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
Related
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
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 wish to make a formula to sum up the value with 2 criteria, example show as below:-
A B C D E
1 1-Apr 2-Apr 3-Apr 4-Apr
2 aa 1 4 7 10
3 bb 2 5 8 11
4 cc 3 6 9 12
5
6 Criteria 1 bb
7 Range start 2-Apr-16
8 Range End 4-Apr-16
9 Total sum #VALUE!
tried formula
1 SUMIF(A2:A4,C6,INDEX(B2:E4,0,MATCH(C7,B1:E1,0)))
* Only return 1 cell value
2 SUMIF(A2:A4,C6,INDEX(B2:E4,0,MATCH(">="&C7,B1:E1,0)))
* Showed N/A error
3 SUMIFS(B2:E4,A2:A4,C6,B1:E1,">="&C7,B1:E1,"<="&C8)
* Showed #Value error
Hereby I attached a link of picture for better understanding :
Can anyone help me on the formula?
I figured out the solution with step evaluation:
=SUMIF(B1:F1,">="&C7,INDEX(B2:F4,MATCH(C6,A2:A4,0),0)) -
SUMIF(B1:F1,">"&C8,INDEX(B2:F4,MATCH(C6,A2:A4,0),0))
I have two dataframes df1 and df2 I want to join. Their indexes are not the same and they don't have any common columns. What I want is to join them based on the order of the rows, i.e. join the first row of df1 with the first row of df2, the second row of df1 with the second row of df2, etc.
Example:
df1:
'A' 'B'
0 1 2
1 3 4
2 5 6
df2:
'C' 'D'
0 7 8
3 9 10
5 11 12
Should give
'A' 'B' 'C' 'D'
0 1 2 7 8
3 3 4 9 10
5 5 6 11 12
I don't care about the indexes in the final dataframe. I tried reindexing df1 with the indexes of df2 but could not make it work.
You could assign to df1 index of df2 and then use join:
df1.index = df2.index
res = df1.join(df2)
In [86]: res
Out[86]:
'A' 'B' 'C' 'D'
0 1 2 7 8
3 3 4 9 10
5 5 6 11 12
Or you could do it in one line with set_index:
In [91]: df1.set_index(df2.index).join(df2)
Out[91]:
'A' 'B' 'C' 'D'
0 1 2 7 8
3 3 4 9 10
5 5 6 11 12
Try concat:
pd.concat([df1.reset_index(), df2.reset_index()], axis=1)
The reset_index() calls make the indices the same, then, concat with axis=1 simply joins horizontally.
I guess you can try to join them (doing this it performs the join on the index, which is the same for the two DataFrame due to reset_index):
In [18]: df1.join(df2.reset_index(drop=True))
Out[18]:
'A' 'B' 'C' 'D'
0 1 2 7 8
1 3 4 9 10
2 5 6 11 12
I have written this formula below. I do not know the correct part of this formula that will add the numbers I have in Column AB2:AB552. As it is, this formula is counting the number of cells in that range that has numbers in it, but I need it to total those numbers as my final result. Any help would be great.
=COUNTIFS(Cases!B2:B552,"1",Cases!G2:G552,"c*",Cases!X2:X552,"No",**Cases!AB2:AB552,">0"**)
Assuming you don't actually need the intermediate counts, the sumifs function should give you the final result:
=SUMIFS(Cases!AB2:AB552,Cases!B2:B552,1,Cases!G2:G552,"c",Cases!X2:X552,"No",Cases!AB2:AB552,">0")
Testing this with some limited data:
Row B G X AB
2 2 a No 10
3 1 c No 24
4 2 c No 4
5 1 c No 0
6 1 a Yes 9
7 2 c No 12
8 2 c No 6
9 2 b No 0
10 1 b No 0
11 1 a No 10
12 2 c No 6
13 1 c No 20
14 1 c No 4
15 1 b Yes 22
16 1 b Yes 22
the formula above returned 48, the sum of AB3, AB13, and AB14, which were the only rows matching all 4 criteria