I have a table of rows which consist of zeros and numbers like this:
A B C D E F G H I J K L M N
0 0 0 4 3 1 0 1 0 2 0 0 0 0
0 1 0 1 4 0 0 0 0 0 1 0 0 0
9 5 7 9 10 7 2 3 6 4 4 0 1 0
I want to calculate an average of the numbers including zeros, but starting from the first nonzero value and put it into column after tables end. E.g. for the first row first value is 4, so average - 11/11; for the second - 7/13; the last one is 67/14.
How could I using excel formulas do this? Probably OFFSET with nested IF?
This still needs to be entered as an array formula (ctrl-shift-enter) but it isn't volatile:
=AVERAGE(INDEX(($A2:$O2),MATCH(TRUE,$A2:$O2<>0,0)):$O2)
or, depending on location:
=AVERAGE(INDEX(($A2:$O2);MATCH(TRUE;$A2:$O2<>0;0)):$O2)
The sum is the same no matter how many 0's you include, so all you need to worry about is what to divide it by, which you could determine using nested IFs, or take a cue from this: https://superuser.com/questions/671435/excel-formula-to-get-first-non-zero-value-in-row-and-return-column-header
Thank you, Scott Hunter, for good reference.
I solved the problem using a huge formula, and I think it's a bit awkward.
Here it is:
=AVERAGE(INDIRECT(CELL("address";INDEX(A2:O2;MATCH(TRUE;INDEX(A2:O2<>0;;);0)));TRUE):O2)
Related
2D array of 1s and 0s. How to label every group of 1s with a unique number?
I’m stuck on this problem for a while now. 1s can be grouped vertically, horizontally and diagonally. How can you go about solving this? For example,
0 0 1 1 0
0 1 1 0 0
0 0 0 0 1
0 0 0 1 0
Should be transformed to
0 0 x x 0
0 x x 0 0
0 0 0 0 y
0 0 0 y 0
x, y can be any unique numbers.
Appreciate it.
Here is what I have so far for iterative: https://i.imgur.com/oCmYC02.png
But the result is a bit off because it only checks for immediate adjacent 1's: https://i.imgur.com/DAtTBmM.png
Anyone have any idea how to fix this?
I'd do it like this:
Scan 2D array sequentially, row by row, column by column
If 1 found, use variation of the flood fill algorithm, which moves in 8 directions instead of 4, from that starting point (see normal 4-direction algorithm at https://en.wikipedia.org/wiki/Flood_fill), since you have diagonal example with "y", each time using new filler number.
Repeat 1 and 2 until no more ones left.
I want to pick out the elements which are
(2*pi*k),
where k=0,1,2,3... which means integer, and fill them (i1) into the other matrix.
But my problem is, I don't know how to make "k" be row. (By the way, dividends and divisors are float, so I need to find the approximations and see them as 2*pi*k).
My code, only can find the elements which are (2*pi*k), but can't order them like if k=1, then it will be put into k=1 row; if k=2, then the element should be put into k=2 row.
For example,
A = [2*pi 6 3 4;0.5*pi 0 2;3.1 7 4 8;2*pi 7 2 9;2.6 4*pi 6*pi 0]
I want the output to be
B = [0 2*pi 4*pi 6*pi;0 2*pi NaN NaN;NaN 2*pi NaN NaN]
This is my code:
k=0;
for m=380:650;
for n=277:600;
if abs((rem(abs(i(m,n)),(2*PI)))-(PI))>=3.11;
k=k+1;
B(m,k)=i1(m,n);
end
end
k=0;
end
It can find what I want but they seem not to be ordered the way I want.
As others, I'm a bit unsure what you want. Here's how I understood it and would code it:
check whether (2*pi*k) is contained in A, you want a numerical approach
output binary result
here's the code:
testPI=#(k) (2*pi*k); %generates 2*pi*k, where k is up to the user
A = [2*pi 6 3 4;0.5*pi 0 2 0;3.1 7 4 8;2*pi 7 2 9;2.6 4*pi 6*pi 0]; %A from example (fixed dimension error)
ismember(A,f(1:10)) %test if k=1:10 is contained in A
ans =
5×4 logical array
1 0 0 0
0 0 0 0
0 0 0 0
1 0 0 0
0 1 1 0
Adapt 1:10 to any value you'd like. Of course this only works if k is within reasonable range, otherwise this approach is suboptimal
I'm trying to remove the rows which has duplicates in sequence. I have only 2 possible values which are 0 and 1. I have nXm which n shows possible number of bits and m is not important for my question. My goal is to find an matrix which is nX(m-a). The rows a which has the property which includes duplicates in sequence. For example:
My matrix is :
A=[0 1 0 1 0 1;
0 0 0 1 1 1;
0 0 1 0 0 1;
0 1 0 0 1 0;
1 0 0 0 1 0]
I want to remove the rows has t duplicates in sequence for 0. In this question let's assume t is 3. So I want the matrix which:
B=[0 1 0 1 0 1;
0 0 1 0 0 1;
0 1 0 0 1 0]
2nd and 5th rows are removed.
I probably need to use diff.
So you want to remove rows of A that contain at least t zeros in sequence.
How about a single line?
B = A(~any(conv2(1,ones(1,t),2*A-1,'valid')==-t, 2),:);
How this works:
Transform A to bipolar form (2*A-1)
Convolve each row with a sequence of t ones (conv2(...))
Keep only rows for which the convolution does not contain -t (~any(...)). The presence of -t indicates a sequence of t zeros in the corresponding row of A.
To remove rows that contain at least t ones, just change -t to t:
B = A(~any(conv2(1,ones(1,t),2*A-1,'valid')==t, 2),:);
Here is a generalized approach which removes any rows which has given number of consecutive duplicates (not just zero. could be any number).
t = 3;
row_mask = ~any(all(~diff(reshape(im2col(A,[1 t],'sliding'),t,size(A,1),[]))),3);
out = A(row_mask,:)
Sample Run:
>> A
A =
0 1 0 1 0 1
0 0 1 5 5 5 %// consecutive 3 5's
0 0 1 0 0 1
0 1 0 0 1 0
1 1 1 0 0 1 %// consecutive 3 1's
>> out
out =
0 1 0 1 0 1
0 0 1 0 0 1
0 1 0 0 1 0
How about an approach using strings? This is certainly not as fast as Luis Mendo's method where you work directly with the numerical array, but it's thinking a bit outside of the box. The basis of this approach is that I consider each row of A to be a unique string, and I can search each string for occurrences of a string of 0s by regular expressions.
A=[0 1 0 1 0 1;
0 0 0 1 1 1;
0 0 1 0 0 1;
0 1 0 0 1 0;
1 0 0 0 1 0];
t = 3;
B = sprintfc('%s', char('0' + A));
ind = cellfun('isempty', regexp(B, repmat('0', [1 t])));
B(~ind) = [];
B = double(char(B) - '0');
We get:
B =
0 1 0 1 0 1
0 0 1 0 0 1
0 1 0 0 1 0
Explanation
Line 1: Convert each line of the matrix A into a string consisting of 0s and 1s. Each line becomes a cell in a cell array. This uses the undocumented function sprintfc to facilitate this cell array conversion.
Line 2: I use regular expressions to find any occurrences of a string of 0s that is t long. I first use repmat to create a search string that is full of 0s and is t long. After, I determine if each line in this cell array contains this sequence of characters (i.e. 000....). The function regexp helps us perform regular expressions and returns the locations of any matches for each cell in the cell array. Alternatively, you can use the function strfind for more recent versions of MATLAB to speed up the computation, but I chose regexp so that the solution is compatible with most MATLAB distributions out there.
Continuing on, the output of regexp/strfind is a cell array of elements where each cell reports the locations of where we found the particular string. If we have a match, there should be at least one location that is reported at the output, so I check to see if any matches are empty, meaning that these are the rows we don't want to remove. I want to turn this into a logical array for the purposes of removing rows from A, and so this is wrapped with a cellfun call to determine the cells that are empty. Therefore, this line returns a logical array where a 0 means that remove this row and a 1 means that we don't.
Line 3: I take the logical array from Line 2 and invert it because that's what we really want. We use this inverted array to index into the cell array and remove those strings.
Line 4: The output is still a cell array, so I convert it back into a character array, and finally back into a numerical array.
I would like to enumerate all the combinations (tuples of values) of 3 or more finite-valued variables which satisfy a given condition. In math notation:
For example (inspired by Project Euler problem 9):
The truth tables for two variables at a time are easy enough:
a ∘.≤ b
1 1 1 1
0 1 1 1
0 0 1 1
b ∘.≤ c
1 1 1 1 1
0 1 1 1 1
0 0 1 1 1
0 0 0 1 1
After much head-scratching, I managed to combine them, by computing the ∧ of every 4-valued row of the former with each 4-valued column of the latter, and disclosing (⊃) on the correct axis, between 1 and 2:
⎕← tt ← ⊃[1.5] (⊂[2] a ∘.≤ b) ∘.∧ (⊂[1] b ∘.≤ c)
1 1 1 1 1
0 1 1 1 1
0 0 1 1 1
0 0 0 1 1
0 0 0 0 0
0 1 1 1 1
0 0 1 1 1
0 0 0 1 1
0 0 0 0 0
0 0 0 0 0
0 0 1 1 1
0 0 0 1 1
Then I could use its ravel to filter all possible tuples of values:
⊃ (,tt) / , a ∘., b ∘., c
1 1 1
1 1 2
1 1 3
1 1 4
1 1 5
1 2 2
1 2 3
...
3 3 5
3 4 4
3 4 5
Is this the best approach to this particular class of problems in APL?
Is there an easier or faster formula for this example, or for the general case?
More generally, comparing my (naïve?) array approach above to traditional scalar languages, I can see that I'm translating each loop into an additional dimension: 3 nested loops become a 3-rank truth table:
for c in 1..NC:
for b in 1..min(c, NB):
for a in 1..min(b, NA):
collect (a,b,c)
But in a scalar language one can effect optimizations along the way, for example breaking loops as soon as possible, or choosing the loop boundaries dynamically. In this case I don't even need to test for a ≤ b ≤ c, because it's implicit in the loop boundaries.
In this example both approaches have O(N³) complexity, so their runtime will only differ by a factor. But I'm wondering: how could I write the array solution in a more optimized way, if I needed to do so?
Are there any good books or online resources that address algorithmic issues or best practices in APL?
Here's an alternative approach. I'm not sure if it would run faster.
Following your algorithm for scalar languages, the possible values of c are
⎕IO←0
c←1+⍳NC
In the inner loops the values for b and a are
b←1+⍳¨NB⌊c
a←1+⍳¨¨NA⌊b
If we combine those
r←(⊂¨¨¨a,¨¨¨b),¨¨¨c
we get a nested array of (a,b,c) triplets which can be flattened and rearranged in a matrix
r←∊r
(((⍴r)÷3),3)⍴r
ADD:
Morten Kromberg sent me the following solution. On Dyalog APL it's ~ 30 times more efficient than the one above:
⎕IO←1
AddDim←{0≡⍵:⍪⍳⍺ ⋄ n←0⌈⍺-x←¯1+⊢/⍵ ⋄ (n⌿⍵),∊x+⍳¨n}
TTable←{⊃AddDim/⌽0,⍵}
TTable 3 4 5
I would like to find out the number of different combinations of non-negative numbers(can be any number, it is not fixed) such that its total equals to the sum that is provided.
for example : I have 3 numbers and i want to find the different combinations of numbers such that the sum is 4. the value of num starts from 0. no negative numbers.
For 3 numbers that sum to 4, the combinations are
2 0 2
2 2 0
0 2 2
0 1 3
3 1 0
0 3 1
1 0 3
1 3 0
3 0 1
0 0 4
4 0 0
0 4 0
2 1 1
1 2 1
1 1 2
I saw this as an example : Finding the total number combinations for an integer using three numbers
But the problem is it only uses three numbers.
Any algorithm or code will be useful. Thanks.
You can view this as the number of ways to put s indistinguishable coins in n distinguishable jars. (In the example, s=4 and n=3).
As explained here, that is C(n+s-1,s-1), which gives 15 in the example.
If order does not matter and 0 counts too, like in example link, then
n=total+1
k=number-1
binomial(k+n-1,k) #combinations whith reptetitions
or
binomial(number+total-1,number-1)
If you represent number 5 as
1+1+1+1+1
and have to find number of sums sums whith 3 integers
You can see that you have to do 2 slices out of 6 calculating combinations whit repetitions.