I have this sample of a data frame. I would like to find an average value for each assignment type value according to the target value. For example, for rows with pass result, I want to calculate their average C and T. There are about 5 Cs, and 3 Ts based on time. id_assigment differentiate between them you can see, but I want to find the average for each C and T for each class value. For example, average C with id 45 for Pass rows, for Fail rows,etc. I wonder how to calculate the averages?
Try pd.pivot_table
pd.pivot_table(df, columns=['id_assignemnt', 'assignemnt_type', 'result'], values='score')
Related
I'm looking for a solution for a problem I'm facing in Excel. This is my table simplified:
Every sale has an unique ID, but more people can have contributed to a sale. the column "name" and "share of sales(%)" show how many people have contributed and what their percentage was.
Sale_ID
Name
Share of sales(%)
1
Person A
100
2
Person B
100
3
Person A
30
3
Person C
70
Now I want to add a column to my table that shows the name of the person that has the highest share of sales percentage per Sales_ID. Like this:
Sale_ID
Name
Share of sales(%)
Highest sales
1
Person A
100
Person A
2
Person B
100
Person B
3
Person A
30
Person C
3
Person C
70
Person C
So when multiple people have contributed the new column shows only the one with the highest value.
I hope someone can help me, thanks in advance!
You can try this on cell D2:
=LET(maxSales, MAXIFS(C2:C5,A2:A5,A2:A5),
INDEX(B2:B5, XMATCH(A2:A5&maxSales,A2:A5&C2:C5)))
or just removing the LET since maxSales is used only one time:
=INDEX(B2:B5, XMATCH(A2:A5&MAXIFS(C2:C5,A2:A5,A2:A5),A2:A5&C2:C5))
On cell E2 I provided another solution via MAP/XLOOKUP:
=LET(maxSales, MAXIFS(C2:C5,A2:A5,A2:A5),
MAP(A2:A5, maxSales, LAMBDA(a,b, XLOOKUP(a&b, A2:A5&C2:C5, B2:B5))))
similarly without LET:
=MAP(A2:A5, MAXIFS(C2:C5,A2:A5,A2:A5),
LAMBDA(a,b, XLOOKUP(a&b, A2:A5&C2:C5, B2:B5)))
and here is the output:
Explanation
The trick here is to identify the max share of sales per each group and this can be done via MAXIFS(max_range, criteria_range1, criteria1, [criteria_range2, criteria2], ...). The size and shape of the max_range and criteria_rangeN arguments must be the same.
MAXIFS(C2:C5,A2:A5,A2:A5)
it produces the following output:
maxSales
100
100
70
70
MAXIFS will provide an output of the same size as criteria1, so it returns for each row the corresponding maximum sales for each Sale_ID column value.
It is the array version equivalent to the following formula expanding it down:
MAXIFS($C$2:$C$5,$A$2:$A$5,A2)
INDEX/XMATCH Solution
Having the array with the maximum Shares of sales, we just need to identify the row position via XMATCH to return the corresponding B2:B5 cell via INDEX. We use concatenation (&) to consider more than one criteria to find as part of the XMATCH input arguments.
MAP/XLOOKUP Solution
We use MAP to find for each pair of values (a,b) per row, of the first two MAP input arguments where is the maximum value found for that group and returns the corresponding Name column value. In order to make a lookup based on an additional criteria we use concatenation (&) in XLOOKUP first two input arguments.
Suppose I have the following datapoints. I would like to extract the cumulative percentage distribution of this set of the three largest values.
So first step would be to transform to 100% distribution and secondly summarise the three largest values of the new distribution.
Data
0.00
1.35
11.05
24.85
37.85
15.40
6.95
1.65
0.25
I can calculate the individual percentage point with a simple datapoint / sum of datapoints per row and use =LARGE 1,2,3 on the new column to sum up the values. However, the challenge is to make all calculations in a single cell and just return just the final value.
In this case, the target value would be: 0.2494 + 0.3804 + 0.1548 = 0.7849 or 78.48%
Thanks for the help
Wrap a LARGE in SUMPRODUCT:
=SUMPRODUCT(LARGE(A2:A10,{1,2,3}))/SUM(A2:A10)
I want to find the row and column names of the n-th highest values in a 2D array in Excel.
My array has a header row (the Coins) and a header column (the Markets). The data itself displays if a coin is supported on the market and if so what the approximate return of investment (ROI) will be in percent.
Example
An example of the array could look like this:
ROI
Coin A
Coin B
Coin C
Market 1
N/A
7.8%
5.7%
Market 2
0.4%
6.8%
N/A
Market 3
0.45%
7.6%
12.3%
Pay attention: So some values are also set to N/A (or is there a better way to display that a market doesn't support a specific coin? I don't want to enter 0% as it makes it harder to spot is a coin is supported by the market. I also don't want to leave that field blank because then I don't know if I already checked that market for that coin.)
Preferred output
The output for the example table from above with n=3 should then look like this (from high ROI to low):
Coin
Market
ROI
C
3
12.3%
B
1
7.8%
A
3
0.45%
Requirements
Each coin must only be shown once. So, for example, Coin B must not be listed twice in the Top3 output (once for Market 1: 7.8% and once for Market 3: 7.6%)
What I tried
So I thought about how to split up that problem into smaller parts. I think, it will come to these main parts:
find header/row name
here I found something to find the column name for the highest value per row but I wasn't able to adapt it to a working solution for a 2D array
find max in 2D array
here they describe to find the max value in a 2D array but not how to find the n-th highest values
find n-th highest values
here is a good explanation on how to find the highest n values of a 1D array but not how to apply that for a 2D array
only include each coin once
So I really tried to solve this myself but I struggle with adding these different parts together.
I have a data frame
date object price
190403 a 1
190405 a 23
190506 b -4
190507 d 56
I want to get a date of a column having a maximum Price i.e 190507
expected output
190507
For scalar output, always one max date value use Series.idxmax with convert date to index by DataFrame.set_index:
df.set_index('date')['price'].idxmax()
If want all max values in Series use boolean indexing and compare all values by max, DataFrame.loc is fir also filtering date column:
df.loc[df['price'].eq(df['price'].max()), 'date']
You can subset the df so that you only have the row where the price column is at its maximum, and then choose the date column:
df[df.price==df.price.max()].date
I'd like to accomplish the following task. There are three columns of data. Column A represents price, where the sum needs to be kept under $100,000. Column B represents a value. Column C represents a name tied to columns A & B.
Out of >100 rows of data, I need to find the highest 8 values in column B while keeping the sum of the prices in column A under $100,000. And then return the 8 names from column C.
Can this be accomplished?
EDIT:
I attempted the Solver solution w/ no luck. 200 rows looks to be the max w/ Solver, and that is what I'm using now. Here are the steps I've taken:
Create a column called rank RANK(B2,$B$2:$B$200) (used column D -- what is the purpose of this?)
Create a column called flag just put in zeroes (used column E)
Create 3 total cells total_price (=SUM(A2:A200)), total_value (=SUM(B2:B200)) and total_flag (=(E2:E200))
Use solver to minimize total_value (shouldn't this be maximize??)
Add constraints -Total_price<=100000 -Total_flag=8 -Flag cells are binary
Using Simplex LP, it simply changes the flags for the first 8 values. However, the total price for the first 8 values is >$100,000 ($140k). I've tried changing some options in the Solver Parameters as well as using different solving methods to no avail. I'd like to post an image of the parameter settings, but don't have enough "reputation".
EDIT #2:
The first 5 rows looks like this, price goes down to ~$6k at the bottom of the table.
Price Value Name Rank Flag
$22,538 42.81905675 Blow, Joe 1 0
$22,427 37.36240932 Doe, Jane 2 0
$17,158 34.12127693 Hall, Cliff 3 0
$16,625 33.97654031 Povich, John 4 0
$15,631 33.58212402 Cow, Holy 5 0
I'll give you the solver solution as a starting point. It involves the creation of some extra columns and total cells. Note solver is limited in the amount of cells it can handle but will work with 100 anyway.
Create a column called rank RANK(B2,$B$2:$B$100)
Create a column called flag just put in zeroes
Create 3 total cells total_price, total_value and total_flag
Use solver to minimize total_value
Add constraints
-Total_price<=100000
-Total_flag=8
-Flag cells are binary
This will flag the rows you want and you can grab the names however you want.