Hi All Fortran Lovers,
I am trying to write to a file which outputs three variables as
program main
integer N, u
parameter(u=20)
open (u, FILE='points.dat', STATUS='new')
do 10 i= 1, 100
write(u,100) i, i*2, i*5
10 continue
100 format (I5, I10, 9X, I10)
close(u)
print *,'COMPLETE!!'
end
Which Gives output (points.dat stripped file content):
1 2 5
2 4 10
3 6 15
4 8 20
5 10 25
6 12 30
7 14 35
8 16 40
9 18 45
10 20 50
11 22 55
12 24 60
...
...
...
...
...
99 198 495
100 200 500
|(This line added by the write statement)
But I want something like this:
1 2 5
2 4 10
3 6 15
4 8 20
5 10 25
6 12 30
7 14 35
8 16 40
9 18 45
10 20 50
11 22 55
12 24 60
...
...
...
...
...
99 198 495
100 200 500|(The cursor stop here)
i.e. No space at start of each line. The last line stops after printing '500'
I tried using Horizontal spacing using '1X' specifier but no success.
Add advance='no' in write statement. If the line is not the last one, write EOL:
do 10 i= 1, 100
write(u,100,advance='no') i, i*2, i*5
if (i.ne.100) write(u,*)
10 continue
Edit: I see it now, it seems that the fortran program will add EOL to the end of file anyway. Then you have to use external programs to truncate your file, see for example https://www.quora.com/How-do-I-chop-off-just-the-last-byte-of-a-file-in-Bash .
Related
for a=1:50; %numbers 1 through 50
for b=1:50;
c=sqrt(a^2+b^2);
if c<=50&c(rem(c,1)==0);%if display only if c<=50 and c=c/1 has remainder of 0
pyth=[a,b,c];%pythagorean matrix
disp(pyth)
else c(rem(c,1)~=0);%if remainder doesn't equal to 0, omit output
end
end
end
answer=
3 4 5
4 3 5
5 12 13
6 8 10
7 24 25
8 6 10
8 15 17
9 12 15
9 40 41
10 24 26
12 5 13
12 9 15
12 16 20
12 35 37
14 48 50
15 8 17
15 20 25
15 36 39
16 12 20
16 30 34
18 24 30
20 15 25
20 21 29
21 20 29
21 28 35
24 7 25
24 10 26
24 18 30
24 32 40
27 36 45
28 21 35
30 16 34
30 40 50
32 24 40
35 12 37
36 15 39
36 27 45
40 9 41
40 30 50
48 14 50
This problem involves the Pythagorean theorem but we cannot use the built in function so I had to write one myself. The problem is for example columns 1 & 2 from the first two rows have the same numbers. How do I code it so it only deletes one of the rows if the columns 1 and 2 have the same number combination? I've tried unique function but it doesn't really delete the combinations. I have read about deleting duplicates from previous posts but those have confused me even more. Any help on how to go about this problem will help me immensely!
Thank you
welcome to StackOverflow.
The problem in your code seems to be, that pyth only contains 3 values, [a, b, c]. The unique() funcion used in the next line has no effect in that case, because only one row is contained in pyth. another issue is, that the values idx and out are calculated in each loop cycle. This should be placed after the loops. An example code could look like this:
pyth = zeros(0,3);
for a=1:50
for b=1:50
c = sqrt(a^2 + b^2);
if c<=50 && rem(c,1)==0
abc_sorted = sort([a,b,c]);
pyth = [pyth; abc_sorted];
end
end
end
% do final sorting outside of the loop
[~,idx] = unique(pyth, 'rows', 'stable');
out = pyth(idx,:);
disp(out)
a few other tips for writing MATLAB code:
You do not need to end for or if/else stements with a semicolon
else statements cover any other case not included before, so they do not need a condition.
Some performance reommendations:
Due to the symmetry of a and b (a^2 + b^2 = b^2 + a^2) the b loop could be constrained to for b=1:a, which would roughly save you half of the loop cycles.
if you use && for contencation of scalar values, the second part is not evaluated, if the first part already fails (source).
Regards,
Chris
You can also linearize your algorithm (but we're still using bruteforce):
[X,Y] = meshgrid(1:50,1:50); %generate all the combination
C = (X(:).^2+Y(:).^2).^0.5; %sums of two square for every combination
ind = find(rem(C,1)==0 & C<=50); %get the index
res = unique([sort([X(ind),Y(ind)],2),C(ind)],'rows'); %check for uniqueness
Now you could really optimized your algorithm using math, you should read this question. It will be useful if n>>50.
Example 1)
I have the code below
5#10+1*2
that generates
index value
0 12
1 12
2 12
3 12
4 12
How can I replace the number "1" by the index?
then generating
5#10+index*2
index value
0 10
1 12
2 14
3 16
4 18
update Example 2)
Now, if I have, let's say
mult:5;
t:select from ([]numC:1 3 6 4 1;[]s:50 16 53 6 33);
update lst:(numC#'s) from t
the last update will generate
numC s lst
1 50 50
3 16 16 16 16
6 53 53 53 53 53 53 53
4 6 6 6 6 6
1 33 33
How can I generate the "lst" column as per below?
numC s lst
1 50 50+0*mult
3 16 16+0*mult 16+1*mult 16+2*mult
6 53 53+0*mult 53+1*mult 53+2*mult 53+3*mult 53+4*mult 53+5*mult
4 6 6+0*mult 6+1*mult 6+2*mult 6+3*mult
1 33 33+0*mult
I tried something like
update lst:(numC#'s + (til numC)*mult) from t
but I am getting an error
ERROR: 'type
Thanks vm
Is this what you're looking for:
q)x:5
q)x#10+(til x)*2
10 12 14 16 18
http://code.kx.com/q/ref/arith-integer/#til
You can remove take # and use til to simplify to:
q)10+2*til 5
10 12 14 16 18
Using til will create a list of a list of 5 elements (0->4), so you will not need take 5 elements from the resulting list. Take will only be required if your list of indices is greater than 5.
Update:
For your second example the following should work:
q)update lst:{y+x*til z}'[mult;s;numC] from t
q)update lst:s+mult*til each numC from t
numC s lst
-------------------------
1 50 ,50
3 16 16 21 26
6 53 53 58 63 68 73 78
4 6 6 11 16 21
1 33 ,33
There are many ways with which we can get achieve this:
1) 10+2*til 5
2) (2*til 5) + 10
/ take operator: The dyadic take function creates lists. The left argument specifies the count and shape and the right argument provides the data.
It is useful for selecting from the front or end of a list.
https://code.kx.com/wiki/Reference/NumberSign
q)5#0 1 2 3 4 5 6 7 8 / take the first 5 items
0 1 2 3 4
q)-5#0 1 2 3 4 5 6 7 8 / take the last 5 elements
4 5 6 7 8
use take operator # only when it is required.
say we have 10 elements, of which we need five on output, then we can use:
5#10+2*til 10
/ The til function takes a non-negative integer argument X and returns the first X integers
I got a dataset (azimuth vs time) with measure the compass of an object trough time. So I can see when the object is moving (the compass vary so much), and when it's static, without moving (compass do not vary). My question is how to program this in matlab in order to eliminate the data which show that the object is moving and just filter data that shows the object is static.
For example:
Azimuth (angle) | 30 30 30 15 10 16 19 24 24 24 17 14 12 15 16
Time (s) | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
The output would be:
Azimuth (angle) | 30 30 30 24 24 24
Time (s) | 1 2 3 8 9 10
s=diff(Azumuth)==0
%diff only would skip the values at t=1 and t=8. Modify to include them as well:
s=[s(1),s(2:end)|s(1:end-1),s(end)]
Azumuth(s)
Time(s)
I'm trying to implement the Baker map.
Is there a function that would allow one to divide a 8 x 8 matrix by providing, for example, a sequence of divisors 2, 4, 2 and rearranging pixels in the order as shown in the matrices below?
X = reshape(1:64,8,8);
After applying divisors 2,4,2 to the matrix X one should get a matrix like A shown below.
A=[31 23 15 7 32 24 16 8;
63 55 47 39 64 56 48 40;
11 3 12 4 13 5 14 6;
27 19 28 20 29 21 30 22;
43 35 44 36 45 37 46 38;
59 51 60 52 61 53 62 54;
25 17 9 1 26 18 10 2;
57 49 41 33 58 50 42 34]
The link to the document which I am working on is:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.5132&rep=rep1&type=pdf
This is what I want to achieve:
Edit: a little more generic solution:
%function Z = bakermap(X,divisors)
function Z = bakermap()
X = reshape(1:64,8,8)'
divisors = [ 2 4 2 ];
[x,y] = size(X);
offsets = sum(divisors)-fliplr(cumsum(fliplr(divisors)));
if any(mod(y,divisors)) && ~(sum(divisors) == y)
disp('invalid divisor vector')
return
end
blocks = #(div) cell2mat( cellfun(#mtimes, repmat({ones(x/div,div)},div,1),...
num2cell(1:div)',...
'UniformOutput',false) );
%create index matrix
I = [];
for ii = 1:numel(divisors);
I = [I, blocks(divisors(ii))+offsets(ii)];
end
%create Baker map
Y = flipud(X);
Z = [];
for jj=1:I(end)
Z = [Z; Y(I==jj)'];
end
Z = flipud(Z);
end
returns:
index matrix:
I =
1 1 3 3 3 3 7 7
1 1 3 3 3 3 7 7
1 1 4 4 4 4 7 7
1 1 4 4 4 4 7 7
2 2 5 5 5 5 8 8
2 2 5 5 5 5 8 8
2 2 6 6 6 6 8 8
2 2 6 6 6 6 8 8
Baker map:
Z =
31 23 15 7 32 24 16 8
63 55 47 39 64 56 48 40
11 3 12 4 13 5 14 6
27 19 28 20 29 21 30 22
43 35 44 36 45 37 46 38
59 51 60 52 61 53 62 54
25 17 9 1 26 18 10 2
57 49 41 33 58 50 42 34
But have a look at the if-condition, it's just possible for these cases. I don't know if that's enough. I also tried something like divisors = [ 1 4 1 2 ] - and it worked. As long as the sum of all divisors is equal the row-length and the modulus as well, there shouldn't be problems.
Explanation:
% definition of anonymous function with input parameter: div: divisor vector
blocks = #(div) cell2mat( ... % converts final result into matrix
cellfun(#mtimes, ... % multiplies the next two inputs A,B
repmat(... % A...
{ones(x/div,div)},... % cell with a matrix of ones in size
of one subblock, e.g. [1,1,1,1;1,1,1,1]
div,1),... % which is replicated div-times according
to actual by cellfun processed divisor
num2cell(1:div)',... % creates a vector [1,2,3,4...] according
to the number of divisors, so so finally
every Block A gets an increasing factor
'UniformOutput',false...% necessary additional property of cellfun
));
Have also a look at this revision to have a simpler insight in what is happening. You requested a generic solution, thats the one above, the one linked was with more manual inputs.
I have a loop:
for i=1:size(A,1),
if A(i,4:6) == [0,0,3.4]
K = [K; A(i,:)];
end
end
and I would like to delete the last row in the matrix but I do not know what number row it will be. How do I delete the last row in the matrix in the loop? Or should I do it after the loop?
Why do you have loop? it is a one time action, not something you do several times.
check this out, I delete the last row:
>> a = magic(5);
>> a
a =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
>> a = a(1:end-1,:);
>> a
a =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
you can refer to last row by END keyword:
A= A(1:end-1, :)