I've got these 3 groups of data in range F2:G22 as below
(3 groups as minimal example, in reality many thousands of groups, and recurrent similar datasets expected in the future):
I need to number each group's rows sequentially, starting over at 1 at each new group.
The expected result would be like in range E1:E22.
I tried the following formula n cell C2 , then in cell D3:
=INDEX(IF(A2:A22="",COUNTIFS(B2:B22&A2:A22, B2:B22&A2:A22, ROW(B2:B22), "<="&ROW(B2:B22)),1))
In C2:
In D3:
That fixed partially the sequence issue, but there's still 2 issues I can't find remedy for.
1st remaining issue:
I'd prefer not having to manually do the C2 to D3 step each time I get new similar data (but would accomodate if there's no simple solution to this issue).
Is there a simple way to modify the formula to make it output the correct sequencing from C2 ?
2nd remaining issue:
At rows 7, 14 and 23 there still remain unecessary ending numbering for these intermediary rows in D7 , D14 , and D23:
I could only think of an extra manual step of filtering out the non-blank rows in Column A to fix this 2nd issue (i.e. Highlighting Column A > Data tab > Create Filter > Untick all > Tick Blanks > Copy All > Paste In new Sheet).
But would there be a way to do it in the same formula? I'm not seeing the way to add the proper filter or using another method in the formula.
Any help is greatly appreciated.
EDIT (Sorry for Forgotten Sample):
Formula Input A
Formula Input B
Formula Output 1
Formula Output 2
EXPECTED RESULT
rockinfreakshow
ztiaa
DATA
DATA BY GROUP
7
1
1
7
7
2
1
1
1
2
Element-1
Group-1
7
3
2
2
2
3
Element-2
Group-1
7
4
3
3
3
4
Element-3
Group-1
7
5
4
4
4
5
Element-4
Group-1
8
1
5
6
8
8
2
1
1
1
7
Element-1
Group-2
8
3
2
2
2
8
Element-2
Group-2
8
4
3
3
3
9
Element-3
Group-2
8
5
4
4
4
10
Element-4
Group-2
8
6
5
5
5
11
Element-5
Group-2
8
7
6
6
6
12
Element-6
Group-2
9
1
7
13
9
9
2
1
1
1
14
Element-1
Group-3
9
3
2
2
2
15
Element-2
Group-3
9
4
3
3
3
16
Element-3
Group-3
9
5
4
4
4
17
Element-4
Group-3
9
6
5
5
5
18
Element-5
Group-3
9
7
6
6
6
19
Element-6
Group-3
9
8
7
7
7
20
Element-7
Group-3
9
9
8
8
8
21
Element-8
Group-3
9
Can you try:
=INDEX(LAMBDA(y,z,
IF(LEN(z),COUNTIFS(y,y,ROW(z),"<="&ROW(z)),))
(LOOKUP(ROW(G2:G),FILTER(ROW(G2:G),BYROW(G2:G,LAMBDA(z,IF(z<>OFFSET(z,-1,0),row(z),0))))),G2:G))
You can simply use SCAN.
=SCAN(,G2:G,LAMBDA(a,c,IF(c="",,a+1)))
Sample sheet
Related
Using this table:
A
B
C
D
1
2
3
4
5
6
7
8
9
10
11
12
In Google Sheets if I do this here in column E:
={A1:B3;C1:D3}
Teremos:
E
F
1
2
5
6
9
10
3
4
7
8
11
12
But the result I want is this:
E
F
1
2
3
4
5
6
7
8
9
10
11
12
I tried multiple options with FLATTEN, but none of them returned what I wanted.
Well you can try:
=WRAPROWS(TOCOL(A1:D3),2)
You could try with MAKEARRAY
=MAKEARRAY(ROWS(A1:D3)*2,2,LAMBDA(r,c,INDEX(FLATTEN(A1:D3),c+(r-1)*2)))
GENERAL ANSWER
For you or anyone else: to do something similar but with a variable number of columns of origin or of destination, you can use this formula. Changing the range and amount of columns at the end of LAMBDA:
=LAMBDA(range,cols,MAKEARRAY(ROWS(range)*ROUNDUP(COLUMNS(range)/cols),cols,LAMBDA(r,c,IFERROR(INDEX(FLATTEN(range),c+(r-1)*cols)))))(A1:D3,2)
you can do:
={FLATTEN({A1:A3, C1:C3}), FLATTEN({B1:B3, D1:D3})}
for more columns, it could be automated with MOD
Table 1:
Position
Team
1
MCI
2
LIV
3
MAN
4
CHE
5
LEI
6
AST
7
BOU
8
BRI
9
NEW
10
TOT
Table 2
Position
Team
1
LIV
2
MAN
3
MCI
4
CHE
5
AST
6
LEI
7
BOU
8
TOT
9
BRI
10
NEW
Output I'm looking for is
Position difference = 10 as that is the total of the positional difference. How can I do this in excel/google sheets? So the positional difference is always a positive even if it goes up or down. Think of it as a league table.
Table 2 New (using formula to find positional difference):
Position
Team
Positional Difference
1
LIV
1
2
MAN
1
3
MCI
2
4
CHE
0
5
AST
1
6
LEI
1
7
BOU
0
8
TOT
2
9
BRI
1
10
NEW
1
Try this:
=IFNA(ABS(INDEX(A:B,MATCH(E2,B:B,0),1)-D2),"-")
Assuming that table 1 is at columns A:B:
I am attempting to read Aaron Hsu's thesis on A data parallel compiler hosted on the GPU, where I have landed at some APL code I am unable to fix. I've attached both a screenshot of the offending page (page number 74 as per the thesis numbering on the bottom):
The transcribed code is as follows:
d ← 0 1 2 3 1 2 3 3 4 1 2 3 4 5 6 5 5 6 3 4 5 6 5 5 6 3 4
This makes sense: create an array named d.
⍳≢d
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
This too makes sense. Count the number of elements in d and create a sequence of
that length.
⍉↑d,¨⍳≢d
0 1 2 3 1 2 3 3 4 1 2 3 4 5 6 5 5 6 3 4 5 6 5 5 6 3 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
This is slightly challenging, but let me break it down:
zip the sequence ⍳≢d = 1..27 with the d array using the ,¨ idiom, which zips the two arrays using a catenation.
Then, split into two rows using ↑ and transpose to get columns using ⍉
Now the biggie:
(⍳≢d)#(d,¨⍳≢d)⊢7 27⍴' '
INDEX ERROR
(⍳≢d)#(d,¨⍳≢d)⊢7 27⍴' '
Attempting to break it down:
⍳≢d counts number of elements in d
(d,¨⍳≢d) creates an array of pairs (d, index of d)
7 27⍴' ' creates a 7 x 27 grid: presumably 7 because that's the max value of d + 1, for indexing reasons.
Now I'm flummoxed about how the use of ⊢ works: as far as I know, it just ignores everything to the left! So I'm missing something about the parsing of this expression.
I presume it is parsed as:
(⍳≢d)#((d,¨⍳≢d)⊢(7 27⍴' '))
which according to me should be evaluated as:
(⍳≢d)#((d,¨⍳≢d)⊢(7 27⍴' '))
= (⍳≢d)#((7 27⍴' ')) [using a⊢b = b]
= not the right thing
As I was writing this down, I managed to fix the bug by sheer luck: if we increment d to be d + 1 so we are 1-indexed, the bug no longer manifests:
d ← d + 1
d
1 2 3 4 2 3 4 4 5 2 3 4 5 6 7 6 6 7 4 5 6 7 6 6 7 4 5
then:
(⍳≢d)#(d,¨⍳≢d)⊢7 27⍴' '
1
2 5 10
3 6 11
4 7 8 12 19 26
9 13 20 27
14 16 17 21 23 24
15 18 22 25
However, I still don't understand how this works! I presume the context will be useful
for others attempting to leave the thesis, so I'm going to leave the rest of it up.
Please explain what (⍳≢d)#(d,¨⍳≢d)⊢7 27⍴' ' does!
I've attached the raw screenshot to make sure I didn't miss something:
I'm happy to see that you found the the off-by-one error. It stems from Aaron Hsu working with index origin 0. If you set ⎕IO←0 then his code will work.
Some dyadic operators can take an array operand, giving the sequence OPERATOR operand argument, e.g. in -#(1 2 3)(4 5 6 7). This poses a problem because both the operand and the argument are arrays, and juxtaposition of arrays forms a new array with those arrays as elements by a process known as stranding. Compare:
(1 2 3)(4 5 6 7)
┌─────┬───┐
│1 2 3│4 5│
└─────┴───┘
However, in the case of the operator with its array operand, we want to "break" this strand so the left part can act as operand while the right part acts as argument. One way to break the stranding up is by applying a function to the argument, giving the sequence OPERATOR operand Function argument. Now, we don't actually need any transformation of the argument, so an identity function will do: -#(1 2 3)⊢(4 5 6 7).
As for what (⍳≢d)#(d,¨⍳≢d)⊢7 27⍴' ' actually does:
7 27⍴' ' creates a blank matrix.
(⍳≢d) are indices to insert into specified slots in the matrix.
#(d,¨⍳≢d) indicates at which locations in the matrix the above should replace the existing values
⊢ serves solely to separate (d,¨⍳≢d) from 7 27⍴' '. The code could also have been written as ((⍳≢d)#(d,¨⍳≢d))7 27⍴' ' with parentheses serving to "bind" the operand to the operator.
I have a data in "A" column like as below starting from A2 cell.
7cz2
6789efg
abc890
34l1
78kk
88
63
What I need is
7
2
6
7
8
9
8
9
0
3
4
7
8
8
8
6
3.
I do have applied =VALUE(REGEXREPLACE(A2,"\D+", "")) but is is giving me data like
72
6789
890
34
78.
Is there a way to solve this.Thanks in advance.
Use 2 formulas.
join and replace letters in cell C1:
=REGEXREPLACE(JOIN("",A1:A7),"[A-Za-z]","")
split by symbol:
=TRANSPOSE(REGEXEXTRACT(C1,REPT("(.)",LEN(C1))))
The result:
Currently I have a scope that pulls back my records in the following standard order:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
and it is converted in to a html block that's laid out as so:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
I want a html block that's laid out like so:
1 6 11 16
2 7 12
3 8 13
4 9 14
5 10 15
So I think I need to pull the records back in the following order - (records offset by the number of rows)
1 6 11 16 2 7 12 3 8 13 4 9 14 5 10 15
Any idea what is the neatest way to do this in Rails / ActiveRecord?
You must know how many columns you want to eventually render. I think this must work for you:
columns = 5
MyModel.order(:id).in_groups_of(columns).transpose
And you get an array of lines, each one with an array of records.
If you want an unique array you can add .flatten at the end.
Can't do that with AR ordering! You will have to do that in your controller or the view.