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.
I want to understand what the function coverage does to an IRange. for example the codes below:
ir <- IRanges (1:3, width = 3)
ir
IRanges object with 3 ranges and 0 metadata columns:
start end width
[1] 1 3 3
[2] 2 4 3
[3] 3 5 3
coverage (ir)
integer-Rle of length 5 with 5 runs
Lengths: 1 1 1 1 1
Values : 1 2 3 2 1
why the values repeats itself like 123 then 21
I figured it out.
The right answer is that we count the ranges covering each number starting from 1 till the last number in the last range.
for example
ir <- IRanges (4:6, width = 3)
first, we draw a plot for that IRange staring from 1 which is not included in any range and ending with 8 which is the boundry of the last range
second, we count the ranges of the Ir that covers each of these number from 0 to 8
count = c (0,0,0,1,2,3,2,1)
Rle (count)
numeric-Rle of length 8 with 6 runs
Lengths: 3 1 1 1 1 1
Values : 0 1 2 3 2 1
Here's the table:
Should not they have the same result mathematically? (the average score of the per column and per row average)
The missing cells mean that your cells aren't all weighted evenly.
For example, row 11 has only two cells 82.67 and 90. So for your row average for row 11 they are weighted much more heavily than in your column averages where they are 1/13 and 1/14 of a column instead of 1/2 of a row.
Try filling up all the empty cells with 0 and the averages should match.
Taking a more extreme version of Ruslan Karaev's example:
5 5 5 | 5
1 | 1 Average of Average of Rows = (5 + 1 + 0) / 3 = 2
0 | 0
-----
2 5 5
Average of Average of Columns = (2 + 5 + 5) / 3 = 4
Yes, for example, the following two expressions:
/ a + b X + Y \ / a + X b + Y \
( ----- + ----- ) ( ----- + ----- )
\ 2 2 / \ 2 2 /
------------------- -------------------
2 2
are indeed mathematically equivalent, both coming out to be (a + b + X + Y) / 4.
However, short of having enough sufficient precision to store values, you may find that rounding errors accumulate differently depending on the order of operations.
You can see this sort of effect in a much simpler example if you assume a 3-digit precision and divide one by three, then multiply the result of that by three again:
1 / 3 -> 0.333, 0.333 x 3 -> 0.999
Contrast that with doing the operations in the oppisite order:
1 x 3 = 3, 3 / 1 = 1
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
Suppose you have given dp table for string X = "AGGGCT" and string Y = "AGGCA"
m = length of X + 1
n = length of Y + 1
0 1 2 3 4 5
1 0 1 2 3 4
2 1 0 1 2 3
dp[m][n] = 3 2 1 0 1 2
4 3 2 1 1 2
5 4 3 2 1 2
6 5 4 3 2 2
and you want to reconstruct three strings as follows
string row1 = "AGGGCT" ;
string row2 = "||| | " ;
string row3 = "AGG-CA" ;
How to recontruct strings row1, row2 and row3, if possible post code in C/C++/Java.
I think this page can be a good starting point:
http://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Levenshtein_distance#Java
You have to make a few modifications, but the core idea should be to store in the "min" which case was choosed for a given (i,j), and before the return you can walk through the matrix backwards starting by distance[str1.length()][str2.length()] step-by-step. If in the steps the distances are the same you show a |, if they differ but stepping diagonal then it was a change step, otherwise if vertical/horizontal a remove/add.
You can store this "backwards" information in a string and later display it in a reverse order.