Here is the code for sample simulated data. Actual data can have varying start and end dates.
import pandas as pd
import numpy as np
dates = pd.date_range("20100121", periods=3653)
df = pd.DataFrame(np.random.randn(3653, 1), index=dates, columns=list("A"))
dfb=df.resample('B').apply(lambda x:x[-1])
From the dfb, I want to select the rows that contain values for all the days of the month.
In dfb, 2010 January and 2020 January have incomplete data. So I would like data from 2010 Feb till 2019 December.
For this particular dataset, I could do
df_out=dfb['2010-02':'2019-12']
But please help me with a better solution
Edit-- Seems there is plenty of confusion in the question. I want to omit rows that does not begin with first day of the month and rows that does not end on last day of the month. Hope that's clear.
When you say "better" solution - I assume you mean make the range dynamic based on input data.
OK, since you mention that your data is continuous after the start date - it is a safe assumption that dates are sorted in increasing order. With this in mind, consider the code:
import pandas as pd
import numpy as np
from datetime import date, timedelta
dates = pd.date_range("20100121", periods=3653)
df = pd.DataFrame(np.random.randn(3653, 1), index=dates, columns=list("A"))
print(df)
dfb=df.resample('B').apply(lambda x:x[-1])
# fd is the first index in your dataframe
fd = df.index[0]
first_day_of_next_month = fd
# checks if the first month data is incomplete, i.e. does not start with date = 1
if ( fd.day != 1 ):
new_month = fd.month + 1
if ( fd.month == 12 ):
new_month = 1
first_day_of_next_month = fd.replace(day=1).replace(month=new_month)
else:
first_day_of_next_month = fd
# ld is the last index in your dataframe
ld = df.index[-1]
# computes the next day
next_day = ld + timedelta(days=1)
if ( next_day.month > ld.month ):
last_day_of_prev_month = ld # keeps the index if month is changed
else:
last_day_of_prev_month = ld.replace(day=1) - timedelta(days=1)
df_out=dfb[first_day_of_next_month:last_day_of_prev_month]
There is another way to use dateutil.relativedelta but you will need to install python-dateutil module. The above solution attempts to do it without using any extra modules.
I assume that in the general case the table is chronologically ordered (if not use .sort_index). The idea is to extract the year and month from the date and select only the lines where (year, month) is not equal to the first and last lines.
dfb['year'] = dfb.index.year # col#1
dfb['month'] = dfb.index.month # col#2
first_month = (dfb['year']==dfb.iloc[0, 1]) & (dfb['month']==dfb.iloc[0, 2])
last_month = (dfb['year']==dfb.iloc[-1, 1]) & (dfb['month']==dfb.iloc[-1, 2])
dfb = dfb.loc[(~first_month) & (~last_month)]
dfb = dfb.drop(['year', 'month'], axis=1)
Related
i have this kind of dataframe
These data represents the value of an consumption index generally encoded once a month (at the end or at the beginning of the following month) but sometimes more. This value can be resetted to "0" if the counter is out and be replaced. Moreover some month no data is available.
I would like select only one entry per month but this entry has to be the nearest to the first day of the month AND inferior to the 15th day of the month (because if the day is higher it could be the measure of the end of the month). Another condition is that if the difference between two values is negative (the counter has been replaced), the value need to be kept even if the date is not the nearest day near the first day of month.
For example, the output data need to be
The purpose is to calculate only a consumption per month.
A solution is to parse the dataframe (as a array) and perform some if conditions statements. However i wonder if there is "simple" alternative to achieve that.
Thank you
You can normalize the month data with MonthEnd and then drop duplicates based off that column and keep the last value.
from pandas.tseries.offsets import MonthEnd
df.New = df.Index + MonthEnd(1)
df.Diff = abs((df.Index - df.New).dt.days)
df = df.sort_values(df.New, df.Diff)
df = df.drop_duplicates(subset='New', keep='first').drop(['New','Diff'], axis=1)
That should do the trick, but I was not able to test, so please copy and past the sample data into StackOverFlow if this isn't doing the job.
Defining dataframe, converting index to datetime, defining helper columns,
using them to run shift method to conditionally remove rows, and finally removing the helper columns:
from pandas.tseries.offsets import MonthEnd, MonthBegin
import pandas as pd
from datetime import datetime as dt
import numpy as np
df = pd.DataFrame([
[1254],
[1265],
[1277],
[1301],
[1345],
[1541]
], columns=["Value"]
, index=[dt.strptime("05-10-19", '%d-%m-%y'),
dt.strptime("29-10-19", '%d-%m-%y'),
dt.strptime("30-10-19", '%d-%m-%y'),
dt.strptime("04-11-19", '%d-%m-%y'),
dt.strptime("30-11-19", '%d-%m-%y'),
dt.strptime("03-02-20", '%d-%m-%y')
]
)
early_days = df.loc[df.index.day < 15]
early_month_end = early_days.index - MonthEnd(1)
early_day_diff = early_days.index - early_month_end
late_days = df.loc[df.index.day >= 15]
late_month_end = late_days.index + MonthBegin(1)
late_day_diff = late_month_end - late_days.index
df["day_offset"] = (early_day_diff.append(late_day_diff) / np.timedelta64(1, 'D')).astype(int)
df["start_of_month"] = df.index.day < 15
df["month"] = df.index.values.astype('M8[D]').astype(str)
df["month"] = df["month"].str[5:7].str.lstrip('0')
# df["month_diff"] = df["month"].astype(int).diff().fillna(0).astype(int)
df = df[df["month"].shift().ne(df["month"].shift(-1))]
df = df.drop(columns=["day_offset", "start_of_month", "month"])
print(df)
Returns:
Value
2019-10-05 1254
2019-10-30 1277
2019-11-04 1301
2019-11-30 1345
2020-02-03 1541
date['Maturity_date'] = data.apply(lambda data: relativedelta(months=int(data['TRM_LNTH_MO'])) + data['POL_EFF_DT'], axis=1)
Tried this also:
date['Maturity_date'] = date['POL_EFF_DT'] + date['TRM_LNTH_MO'].values.astype("timedelta64[M]")
TypeError: 'type' object does not support item assignment
import pandas as pd
import datetime
#Convert the date column to date format
date['date_format'] = pd.to_datetime(date['Maturity_date'])
#Add a month column
date['Month'] = date['date_format'].apply(lambda x: x.strftime('%b'))
If you are using Pandas, you may use a resource called: "Frequency Aliases". Something very out of the box:
# For "periods": 1 (is the current date you have) and 2 the result, plus 1, by the frequency of 'M' (month).
import pandas as pd
_new_period = pd.date_range(_existing_date, periods=2, freq='M')
Now you can get exactly the period you want as the second element returned:
# The index for your information is 1. Index 0 is the existing date.
_new_period.strftime('%Y-%m-%d')[1]
# You can format in different ways. Only Year, Month or Day. Whatever.
Consult this link for further information
I am using spyder and python 3.8.
I would like to filter a list of dates and only return if it is from and even month and the end of the business month, incorporating if it is a holiday.
I have created my date_range as following:
from pandas.tseries.holiday import USFederalHolidayCalendar
from pandas.tseries.offsets import CustomBusinessDay
import pandas as pd
us_bd = CustomBusinessDay(calendar=USFederalHolidayCalendar())
start = '2009-12-31'
end = '2020-01-17'
rng = pd.Series(pd.to_datetime(pd.date_range(start,end, freq=us_bd).date))
Then I know I can test if the month is even with
mth = datetime.datetime.strptime(str(start),"%Y-%m-%d %H:%M:%S").month
if (mth % 2 == 0):
# keep date
How do I check if the day is the last day of the month and a business day and not a holiday. If True, then keep date, otherwise remove date?
So I was able to figure it out, but it could be cleaner.
rng = pd.Series(pd.to_datetime(pd.date_range(start,end, freq=us_bd).date))
rng = [date for date in rng if datetime.datetime.strptime(str(date),"%Y-%m-%d %H:%M:%S").month % 2 == 0]
rng_df = pd.DataFrame({'date':rng})
rng_df['Year']= rng_df['date'].dt.year
rng_df['Month']=rng_df['date'].dt.month
rng_df['Day']=rng_df['date'].dt.day
rng_df['ym_cuml'] = rng_df.groupby(['Year','Month']).cumcount()
rng_df['group'] = pd.to_datetime(rng_df[['Year', 'Month']].assign(Day=1))
groups = pd.Series(np.unique(rng_df['group']))
groups = groups.apply(lambda x : x.strftime('%Y-%m-%d'))
dates = []
for group in groups:
sub_rng_df = rng_df.loc[rng_df.group == group]
mx = sub_rng_df['ym_cuml'].max()
date_to_append = sub_rng_df.iloc[mx]['date']
date_to_append = date_to_append.strftime('%Y-%m-%d')
dates.append(date_to_append)
I have a dataframe with the date and month_diff variable. I would like to get a new date (name it as Target_Date) based on the following logic:
For example, the date is 2/13/2019, month_diff is 3, then the target date should be the month-end of the original date plus 3 months, which is 5/31/2019
I tried the following method to get the traget date first:
df["Target_Date"] = df["Date"] + pd.DateOffset(months = df["month_diff"])
But it failed, as I know, the parameter in the dateoffset should be a varaible or a fixed number.
I also tried:
df["Target_Date"] = df["Date"] + relativedelta(months = df["month_diff"])
It failes too.
Anyone can help? thank you.
edit:
this is a large dataset with millions rows.
You could try this
import pandas as pd
from dateutil.relativedelta import relativedelta
df = pd.DataFrame({'Date': [pd.datetime(2019,1,1), pd.datetime(2019,2,1)], 'month_diff': [1,2]})
df.apply(lambda row: row.Date + relativedelta(months=row.month_diff), axis=1)
Or list comprehension
[date + relativedelta(months=month_diff) for date, month_diff in df[['Date', 'month_diff']].values]
I would approach in the following method to compute your "target_date".
Apply the target month offset (in your case +3months), using your pd.DateOffset.
Get the last day of that target month (using for example calendar.monthrange, see also "Get last day of the month"). This will provide you with the "flexible" part of that date" offset.
Apply the flexible day offset, when comparing the result of step 1. and step 2. This could be a new pd.DateOffset.
A solution could look something like this:
import calendar
from dateutil.relativedelta import relativedelta
for ii in df.index:
new_ = df.at[ii, 'start_date'] + relativedelta(months=df.at[ii, 'month_diff'])
max_date = calendar.monthrange(new_.year, new_.month)[1]
end_ = new_ + relativedelta(days=max_date - new_.day)
print(end_)
Further "cleaning" into a function and / or list comprehension will probably make it much faster
import pandas as pd
from datetime import datetime
from datetime import timedelta
This is my approach in solving your issue.
However for some reason I am getting a semantic error in my output even though I am sure it is the correct way. Please everyone correct me if you notice something wrong.
today = datetime.now()
today = today.strftime("%d/%m/%Y")
month_diff =[30,5,7]
n = 30
for i in month_diff:
b = {'Date': today, 'month_diff':month_diff,"Target_Date": datetime.now()+timedelta(days=i*n)}
df = pd.DataFrame(data=b)
Output:
For some reason the i is not getting updated.
I was looking for a solution I can write in one line only and apply does the job. However, by default apply function performs action on each column, so you have to remember to specify correct axis: axis=1.
from datetime import datetime
from dateutil.relativedelta import relativedelta
# Create a new column with date adjusted by number of months from 'month_diff' column and later adjust to the last day of month
df['Target_Date'] = df.apply(lambda row: row.Date # to current date
+ relativedelta(months=row.month_diff) # add month_diff
+ relativedelta(day=+31) # and adjust to the last day of month
, axis=1) # 1 or ‘columns’: apply function to each row.
I created an hourly dates dataframe, and now I would like to create a column that flags whether each row (hour) is in Daylight Saving Time or not. For example, in summer hours, the flag should == 1, and in winter hours, the flag should == 0.
# Localized dates dataframe
dates = pd.DataFrame(data=pd.date_range('2018-1-1', '2019-1-1', freq='h', tz='America/Denver'), columns=['date_time'])
# My failed attempt to create the flag column
dates['dst_flag'] = np.where(dates['date_time'].dt.daylight_saving_time == True, 1, 0)
There's a nice link in the comments that at least let you do this manually. AFAIK, there isn't a vectorized way to do this.
import pandas as pd
import numpy as np
from pytz import timezone
# Generate data (as opposed to index)
date_range = pd.to_datetime(pd.date_range('1/1/2018', '1/1/2019', freq='h', tz='America/Denver'))
date_range = [date for date in date_range]
# Localized dates dataframe
df = pd.DataFrame(data=date_range, columns=['date_time'])
# Map transition times to year for some efficiency gain
tz = timezone('America/Denver')
transition_times = tz._utc_transition_times[1:]
transition_times = [t.astimezone(tz) for t in transition_times]
transition_times_by_year = {}
for start_time, stop_time in zip(transition_times[::2], transition_times[1::2]):
year = start_time.year
transition_times_by_year[year] = [start_time, stop_time]
# If the date is in DST, mark true, else false
def mark_dst(dates):
for date in dates:
start_dst, stop_dst = transition_times_by_year[date.year]
yield start_dst <= date <= stop_dst
df['dst_flag'] = [dst_flag for dst_flag in mark_dst(df['date_time'])]
# Do a quick sanity check to make sure we did this correctly for year 2018
dst_start = df[df['dst_flag'] == True]['date_time'][0] # First dst time 2018
dst_end = df[df['dst_flag'] == True]['date_time'][-1] # Last dst time 2018
print(dst_start)
print(dst_end)
this outputs:
2018-03-11 07:00:00-06:00
2018-11-04 06:00:00-07:00
which is likely correct. I didn't do the UTC conversions by hand or anything to check that the hours are exactly right for the given timezone. You can at least verify the dates are correct with a quick google search.
Some gotchas:
pd.date_range generates an index, not data. I changed your original code slightly to make it be data as opposed to the index. I assume you have the data already.
There's something goofy about how tz._utc_transition_times is structured. It's start/stop utc DST transition times, but there is some goofy stuff in the early dates. It should be good from 1965 onward though. If you are doing dates earlier than that change tz._utc_transition_times[1:] to tz._utc_transition_times. Note not all years before 1965 are present.
tz._utc_transition_times is "Python private". It is liable to change without warning or notice, and may or may not work for future or past versions of pytz. I'm using pytz verion 2017.3. I recommend you run this code to make sure the output matches, and if not, make sure to use version 2017.3.
HTH, good luck with your research/regression problem!
If you are looking for a vectorized way of doing this (which you probably should be), you can use something like the code below.
The fundamental idea behind this is to find the difference between the current time in your timezone and the UTC time. In the winter months, the difference will be one extra hour behind UTC. Whatever the difference is, add what is needed to get to the 1 or 0 for the flag.
In Denver, summer months are UTC-6 and winter months are UTC-7. So, if you take the difference between the tz-aware time in Denver and UTC time, then add 7, you'll get a value of 1 for summer months and a value of 0 for winter months.
import pandas as pd
start = pd.to_datetime('2020-10-30')
end = pd.to_datetime('2020-11-02')
dates = pd.date_range(start=start, end=end, freq='h', tz='America/Denver')
df1 = pd.DataFrame({'dst_flag': 1, 'date1': dates.tz_localize(None)}, index=dates)
# add extra day on each end so that there are no nan's after the join
dates = pd.to_datetime(pd.date_range(start=start - pd.to_timedelta(1, 'd'), end=end + pd.to_timedelta(1, 'd'), freq='h'), utc=True)
df2 = pd.DataFrame({'date2': dates.tz_localize(None)}, index=dates)
out = df1.join(df2)
out['dst_flag'] = (out['date1'] - out['date2']) / pd.to_timedelta(1, unit='h') + 7
out.drop(columns=['date1', 'date2'], inplace=True)
Here is what I ended up doing, and it works for my purposes:
import pandas as pd
import pytz
# Create dates table and flag Daylight Saving Time dates
dates = pd.DataFrame(data=pd.date_range('2018-1-1', '2018-12-31-23', freq='h'), columns=['date_time'])
# Create a list of start and end dates for DST in each year, in UTC time
dst_changes_utc = pytz.timezone('America/Denver')._utc_transition_times[1:]
# Convert to local times from UTC times and then remove timezone information
dst_changes = [pd.Timestamp(i).tz_localize('UTC').tz_convert('America/Denver').tz_localize(None) for i in dst_changes_utc]
flag_list = []
for index, row in dates['date_time'].iteritems():
# Isolate the start and end dates for DST in each year
dst_dates_in_year = [date for date in dst_changes if date.year == row.year]
spring = dst_dates_in_year[0]
fall = dst_dates_in_year[1]
if (row >= spring) & (row < fall):
flag = 1
else:
flag = 0
flag_list.append(flag)
print(flag_list)
dates['dst_flag'] = flag_list
del(flag_list)
the following vectorized way seem to work fine.
The idea behind is the same as Nick Klavoht's idea : find the difference between the current time in your timezone and the utc time.
# Localized dates dataframe
df = pd.DataFrame(data=pd.date_range('2018-1-1', '2019-1-1', freq='h', tz='America/Denver'), columns=['date_time'])
df['utc_offset'] = df['date_time'].dt.strftime('%z').str[0:3].astype(float)
df['utc_offset_shifted'] = df['utc_offset'].shift(-1)
df['dst'] = df['utc_offset'] - df['utc_offset_shifted']
df_dst = df[(df['dst'] != 0) & (df['dst'])]
df_dst = df_dst.drop(['utc_offset', 'utc_offset_shifted'], axis=1).reset_index(drop=True)
print(df_dst)
This outputs :
date_time dst
0 2018-03-11 01:00:00-07:00 -1.0
1 2018-11-04 01:00:00-06:00 1.0
If you know what time zone you are dealing with you could use:
dates['dst_flag'] = dates['date_time'].apply(lambda x: x.tzname() == 'CEST')
This would flag the all hours in CET as False and in CEST as True. I'm not sure if I'd want to do that on a huge column.