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
Good morning all!
I have a Pandas df and Im trying to create a monthly box and whisker of 30 years ofdata.
DataFrame
datetime year month day hour lon lat
0 3/18/1986 10:17 1986 3 18 10 -124.835 46.540
1 6/7/1986 13:38 1986 6 7 13 -121.669 46.376
2 7/17/1986 20:56 1986 7 17 20 -122.436 48.044
3 7/26/1986 2:46 1986 7 26 2 -123.071 48.731
4 8/2/1986 19:54 1986 8 2 19 -123.654 48.480
Trying to see the mean amount of occurrences in X month, the median, and the max/min occurrence ( and date of max and min)..
Ive been playing around with pandas.DataFrame.groupby() but dont fully understand it.
I have grouped the date by month and day occurrences. I like this format:
Code:
df = pd.read_csv(masterCSVPath)
months = df['month']
test = df.groupby(['month','day'])['day'].count()
output: ---->
month day
1 1 50
2 103
3 97
4 29
5 60
...
12 27 24
28 7
29 17
30 18
31 9
So how can i turn that df above into a box/whisker plot?
The x-axis i want to be months..
y axis == occurrences
Try this (without doing groupby):
import matplotlib.pyplot as plt
import seaborn as sns
sns.boxplot(x = 'month', y = 'day', data = df)
In case you want the months to be in Jan, Feb format then try this:
import matplotlib.pyplot as plt
import seaborn as sns
import datetime as dt
df['month_new'] = df['datetime'].dt.strftime('%b')
sns.boxplot(x = 'month_new', y = 'day', data = df)
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))
My query is regrading getting the data, given two timestamp in python.
I need to have a input field, where i can enter the two timestamp, then from the CSV read, i need to retrieve for that particular input.
Actaul Data(CSV)
Daily_KWH_System PowerScout Temperature Timestamp Visibility Daily_electric_cost kW_System
0 4136.900384 P371602077 0 07/09/2016 23:58 0 180.657705 162.224216
1 3061.657187 P371602077 66 08/09/2016 23:59 10 133.693074 174.193804
2 4099.614033 P371602077 63 09/09/2016 05:58 10 179.029562 162.774013
3 3922.490275 P371602077 63 10/09/2016 11:58 10 171.297701 169.230047
4 3957.128982 P371602077 88 11/09/2016 17:58 10 172.806125 164.099307
Example:
Input:
start date : 2-1-2017
end date :10-1-2017
Output
Timestamp Value
2-1-2017 10
3-1-2017 35
.
.
.
.
10-1-2017 25
The original CSV would contain all the data
Timestamp Value
1-12-2016 10
2-12-2016 25
.
.
.
1-1-2017 15
2-1-2017 10
.
.
.
10-1-2017 25
.
.
31-1-2017 50
use pd.read_csv to read the file
df = pd.read_csv('my.csv', index_col='Timestamp', parse_dates=[0])
Then use your inputs to slice
df[start_date:end_date]
It seems you need dayfirst=True in read_csv with select by [] if all start and end dates are in df.index:
import pandas as pd
from pandas.compat import StringIO
temp=u"""Timestamp;Value
1-12-2016;10
2-12-2016;25
1-1-2017;15
2-1-2017;10
10-1-2017;25
31-1-2017;50"""
#after testing replace 'StringIO(temp)' to 'filename.csv'
#if necessary add sep
#index_col=[0] convert first column to index
#parse_dates=[0] parse first column to datetime
df = pd.read_csv(StringIO(temp), sep=";", index_col=[0], parse_dates=[0], dayfirst=True)
print (df)
Value
Timestamp
2016-12-01 10
2016-12-02 25
2017-01-01 15
2017-01-02 10
2017-01-10 25
2017-01-31 50
print (df.index.dtype)
datetime64[ns]
print (df.index)
DatetimeIndex(['2016-12-01', '2016-12-02', '2017-01-01', '2017-01-02',
'2017-01-10', '2017-01-31'],
dtype='datetime64[ns]', name='Timestamp', freq=None)
start_date = pd.to_datetime('2-1-2017', dayfirst=True)
end_date = pd.to_datetime('10-1-2017', dayfirst=True)
print (df[start_date:end_date])
Value
Timestamp
2017-01-02 10
2017-01-10 25
If some dates are not in index you need boolean indexing:
start_date = pd.to_datetime('3-1-2017', dayfirst=True)
end_date = pd.to_datetime('10-1-2017', dayfirst=True)
print (df[(df.index > start_date) & (df.index > end_date)])
Value
Timestamp
2017-01-31 50