I was wondering how modulo works. I know how it works when the bigger number comes first, but not the opposite. I know that 7 % 3 = 1 as 3 goes up to 7 2 times and the remaining is 1. However, when it's 3 % 7. I have used the calculator it shows 3. Is this because 7 goes up to 3 zero times and the remaining is 3? Is that how it works?
Your reasoning is correct. Any time the divisor is larger than the dividend, the result of the modulo operation equals the dividend.
7*x + y = 3, x and y are int, and x >= 0,
what y = ?
yes, y = 3.
3 MOD 7 = 0R3
This is so because 3/7 is >0 but <1
Mod just means you take the remainder after performing the division. When you divide 3 by 7 you get 3= 0*7 + 3 which means that the remainder is 3.
7/3 -> 3 goes into 7 2 times, and 7 - (3x2) = 1, so your modulus is 1
3/7 -> 7 goes into 3 zero times, and 3 - 0 = 3, so here your modulus is 3
Related
I have a project I need to do for my online class and I can not figure out how to do this. So it is suppose to be a program where you put a base and an exponent.
Let's say your base was 4 and exponent was 3 it should give an answer that looks like this. 4 exponent 0 = 1 4 exponent 1 = 4 4 exponent 2 = 16 4 exponent 3 = 64.
I have searched for an Octave function that facilitates conditional merging of matrices but haven't one so far. My goal is to do this using vectors without looping. Here is an example of what I am trying to do.
A= [1 1
2 2
3 1
5 2];
B= [1 9
2 10];
I would like to get C as
C= [1 1 9
2 2 10
3 1 9
5 2 10];
Is there a function that takes A, B and the list of column(s) to join on and then produce C?
You can use the second output of ismember to find the occurrences of the second column of A in the first column of B and then use that to grab specific entries from the second column of B to construct C.
[~, inds] = ismember(A(:,2), B(:,1));
C = [A, B(inds,2)];
%// 1 1 9
%// 2 2 10
%// 3 1 9
%// 5 2 10
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 know that tensors have an apply method, but this only applies a function to each element. Is there an elegant way to do row-wise operations? For example, can I multiply each row by a different value?
Say
A =
1 2 3
4 5 6
7 8 9
and
B =
1
2
3
and I want to multiply each element in the ith row of A by the ith element of B to get
1 2 3
8 10 12
21 24 27
how would I do that?
See this link: Torch - Apply function over dimension
(Thanks to Alexander Lutsenko for providing it. I just moved it to the answer.)
One possibility is to expand B as follow:
1 1 1
2 2 2
3 3 3
[torch.DoubleTensor of size 3x3]
Then you can use element-wise multiplication directly:
local A = torch.Tensor{{1,2,3},{4,5,6},{7,8,9}}
local B = torch.Tensor{1,2,3}
local C = A:cmul(B:view(3,1):expand(3,3))
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