Hoping to sort (below left) by sector but distribute evenly (below right):
Name
Sector.
Name.
Sector
A
1
A
1
B
1
E
2
C
1
H
3
D
4
D
4
E
2
B
1
F
2
F
2
G
2
J
3
H
3
I
4
I
4
C
1
J
3
G
2
Real data is 70+ rows with 4 sectors.
I've worked around it manually but would love to figure out how to do it with a formula in excel.
Here's a more complete (and hopefully more accurate) idea - the carouselOrder is the column I'd like to generate via a formula.
guestID
guestSector
carouselOrder
1
1
1
2
1
5
3
1
9
4
1
13
5
2
2
6
2
6
7
2
10
8
2
14
9
3
3
10
3
7
11
3
11
12
2
18
13
1
17
14
1
20
15
1
23
16
2
21
17
2
24
18
2
27
19
1
26
20
1
29
21
1
30
22
1
31
23
3
15
24
3
19
25
3
22
26
3
25
27
3
28
28
1
32
29
4
4
30
4
8
31
4
12
32
4
16
When using Office 365 you can use the following in D2: =MOD(SEQUENCE(COUNTA(A2:A11),,0),4)+1
This create the repetitive counter of the sectors 1 to 4 to the total count of rows in your data.
In C2 use the following:
=BYROW(D2#,LAMBDA(x,
INDEX(
FILTER($A$2:$A$11,$B$2:$B$11=x),
SUM(--(D$2:x=x)))))
This filters the Names that equal the sector of mentioned row and indexes it to show only the result where the row in the filter result equals the count of the same sector (D2#) up to current row.
Let's try the following approach that doesn't require to create a helper column. I would like to explain first the logic to build the recurrence, then the excel formula that builds such recurrence.
If we sort the input data Name and Sector. by Sector. in ascending order, the new positions of the Name values (letters) can be calculated as follow (Table 1):
Name
Sector.Sorted
Position
A
1
1+4*0=1
B
1
1+4*1=5
C
1
1+4*2=9
E
2
2+4*0=2
F
2
2+4*1=6
G
2
2*4*2=10
H
3
3+4*0=3
J
3
3+4*1=7
D
4
4+4*0=4
I
4
4+4*1=8
The new positions of Name (letters) follows this pattern (Formula 1):
position = Sector.Sorted + groupSize * factor
where groupSize is 4 in our case and factor counts how many times the same Sector.Sorted value is repeated, starting from 0. Think about Sector.Sorted as groups, where each set of repeated values represents a group: 1,2,3 and 4.
If we are able to build the Position values we can sort Name, based on the new positions via SORTBY(array, by_array1) function. Check SORTBY documentation for more information how this function works.
Here is the formula to get the Name sorted in cell E2:
=LET(groupSize, 4, sorted, SORT(A2:B11,2), sName,
INDEX(sorted,,1),sSector, INDEX(sorted,,2),
seq0, SEQUENCE(ROWS(sSector),,0), mapResult,
MAP(sSector, seq0, LAMBDA(a,b, IF(b=0, "SAME",
IF(a=INDEX(sSector,b), "SAME", "NEW")))), factor,
SCAN(-1,mapResult, LAMBDA(aa,c,IF(c="SAME", aa+1,0))),
pos,MAP(sSector, factor, LAMBDA(m,n, m + groupSize*n)),
SORTBY(sName,pos)
)
Here is the output:
Explanation
The name sorted represents the input data sorted by Sector. in ascending order, i.e.: SORT(A2:B11,2). The names sName and sSector represent each column of sorted.
To identify each group we need the following sequence (seq0) starting from 0, i.e. SEQUENCE(ROWS(sSector),,0).
Now we need to identify when a new group starts. We use MAP function for that and the result is represented by the name mapResult:
MAP(sSector, seq0, LAMBDA(a,b, IF(b=0, "SAME",
IF(a=INDEX(sSector,b), "SAME", "NEW"))))
The logic is the following: If we are at the beginning of the sequence (first value of seq0), then returns SAME otherwise we check current value of sSector (a) against the previous one represented by INDEX(sSector,b) if they are the same, then we are in the same group, otherwise a new group started.
The intermediate result of mapResult is:
Name
Sector Sorted
mapResult
A
1
SAME
B
1
SAME
C
1
SAME
E
2
NEW
F
2
SAME
G
2
SAME
H
3
NEW
J
3
SAME
D
4
NEW
I
4
SAME
The first two columns are shown just for illustrative purpose, but mapResult only returns the last column.
Now we just need to create the counter based on every time we find NEW. In order to do that we use SCAN function and the result is stored under the name factor. This value represents the factor we use to multiply by 4 within each group (see Table 1):
SCAN(-1,mapResult, LAMBDA(aa,c,IF(c="SAME", aa+1,0)))
The accumulator starts in -1, because the counter starts with 0. Every time we find SAME, it increments by 1 the previous value. When it finds NEW (not equal to SAME), the accumulator is reset to 0.
Here is the intermediate result of factor:
Name
Sector Sorted
mapResult
factor
A
1
SAME
0
B
1
SAME
1
C
1
SAME
2
E
2
NEW
0
F
2
SAME
1
G
2
SAME
2
H
3
NEW
0
J
3
SAME
1
D
4
NEW
0
I
4
SAME
1
The first three columns are shown for illustrative purpose.
Now we have all the elements to build our pattern for the new positions represented with the name pos:
MAP(sSector, factor, LAMBDA(m,n, m + groupSize*n))
where m represents each element of Sector.Sorted and factor the previous calculated values. As you can see the formula in Excel represents the generic formula (Formula 1 see above). The intermediate result will be:
Name
Sector Sorted
mapResult
factor
pos
A
1
SAME
0
1
B
1
SAME
1
5
C
1
SAME
2
9
E
2
NEW
0
2
F
2
SAME
1
6
G
2
SAME
2
10
H
3
NEW
0
3
J
3
SAME
1
7
D
4
NEW
0
4
I
4
SAME
1
8
The previous columns are shown just for illustrative purpose. Now we have the new positions, so we are ready to sort based on the new positions for Name via:
SORTBY(sName,pos)
Update
The first MAP can be removed creating an array as input for SCAN that has the information of sSector and the index position to be used for finding the previous element. SCAN only allows a single array as input argument, so we can combine both information in a new array. This is the formula can be used instead:
=LET(groupSize, 4, sorted, SORT(A2:B11,2), sName,
INDEX(sorted,,1),sSector, INDEX(sorted,,2),
factor, SCAN(-1,sSector&"-"&SEQUENCE(ROWS(sSector),,0),
LAMBDA(aa,b, LET(s, TEXTSPLIT(b,"-"),item, INDEX(s,,1),
idx, INDEX(s,,2), IF(aa=-1, 0, IF(1*item=INDEX(sSector, idx), aa+1,0))))),
pos,MAP(sSector, factor, LAMBDA(m,n, m + groupSize*n)),
SORTBY(sName,pos)
)
We use inside of SCAN a LET function to calculate all required elements for doing the comparison as part of the calculation of the corresponding LAMBDA function. We extract the item and the idx position used to find previous element of sSector via:
1*item=INDEX(sSector, idx)
we are able to compare each element of sSector with previous one, starting from the second element of sSector. We multiply item by 1, because TEXTSPLIT converts the result to text, otherwise the comparison will fail.
Bacteria B replicates itself each 2 minutes, write a program that asks users to enter two numbers: the initial B bacteria number and a period of time (in minutes). Calculate and print out the total number of B bacteria after this period.
How to do. I did like below but the result is wrong. Because bateria replicates itself each 2 minutes then I multiply it with time.
b = int(input('How many B Bacterias are there? '))
t = int(input('How much time will we waits (minutes)? '))
r = b * 2 * t
print ('After',t,'minutes(s) we would have',r,'B Bacterias')
See the image here
You have b bacterias at the t time like this:
t b
0 b
1 b
2 b * 2
3 b * 2
4 b * 4
5 b * 4
6 b * 8
7 b * 8
8 b * 16
and so on.
So your formula is r = b * 2 ** math.floor(t/2) and the code will be:
import math
b = int(input('How many B Bacterias are there? '))
t = int(input('How much time will we waits (minutes)? '))
r = b * 2 ** math.floor(t/2)
print ('After',t,'minutes(s) we would have',r,'B Bacterias')
Your function to calculate the number of bacteria is wrong, this is a textbook example of an exponential function.
should be r = b * 2 ** (t/2)
You need to take the time t and divide it by the division time of 2 minutes to get the number of divisions that occurred.
By raising 2 to this number you get the total offspring of one bacterial cell and this multiplied by the initial number of bacteria is your answer
If have the following two columns with the values:
A B
1 1
2 2
3 3
4 4
How can I multiply and return the sum:
1* 1 + 1 * 2 + 1* 3 + 1 * 4 + 2 * 1 + 2 * 2 + ... 4 * 4
exactly like a nested FOR loop?
Similarly, how can I multiply and sum only when first column value is bigger than second:
2 * 1 + 3 * 1 + 3 * 2 + 4 * 1 + 4 * 2 + 4 * 3 ?
With formula I think what is required, assuming A1:E1 and A1:A4 as shown, may be:
(1) The sum of B1:E4 where B2 copied across and down is:
=$A2*B$1
Alternatively the array formula with layout as you show:
=SUM(MMULT(A2:A5,TRANSPOSE(B2:B5)))
answer 100, and
(2) in B2 copied across and down and then B2:E4 summed:
=IF($A2>B$1,$A2*B$1,0)
answer 35.
Once, one boy's teacher asked him to calculate the sum of numbers 1 through n.
the boy quickly answered, and his teacher made him another challenge. He asked him to calculate the sum of the digits of numbers 1 through n.
Input
Two space-separated integers 0 <= a <= b <= 109.
Output
The sum of the digits of numbers a through b.
Example
Input:
1 10
Output: 46
can someone explain what is meant by sum of the digits of numbers a to b?
from above, sum of {1 2 3 4 5 6 7 8 9 10 } is 55 , it is a well known Gaussian formula
but the output is 46!
if i count from 2 to 9, excluding the border numbers 1 and 10, the answer is 44 , still not 46
So what is meant by sum of digits of numbers?
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + (1 + 0)
Don't treat the 10 as the number 10, rather the digits 1 and 0
Say I have a list of 2 colours, Black and White. It's only possible to have 1 combination using these colours, because you can't have two of the same.
If I have 3 colours (Black, White and Red), there are 3 possible combination (Black+White, Black+Red, White+Red).
If I have 4 colours, there are 5 possible combinations and if I have 5 colours there are 10 possible combinations.
I'm trying to work out the relationship between the number of colours and the possible combinations. Here is some data:
Colours Combinations
0 0
1 0
2 1
3 3
4 5
5 10
6 14
You want the binomial coefficients.
The formula for the number of pairs from a set of size n is n * (n - 1) / 2.
Your values are incorrect. The correct values are:
n nCr (r=2)
2 1
3 3
4 6
5 10
6 15
This sequence is also known as the triangular numbers.
Read about Combinations
The formula to calculate the value is:
(n!)/(k!(n-k)!)
Where n is the total amount of possible colors and k is how many colors will you pick, so
"1 out of 3" = 3! / 1!*2! => 3*2*1/1*2*1 = 3
and so on...
Combinations of num_colors taken 2 at a time:
C(n, k) = n!/(k!*(n - k!))
C(0, 2) = C(1, 2) = 0 by definition in your case
C(2, 2) = 2!/(2!*0!) = 2!/2! = 1 (0! is usually 1)
C(6, 2) = 6!/(2!*4!) = 15 (is your 14 a mistake?)
This simplifies to n*(n - 1) / 2 when k = 2.