dfply | dplyr-style piping operations for pandas dataframes
kandi X-RAY | dfply Summary
kandi X-RAY | dfply Summary
dplyr-style piping operations for pandas dataframes
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dfply Key Features
dfply Examples and Code Snippets
mask = md.groupby('COMB', group_keys=False).apply(lambda x: x['FROM'].isin(x['TO']))
md = md.assign(Type=np.where(mask,"R","O"))
print (md)
COMB FROM TO Type
0 PNR1 MAA BLR R
1 PNR1 BLR MAA R
2 PNR11 DEL MAA O
3
df.groupby('day_month').apply(mp_va)
pd.concat([mp_va(group) for name, group in df.groupby('day_month')])
@dfpipe
def woe_iv(df, variable):
return df >> group_by(X[variable]) >> summarize(COUNT=X[variable].count())
banks = pd.read_excel('banks.xlsx')
>> print(banks >> woe_iv('marital'))
marital COUNT
0 marri
import dfply
from dfply import *
@dfpipe
def woe_iv(df,variable):
step1 = df>>group_by(X.variable)>>summarize(COUNT=X.variable.count())
return step1
banks>>woe_iv(X.job)
@dfpipe
def woe_
def df_apply(col_fn, *col_names):
def inner_fn(df):
cols = [df[col] for col in col_names]
return col_fn(*cols)
return inner_fn
new_table = (
raw_data
.assign(area=df_apply(calc_c
data_test = data[data.year == 1997]
df['AVG_SALDO'] = df['SALDO'] / df.groupby('NROCUENTA').SALDO.transform('count')
Out[1112]:
NROCUENTA SALDO AVG_SALDO
0 210-1-388 159.20 79.60
1 210-1-388 159.20 79.60
2 210-1-1219 0.93 0.93
3 210-1-1
__all__ = ["echo", "print" #Modules name go here
from module import *
df['newCol'] = df.apply(lambda X: X.col1 - 2 if X.col2 == 'c' else X.col1 + 4, axis=1)
df
col1 col2 newCol
0 1 a 5
1 2 b 6
2 3 c 1
3 4 d 8
4 5 e 9
Community Discussions
Trending Discussions on dfply
QUESTION
I want to group by day so I can create a market profile for every day. This is my data frame. It's a usual OHLC data that goes from 2006-04-13 to 2021-06-14 and I add a day_month variable.
Python
...ANSWER
Answered 2021-Jun-21 at 10:40I think a good reference to understand this is the Group by: split-apply-combine user guide from the pandas doc.
There are several steps:
Splitting an object into groups
You already have this right:
df.groupby('day_month')
allows to split your global dataframe into smaller dataframes that have the sameday_month
value.Applying a function to each group independently
Here there are a number of options. You could
- aggregate the data: reduce it to a sum of the values in the group for example,
- filter the data: select some of the groups,
- transform the data: which operates on columns in your group and returns columns of the same shape, for example normalize values within each group by subtracting the average and dividing by the standard deviation
- apply a function to the data: this is a catch-all (thus often slower than the other operations) that just passes each group of data to a function.
Applying a function is usually done by calling
.agg(op)
,.filter(op)
,.transform(op)
,.apply(op)
on the GroupBy result, whereop
that specifies which operation to perform. This can be a string that names a pre-defined operation(mean
,sum
, etc.) or a function (such asmp_va
in your case). However you can not call directlyGroupBy.mp_va()
. This wouldn’t specify what kind of operation is being done.Combining the data
This is just concatenating all the group-wise results into a single dataframe. This work is already done by
agg
,apply
, and friends.
It seems here that your function applies to each group dataframe, and not to its columns individually. Therefore you need GroupBy.apply
.
Basically you can think of the following code
QUESTION
I have a dataset "banks" where if I do a groupby on a column name "jobs" to check counts in each category,I could find the following:
index jobs count 0 adnin. 478 1 blue-collar 946 2 entrepreneur 168 3 housemaid 112 4 management 969 5 retired 230 6 self-employed 183 7 services 417 8 student 84 9 technician. 768I've also added first 3 lines of the dataset I am using: age,job,marital,education,default,balance,housing,loan,contact,day,month,duration,campaign,pdays,previous,poutcome,y 30,unemployed,married,primary,no,1787,no,no,cellular,19,oct,79,1,-1,0,unknown,no 33,services,married,secondary,no,4789,yes,yes,cellular,11,may,220,1,339,4,failure,no 35,management,single,tertiary,no,1350,yes,no,cellular,16,apr,185,1,330,1,failure,no
My intention is to create a small function which I can use for other columns as well hence I tried to create a function using "dfply" package.
...ANSWER
Answered 2021-Apr-19 at 08:23Shameek Mukherjee, is this a correct interpretation and indentation of your example code? I'm not able to find any difference besides the indentation.
QUESTION
My data set is Churn_Modeling:
...ANSWER
Answered 2020-Dec-15 at 23:04Is this what you are trying to do?
QUESTION
Based on this example data :
...ANSWER
Answered 2020-Apr-21 at 01:51Some points
The example you gave is a bit simple and I believe make it a bit harder to think in a more generic case. I then generated random data for 30 days using numpy.
By seeing the link you sent, I think they're showing us "how many days is the latest day that current day is double of apart from current_day".
To show this explicitly I will use very verbose column names in pandas and
before calculating the metrics you want, I will build in the dataframe a reference list called days_current_day_is_double_of
wich will for each row(day) calculate a list of days which the current deaths_cum is double of the day deaths_cum.
This column later can be substituted for a simple np.where() operation every time you want to find this for a row, if you don't want to keep a reference list in the dataframe. I think it's clearer keeping it.
generating data
QUESTION
I have one requirement:
I have a dataframe "df_input" having 20M rows which includes trip details. columns are "vehicle-no", "geolocation","start","end". For each of the vehicle number there are multiple rows each having different geolocation for different trips.
Now I want to create a new dataframe df_final which will have only the first record for all of the vehicle-no. How can do that in efficient way?
I used something like below which is taking more than 5 hours to complete:
...ANSWER
Answered 2020-Mar-26 at 07:47I think this will work out
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