I need to create a ranking row index as the example below, based on the average on the last three months and the client ID column, the ranking index needs to be unique to each client:
Ranking Index Client ID Month 3 Month 2 Month 1 Avg
1 Client 2 6 5 3 4,66
1 Client 1 4 6 2 4
2 Client 1 5 2 1 2,66
2 Client 2 1 5 2 2,66
3 Client 2 4 2 1 2,33
3 Client 1 1 3 2 2
you need groupby the column client ID and rank the column Avg, using the parameter ascending=False according to your expected output.
with a data example, you have
df = pd.DataFrame({'clientID':list('baabba'), 'Avg':[4.66,4,2.66,2.66,2.33,2]})
# create the column ranking
df['ranking'] = df.groupby('clientID')['Avg'].rank(ascending=False)
print(df)
clientID Avg ranking
0 b 4.66 1.0
1 a 4.00 1.0
2 a 2.66 2.0
3 b 2.66 2.0
4 b 2.33 3.0
5 a 2.00 3.0
Related
I have this df
d={}
d['id']=['1','1','1','1','1','1','1','1','2','2','2','2','2','2','2','2']
d['qty']=[5,5,5,5,5,6,5,5,1,1,2,2,2,3,5,8]
I would like to create a column that is going to have the following non-equal value of column qty. Meaning that if qty is equal to 5 and its next row is 5 I am going to skip it and look until I find next value not equal to 5, In my case it is 6. And all this should be grouped by id
Here is the desired dataframe.
d['id']=['1','1','1','1','1','1','1','1','2','2','2','2','2','2','2','2']
d['qty']=[5,5,5,5,5,6,5,5,1,1,2,2,2,3,5,8]
d['qty2']=[6,6,6,6,6,5,'NAN','NAN',2,2,3,3,3,5,8,'NAN']
Any help is very much appreciated
You can groupby.shift, mask the identical values, and groupby.bfill:
# shift up per group
s = df.groupby('id')['qty'].shift(-1)
# keep only the different values and bfill per group
df['qty2'] = s.where(df['qty'].ne(s)).groupby(df['id']).bfill()
output:
id qty qty2
0 1 5 6.0
1 1 5 6.0
2 1 5 6.0
3 1 5 6.0
4 1 5 6.0
5 1 6 5.0
6 1 5 NaN
7 1 5 NaN
8 2 1 2.0
9 2 1 2.0
10 2 2 3.0
11 2 2 3.0
12 2 2 3.0
13 2 3 5.0
14 2 5 8.0
15 2 8 NaN
Here's the dataset I've got. Basically I would like to create a column containing the sum of the values before the date (which means the sum of the values that is above the row) within the same group. So the first row of each group is supposed to be always 0.
group
date
value
1
10/04/2022
2
1
12/04/2022
3
1
17/04/2022
5
1
22/04/2022
1
2
11/04/2022
3
2
15/04/2022
2
2
17/04/2022
4
The column I want would look like this.
Could you give me an idea how to create such a column?
group
date
value
sum
1
10/04/2022
2
0
1
12/04/2022
3
2
1
17/04/2022
5
5
1
22/04/2022
1
10
2
11/04/2022
3
0
2
15/04/2022
2
3
2
17/04/2022
4
5
You can try groupby.transform and call Series.cumsum().shift()
df['sum'] = (df
# sort the dataframe if needed
.assign(date=pd.to_datetime(df['date'], dayfirst=True))
.sort_values(['group', 'date'])
.groupby('group')['value']
.transform(lambda col: col.cumsum().shift())
.fillna(0))
print(df)
group date value sum
0 1 10/04/2022 2 0.0
1 1 12/04/2022 3 2.0
2 1 17/04/2022 5 5.0
3 1 22/04/2022 1 10.0
4 2 11/04/2022 3 0.0
5 2 15/04/2022 2 3.0
6 2 17/04/2022 4 5.0
I have this dataframe.
from pandas import DataFrame
import pandas as pd
df = pd.DataFrame({'name': ['A','D','M','T','B','C','D','E','A','L'],
'id': [1,1,1,2,2,3,3,3,3,5],
'rate': [3.5,4.5,2.0,5.0,4.0,1.5,2.0,2.0,1.0,5.0]})
>> df
name id rate
0 A 1 3.5
1 D 1 4.5
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
5 C 3 1.5
6 D 3 2.0
7 E 3 2.0
8 A 3 1.0
9 L 5 5.0
df = df.groupby('id')['rate'].mean()
what i want is this:
1) find mean of every 'id'.
2) give the number of ids (length) which has mean >= 3.
3) give back all rows of dataframe (where mean of any id >= 3.
Expected output:
Number of ids (length) where mean >= 3: 3
>> dataframe where (mean(id) >=3)
>>df
name id rate
0 A 1 3.0
1 D 1 4.0
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
5 L 5 5.0
Use GroupBy.transform for means by all groups with same size like original DataFrame, so possible filter by boolean indexing:
df = df[df.groupby('id')['rate'].transform('mean') >=3]
print (df)
name id rate
0 A 1 3.5
1 D 1 4.5
2 M 1 2.0
3 T 2 5.0
4 B 2 4.0
9 L 5 5.0
Detail:
print (df.groupby('id')['rate'].transform('mean'))
0 3.333333
1 3.333333
2 3.333333
3 4.500000
4 4.500000
5 1.625000
6 1.625000
7 1.625000
8 1.625000
9 5.000000
Name: rate, dtype: float64
Alternative solution with DataFrameGroupBy.filter:
df = df.groupby('id').filter(lambda x: x['rate'].mean() >=3)
I have a pandas dataframe containing retail sales data which shows the total number of a product sold each week and the stock left at the end of the week. Unfortunately, the dataset only shows a row when a product has been sold and the stock left changes.
I would like to bulk out the dataset so that for each week there is a line for each product being sold. I've shown an example of this below - how can this be done?
As-Is:
Week Product Sold Stock
1 1 1 10
1 2 1 10
1 3 1 10
2 1 2 8
2 3 3 7
To-Be:
Week Product Sold Stock
1 1 1 10
1 2 1 10
1 3 1 10
2 1 2 8
2 2 0 10
2 3 3 7
Create a dataframe using product from itertools with all the combinations of both columns 'Week' and 'Product' and use merge with your original data. Let's say your dataframe is called dfp:
from itertools import product
new_dfp = (pd.DataFrame(list(product(dfp.Week.unique(), dfp.Product.unique())),columns=['Week','Product'])
.merge(dfp,how='left'))
You get the missing row in new_dfp:
Week Product Sold Stock
0 1 1 1.0 10.0
1 1 2 1.0 10.0
2 1 3 1.0 10.0
3 2 1 2.0 8.0
4 2 2 NaN NaN
5 2 3 3.0 7.0
Now you fillna on both column with different values:
new_dfp['Sold'] = new_dfp['Sold'].fillna(0).astype(int) # because no sold in missing rows
new_dfp['Stock'] = new_dfp.groupby('Product')['Stock'].fillna(method='ffill').astype(int)
To fill 'Stock', you need to groupby product and use the method 'ffill' to put the same value than last 'week'. At the end, you get:
Week Product Sold Stock
0 1 1 1 10
1 1 2 1 10
2 1 3 1 10
3 2 1 2 8
4 2 2 0 10
5 2 3 3 7
I am looking to create a new column in panda based on the value in the row. My sample data:
df=pd.DataFrame({"A":['a','a','a','a','a','a','b','b','b'],
"Sales":[2,3,7,1,4,3,5,6,9,10,11,8,7,13,14],
"Week":[1,2,3,4,5,11,1,2,3,4])
I want a new column "Last3WeekSales" corresponding to each week, having the sum of sales for the previous 3 weeks.
NOTE: Shift() won't work here as data for some weeks is missing.
Logic which I thought:
Checking the week no. in each row, then summing up the data from w-1, w-2, w-3.
Output required:
A Week Last3WeekSales
0 a 1 0
1 a 2 2
2 a 3 5
3 a 4 12
4 a 5 11
5 a 11 0
6 b 1 0
7 b 2 5
8 b 3 11
9 b 4 20
Use groupby, shift and rolling:
df['Last3WeekSales'] = df.groupby('A')['Sales']\
.apply(lambda x: x.shift(1)
.rolling(3, min_periods=1)
.sum())\
.fillna(0)
Output:
A Sales Week Last3WeekSales
0 a 2 1 0.0
1 a 3 2 2.0
2 a 7 3 5.0
3 a 1 4 12.0
4 a 4 5 11.0
5 a 3 6 12.0
6 b 5 1 0.0
7 b 6 2 5.0
8 b 9 3 11.0
you can use pandas.rolling_sum to sum over 3 last values, and shift(n) to shift your column by n times (1 in your case).
if we suppose you a column 'sales' with the sales of each week, the code would be :
df["Last3WeekSales"] = df.groupby("A")["sales"].apply(lambda x: pd.rolling_sum(x.shoft(1),3))