageDate').max(1).sort_values(ascending=False).head(3)
# print('Top 3:')
# print(top_3_Usage_hours)
# print()
# use set_index with min and tail
bottom_3_Usage_hours = df.set_index('UsageDate').min(1).sort_values(ascending=False).tail(3)
# print('Bottom 3:')
# print(bottom_3_Usage_hours)
out:
Sample Data:
UsageDate ID_0 ID_1 ID_2 ID_3 ID_4
0 2018-01-01 00:00:00 0.417022 0.720324 0.000114 0.302333 0.146756
1 2018-01-01 01:00:00 0.092339 0.186260 0.345561 0.396767 0.538817
2 2018-01-01 02:00:00 0.419195 0.685220 0.204452 0.878117 0.027388
3 2018-01-01 03:00:00 0.670468 0.417305 0.558690 0.140387 0.198101
4 2018-01-01 04:00:00 0.800745 0.968262 0.313424 0.692323 0.876389
Mean Data:
ID_0 ID_1 ID_2 \
UsageDate_year UsageDate_month UsageDate_hour
2018 1 0 0.250716 0.546475 0.202093
1 0.414400 0.264330 0.535928
2 0.335119 0.877191 0.380688
3 0.577429 0.599707 0.524876
4 0.702336 0.654344 0.376141
ID_3 ID_4
UsageDate_year UsageDate_month UsageDate_hour
2018 1 0 0.244185 0.598238
1 0.400003 0.578867
2 0.623516 0.477579
3 0.429835 0.510685
4 0.503908 0.595140
Top 3:
UsageDate
2018-01-01 21:00:00 0.997323
2018-01-03 23:00:00 0.990472
2018-01-01 08:00:00 0.988861
dtype: float64
Bottom 3:
UsageDate
2018-01-01 19:00:00 0.002870
2018-01-03 02:00:00 0.000402
2018-01-01 00:00:00 0.000114
dtype: float64
For top and bottom 3 if you want to find the min sum across rows then:
df.set_index('UsageDate').sum(1).sort_values(ascending=False).tail(3)