i have dataframe like this :
trx_date
trx_amount
2013-02-11
35
2014-03-10
26
2011-02-9
10
2013-02-12
5
2013-01-11
21
how do i filter that into month and year? so that i can sum the trx_amount
example expected output :
trx_monthly
trx_sum
2013-02
40
2013-01
21
2014-02
35
You can convert values to month periods by Series.dt.to_period and then aggregate sum:
df['trx_date'] = pd.to_datetime(df['trx_date'])
df1 = (df.groupby(df['trx_date'].dt.to_period('m').rename('trx_monthly'))['trx_amount']
.sum()
.reset_index(name='trx_sum'))
print (df1)
trx_monthly trx_sum
0 2011-02 10
1 2013-01 21
2 2013-02 40
3 2014-03 26
Or convert datetimes to strings in format YYYY-MM by Series.dt.strftime:
df2 = (df.groupby(df['trx_date'].dt.strftime('%Y-%m').rename('trx_monthly'))['trx_amount']
.sum()
.reset_index(name='trx_sum'))
print (df2)
trx_monthly trx_sum
0 2011-02 10
1 2013-01 21
2 2013-02 40
3 2014-03 26
Or convert to month and years, then output is different - 3 columns:
df2 = (df.groupby([df['trx_date'].dt.year.rename('year'),
df['trx_date'].dt.month.rename('month')])['trx_amount']
.sum()
.reset_index(name='trx_sum'))
print (df2)
year month trx_sum
0 2011 2 10
1 2013 1 21
2 2013 2 40
3 2014 3 26
You can try this -
df['trx_month'] = df['trx_date'].dt.month
df_agg = df.groupby('trx_month')['trx_sum'].sum()
There is Pandas Dataframe as:
year month count
0 2014 Jan 12
1 2014 Feb 10
2 2015 Jan 12
3 2015 Feb 10
How to create DateTime index from 'year' and 'month',so result would be :
count
2014.01.31 12
2014.02.28 10
2015.01.31 12
2015.02.28 10
Use to_datetime with DataFrame.pop for use and remove columns and add offsets.MonthEnd:
dates = pd.to_datetime(df.pop('year').astype(str) + df.pop('month'), format='%Y%b')
df.index = dates + pd.offsets.MonthEnd()
print (df)
count
2014-01-31 12
2014-02-28 10
2015-01-31 12
2015-02-28 10
Or:
dates = pd.to_datetime(df.pop('year').astype(str) + df.pop('month'), format='%Y%b')
df.index = dates + pd.to_timedelta(dates.dt.daysinmonth - 1, unit='d')
print (df)
count
2014-01-31 12
2014-02-28 10
2015-01-31 12
2015-02-28 10
I have a df over 1M rows similar to this
ID Date Amount
x May 1 10
y May 2 20
z May 4 30
x May 1 40
y May 1 50
z May 2 60
x May 1 70
y May 5 80
a May 6 90
b May 8 100
x May 10 110
I have to sort the data based on the date and then create new dataframes depending on the times the value is present in Amount column. So if x has made purchase 3 time then I need it in 3 different dataframes. first_purchase dataframe would have every ID that has purchased even once irrespective of date or amount.
If an ID purchases 3 times, I need that ID to be in first purchase then second and then 3rd with Date and Amount.
Doing it manually is easy with:-
df = df.sort_values('Date')
first_purchase = df.drop_duplicates('ID')
after_1stpurchase = df[~df.index.isin(first_purchase.index)]
second data frame would be created with:-
after_1stpurchase = after_1stpurchase.sort_values('Date')
second_purchase = after_1stpurchase.drop_duplicates('ID')
after_2ndpurchase = after_1stpurchase[~after_1stpurchase.index.isin(second_purchase.index)]
How do I create the loop to provide me with each dataframes?
IIUC, I was able to achieve what you wanted.
import pandas as pd
import numpy as np
# source data for the dataframe
data = {
"ID":["x","y","z","x","y","z","x","y","a","b","x"],
"Date":["May 01","May 02","May 04","May 01","May 01","May 02","May 01","May 05","May 06","May 08","May 10"],
"Amount":[10,20,30,40,50,60,70,80,90,100,110]
}
df = pd.DataFrame(data)
# convert the Date column to datetime and still maintain the format like "May 01"
df['Date'] = pd.to_datetime(df['Date'], format='%b %d').dt.strftime('%b %d')
# sort the values on ID and Date
df.sort_values(by=['ID', 'Date'], inplace=True)
df.reset_index(inplace=True, drop=True)
print(df)
Original Dataframe:
Amount Date ID
0 90 May 06 a
1 100 May 08 b
2 10 May 01 x
3 40 May 01 x
4 70 May 01 x
5 110 May 10 x
6 50 May 01 y
7 20 May 02 y
8 80 May 05 y
9 60 May 02 z
10 30 May 04 z
.
# create a list of unique ids
list_id = sorted(list(set(df['ID'])))
# create an empty list that would contain dataframes
df_list = []
# count of iterations that must be seperated out
# for example if we want to record 3 entries for
# each id, the iter would be 3. This will create
# three new dataframes that will hold transactions
# respectively.
iter = 3
for i in range(iter):
df_list.append(pd.DataFrame())
for val in list_id:
tmp_df = df.loc[df['ID'] == val].reset_index(drop=True)
# consider only the top iter(=3) values to be distributed
counter = np.minimum(tmp_df.shape[0], iter)
for idx in range(counter):
df_list[idx] = df_list[idx].append(tmp_df.loc[tmp_df.index == idx])
for df in df_list:
df.reset_index(drop=True, inplace=True)
print(df)
Transaction #1:
Amount Date ID
0 90 May 06 a
1 100 May 08 b
2 10 May 01 x
3 50 May 01 y
4 60 May 02 z
Transaction #2:
Amount Date ID
0 40 May 01 x
1 20 May 02 y
2 30 May 04 z
Transaction #3:
Amount Date ID
0 70 May 01 x
1 80 May 05 y
Note that in your data, there are four transactions for 'x'. If lets say you wanted to track the 4th iterative transaction as well. All you need to do is change the value if 'iter' to 4 and you will get the fourth dataframe as well with the following value:
Amount Date ID
0 110 May 10 x
I want to transform an integer between 1 and 12 into an abbrieviated month name.
I have a df which looks like:
client Month
1 sss 02
2 yyy 12
3 www 06
I want the df to look like this:
client Month
1 sss Feb
2 yyy Dec
3 www Jun
Most of the info I found was not in python>pandas>dataframe hence the question.
You can do this efficiently with combining calendar.month_abbr and df[col].apply()
import calendar
df['Month'] = df['Month'].apply(lambda x: calendar.month_abbr[x])
Since the abbreviated month names is the first three letters of their full names, we could first convert the Month column to datetime and then use dt.month_name() to get the full month name and finally use str.slice() method to get the first three letters, all using pandas and only in one line of code:
df['Month'] = pd.to_datetime(df['Month'], format='%m').dt.month_name().str.slice(stop=3)
df
Month client
0 Feb sss
1 Dec yyy
2 Jun www
The calendar module is useful, but calendar.month_abbr is array-like: it cannot be used directly in a vectorised fashion. For an efficient mapping, you can construct a dictionary and then use pd.Series.map:
import calendar
d = dict(enumerate(calendar.month_abbr))
df['Month'] = df['Month'].map(d)
Performance benchmarking shows a ~130x performance differential:
import calendar
d = dict(enumerate(calendar.month_abbr))
mapper = calendar.month_abbr.__getitem__
np.random.seed(0)
n = 10**5
df = pd.DataFrame({'A': np.random.randint(1, 13, n)})
%timeit df['A'].map(d) # 7.29 ms per loop
%timeit df['A'].map(mapper) # 946 ms per loop
Solution 1: One liner
df['Month'] = pd.to_datetime(df['Month'], format='%m').dt.strftime('%b')
Solution 2: Using apply()
def mapper(month):
return month.strftime('%b')
df['Month'] = df['Month'].apply(mapper)
Reference:
http://strftime.org/
https://pandas.pydata.org/docs/reference/api/pandas.to_datetime.html
using datetime object methods
I'm surpised this answer doesn't have a solution using strftime
note, you'll need to have a valid datetime object before using the strftime method, use pd.to_datetime(df['date_column']) to cast your target column to a datetime object.
import pandas as pd
dates = pd.date_range('01-Jan 2020','01-Jan 2021',freq='M')
df = pd.DataFrame({'dates' : dates})
df['month_name'] = df['dates'].dt.strftime('%b')
dates month_name
0 2020-01-31 Jan
1 2020-02-29 Feb
2 2020-03-31 Mar
3 2020-04-30 Apr
4 2020-05-31 May
5 2020-06-30 Jun
6 2020-07-31 Jul
7 2020-08-31 Aug
8 2020-09-30 Sep
9 2020-10-31 Oct
10 2020-11-30 Nov
11 2020-12-31 Dec
another method would be to slice the name using dt.month_name()
df['month_name_str_slice'] = df['dates'].dt.month_name().str[:3]
dates month_name month_name_str_slice
0 2020-01-31 Jan Jan
1 2020-02-29 Feb Feb
2 2020-03-31 Mar Mar
3 2020-04-30 Apr Apr
4 2020-05-31 May May
5 2020-06-30 Jun Jun
6 2020-07-31 Jul Jul
7 2020-08-31 Aug Aug
8 2020-09-30 Sep Sep
9 2020-10-31 Oct Oct
10 2020-11-30 Nov Nov
11 2020-12-31 Dec Dec
You can do this easily with a column apply.
import pandas as pd
df = pd.DataFrame({'client':['sss', 'yyy', 'www'], 'Month': ['02', '12', '06']})
look_up = {'01': 'Jan', '02': 'Feb', '03': 'Mar', '04': 'Apr', '05': 'May',
'06': 'Jun', '07': 'Jul', '08': 'Aug', '09': 'Sep', '10': 'Oct', '11': 'Nov', '12': 'Dec'}
df['Month'] = df['Month'].apply(lambda x: look_up[x])
df
Month client
0 Feb sss
1 Dec yyy
2 Jun www
One way of doing that is with the apply method in the dataframe but, to do that, you need a map to convert the months. You could either do that with a function / dictionary or with Python's own datetime.
With the datetime it would be something like:
def mapper(month):
date = datetime.datetime(2000, month, 1) # You need a dateobject with the proper month
return date.strftime('%b') # %b returns the months abbreviation, other options [here][1]
df['Month'].apply(mapper)
In a simillar way, you could build your own map for custom names. It would look like this:
months_map = {01: 'Jan', 02: 'Feb'}
def mapper(month):
return months_map[month]
Obviously, you don't need to define this functions explicitly and could use a lambda directly in the apply method.
Use strptime and lambda function for this:
from time import strptime
df['Month'] = df['Month'].apply(lambda x: strptime(x,'%b').tm_mon)
Suppose we have a DF like this, and Date is already in DateTime Format:
df.head(3)
value
date
2016-05-19 19736
2016-05-26 18060
2016-05-27 19997
Then we can extract month number and month name easily like this :
df['month_num'] = df.index.month
df['month'] = df.index.month_name()
value year month_num month
date
2017-01-06 37353 2017 1 January
2019-01-06 94108 2019 1 January
2019-01-05 77897 2019 1 January
2019-01-04 94514 2019 1 January
Having tested all of these on a large dataset, I have found the following to be fastest:
import calendar
def month_mapping():
# I'm lazy so I have a stash of functions already written so
# I don't have to write them out every time. This returns the
# {1:'Jan'....12:'Dec'} dict in the laziest way...
abbrevs = {}
for month in range (1, 13):
abbrevs[month] = calendar.month_abbr[month]
return abbrevs
abbrevs = month_mapping()
df['Month Abbrev'} = df['Date Col'].dt.month.map(mapping)
You can use Pandas month_name() function. Example:
>>> idx = pd.date_range(start='2018-01', freq='M', periods=3)
>>> idx
DatetimeIndex(['2018-01-31', '2018-02-28', '2018-03-31'],
dtype='datetime64[ns]', freq='M')
>>> idx.month_name()
Index(['January', 'February', 'March'], dtype='object')
For more detail visit this link.
the best way would be to do with month_name() as commented by
Nurul Akter Towhid.
df['Month'] = df.Month.dt.month_name()
First you need to strip "0 " in the beginning (as u might get the exception leading zeros in decimal integer literals are not permitted; use an 0o prefix for octal integers)
step1)
def func(i):
if i[0] == '0':
i = i[1]
return(i)
df["Month"] = df["Month"].apply(lambda x: func(x))
Step2:
df["Month"] = df["Month"].apply(lambda x: calendar.month_name(x))
14 [2018-03-14, 2018-03-13, 2017-03-06, 2017-02-13]
15 [2017-07-26, 2017-06-09, 2017-02-24]
16 [2018-09-06, 2018-07-06, 2018-07-04, 2017-10-20]
17 [2018-10-03, 2018-09-13, 2018-09-12, 2018-08-3]
18 [2017-02-08]
this is my data, every ID has it's own dates that range between 2017-02-05 and 2018-06-30. I need to split dates into 5 time ranges of 4 months each, so that for the first 4 months every ID should have dates only in that time range (from 2017-02-05 to 2017-06-05), like this
14 [2017-03-06, 2017-02-13]
15 [2017-02-24]
16 [null] # or delete empty rows, it doesn't matter
17 [null]
18 [2017-02-08]
then for 2017-06-05 to 2017-10-05 and so on for every 4 month ranges. Also I can't use nested for loops because the data is too big. This is what I tried so far
months_4 = individual_dates.copy()
for _ in months_4['Date']:
_ = np.where(pd.to_datetime(_) <= pd.to_datetime('2017-9-02'), _, np.datetime64('NaT'))
and
months_8 = individual_dates.copy()
range_8 = pd.date_range(start='2017-9-02', end='2017-11-02')
for _ in months_8['Date']:
_ = _[np.isin(_, range_8)]
achieved absolutely no result, data stays the same no matter what
update: I did what you said
individual_dates['Date'] = individual_dates['Date'].str.strip('[]').str.split(', ')
df = pd.DataFrame({
'Date' : list(chain.from_iterable(individual_dates['Date'].tolist())),
'ID' : individual_dates['ClientId'].repeat(individual_dates['Date'].str.len())
})
df
and here is the result
Date ID
0 '2018-06-30T00:00:00.000000000' '2018-06-29T00... 14
1 '2017-03-28T00:00:00.000000000' '2017-03-27T00... 15
2 '2018-03-14T00:00:00.000000000' '2018-03-13T00... 16
3 '2017-12-14T00:00:00.000000000' '2017-03-28T00... 17
4 '2017-05-30T00:00:00.000000000' '2017-05-22T00... 18
5 '2017-03-28T00:00:00.000000000' '2017-03-27T00... 19
6 '2017-03-27T00:00:00.000000000' '2017-03-26T00... 20
7 '2017-12-15T00:00:00.000000000' '2017-11-20T00... 21
8 '2017-07-05T00:00:00.000000000' '2017-07-04T00... 22
9 '2017-12-12T00:00:00.000000000' '2017-04-06T00... 23
10 '2017-05-21T00:00:00.000000000' '2017-05-07T00... 24
For better performance I suggest convert list to column - flatten it and then filtering by isin with boolean indexing:
from itertools import chain
df = pd.DataFrame({
'Date' : list(chain.from_iterable(individual_dates['Date'].tolist())),
'ID' : individual_dates['ID'].repeat(individual_dates['Date'].str.len())
})
range_8 = pd.date_range(start='2017-02-05', end='2017-06-05')
df['Date'] = pd.to_datetime(df['Date'])
df = df[df['Date'].isin(range_8)]
print (df)
Date ID
0 2017-03-06 14
0 2017-02-13 14
1 2017-02-24 15
4 2017-02-08 18