What does asterisk mean in the logical expression of an "IF" function? - arrays

For example, in this formula, what the asterisk does:
=IF ((A2: E <> "") * (A2: E <> "-"); "X"; "Y")
It would be something similar to = IF (AND (... ?
Where can I study this expression?


it means AND and the reason is that ARRAYFORMULA does not support AND nor OR so
AND is *
OR is +
it's just a 0/1 logic
TRUE * TRUE = TRUE as: 1 * 1 = 1
TRUE * FALSE = FALSE as: 1 * 0 = 0
FALSE * TRUE = FALSE as: 0 * 1 = 0
FALSE * FALSE = FALSE as: 0 * 0 = 0

Related

I want to solve this query with c program

Can someone guide me how to put this properly in an if-else statement.
Consider the following if statement, where doesSignificantWork, makesBreakthrough, and nobelPrizeCandidate are all boolean variables:
if (doesSignificantWork) {
if (makesBreakthrough)
nobelPrizeCandidate = true;
else
nobelPrizeCandidate = false;
}
else if (!doesSignificantWork)
nobelPrizeCandidate = false;
First, write a simpler if statement that is equivalent to this one. Then write a single assignment statement that does the same thing.

if (doesSignificantWork) {
if (makesBreakthrough)
nobelPrizeCandidate = true;
else
nobelPrizeCandidate = false;
}
else if (!doesSignificantWork)
nobelPrizeCandidate = false
Is equivalent to
nobelPrizeCandidate = (doesSignificantWork && makesBreakthrough);

You can make a truth table. The first step is to identify inputs, and write down all combinations of their values.
Input Output
d m n
0 0 ?
0 1 ?
1 0 ?
1 1 ?
Then fill the correct output values
Input Output
d m n
0 0 0
0 1 0
1 0 0
1 1 1
You should now see that the output function corresponds to logical AND (&&).

A simpler if statement is:
if (doesSignificantWork && makesBreakthrough)
nobelPrizeCandidate = true;
else
nobelPrizeCandidate = false;
#Blaze's answer gives you the simplest one-liner. An alternative is
nobelPrizeCandidate = (doesSignificantWork && makesBreakthrough) ? true : false;

how to create multi line chart with single datatable in vb.net

I am working on vb.net application where i have to create a multi line chart. The table coming from the database using storing procedure is:-
TagName On Off Trip
P1 0 0 1
P2 0 1 1
P3 0 1 0
Q1 0 1 0
Q2 1 0 1
Q3 2 2 2
W1 4 2 1
W2 2 0 1
W3 1 1 0
W4 0 1 1
W5 2 1 1
And the code in vb.net i used to bind the chart named "chTrend" is:-
ds = ObjTags.GetTrendData()
If (ds.Tables(0).Rows.Count > 0) Then
dt = ds.Tables(0)
chTrend.DataSource = dt
chTrend.Series(0).XValueMember = "TagName"
chTrend.Series(0).YValueMembers = "On"
chTrend.Series(1).XValueMember = "TagName"
chTrend.Series(1).YValueMembers = "Off"
chTrend.Series(1).XValueMember = "TagName"
chTrend.Series(1).YValueMembers = "Trip"
For i = 0 To 2
chTrend.Series(i).ChartType = SeriesChartType.Line
chTrend.Series(i).IsVisibleInLegend = True
chTrend.Series(i).IsValueShownAsLabel = True
chTrend.Series(i).ToolTip = "Data Point Y Value #VALY{G}"
chTrend.Series(i).BorderWidth = 3
Next
End If
When i run the program a Error is coming as following
"Index was out of range. Must be non-negative and less than the size of the collection.Parameter name: index"
At the position
chTrend.Series(0).XValueMember = "TagName"
how to solve this?

Assuming that your DataTable only has the four DataColumns {TagName, On. Off. Trip}, then replace everything in your shown code below the line:
chTrend.DataSource = dt
with this.
chTrend.Series.Clear() ' clears any existing series
Dim s As Series
For columnIndex As Int32 = 1 To dt.Columns.Count - 1
Dim name As String = dt.Columns(columnIndex).ColumnName
s = chTrend.Series.Add(name)
s.XValueMember = dt.Columns(0).ColumnName
s.YValueMembers = name
s.ChartType = SeriesChartType.Line
s.IsVisibleInLegend = True
s.IsValueShownAsLabel = True
s.ToolTip = "Data Point Y Value #VALY{G}"
s.BorderWidth = 3
Next
chTrend.DataBind() ' loads the data from dt to the chart
The Chart databinding is a not a binding in the normal sense of binding. You need to tell the chart to copy the data from the source so that it has something to plot. That is what the DataBind method does.

MATLAB data indexing issue. What is going on here?

Let I be the identity, D an orthonormal projection, and p a vector.
I realized that several of my lines of code combined to be (I-(I-D))(p) and I could just simplify it to D(p). In replacing it, I computed the new method along-side the old to double check I was computing the same thing (Earlier in my code I had a line that was D = I - D. The D you see here is that D.) I wasn't getting the same answer, and traced it to an error in indexing D.
Here you can see I'm using the debugger and checking portions of D and getting the wrong data returned.
The values in the data explorer on the right are what I'd expect them to be. Sometimes I get what I'd expect from D(:,:,k,1), and sometimes I don't, even when I make the queries right after each other.
The vectors those red arrows are pointing to should be the same. Nothing else changed or was computed between those lines, and k = 2 when the first line was run. I've closed MATLAB and restarted it and get the same issue every time. (D depends on random input, but I'm not altering the seed, so I get the same thing every first run after newly opening MATLAB. The way D is computed, I do expect D(:,:,1,1) to be the identity matrix.)
What in the world is going on? Any help is appreciated.
I have wondered if MATLAB is messing with me on purpose. Sometimes when I open it, a pop-up dialog box says I need to update my student license. I click the update button, but nothing ever happens and the dialog box never closes, so I click cancel.
Edit:
K>> whos D P
Name Size Bytes Class Attributes
D 4-D 4608 double
P 4x1x6 192 double
K>> size(D)
ans =
4 4 6 6
I've been playing around with A and B a bit, and I get the same thing. Sometimes it computes correctly and sometimes it doesn't.
K>> B=permute(P,[1,3,2])
B =
0.4155 0.27554 0.52338 0.6991 -0.11346 0.20999
0.53573 -0.83781 0.53182 -0.022364 0.60291 -0.62601
-0.49246 -0.46111 -0.39168 0.45919 0.42377 0.47074
0.54574 0.097595 0.53835 -0.54763 0.66637 0.58516
K>> A=D
A(:,:,1,1) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
A(:,:,2,1) =
0.99071 -0.091198 0.0020814 -0.029755
-0.091198 0.10503 0.020426 -0.292
0.0020814 0.020426 0.99953 0.0066643
-0.029755 -0.292 0.0066643 0.90473
A(:,:,3,1) =
0.46769 0.019281 -0.49725 0.036486
0.019281 0.9993 0.018011 -0.0013215
-0.49725 0.018011 0.53551 0.034083
0.036486 -0.0013215 0.034083 0.9975
A(:,:,4,1) =
0.96774 0.063488 -0.10826 0.12438
0.063488 0.87506 0.21304 -0.24477
-0.10826 0.21304 0.63673 0.41737
0.12438 -0.24477 0.41737 0.52047
A(:,:,5,1) =
0.7542 0.031217 0.42575 0.056052
0.031217 0.99604 -0.054071 -0.0071187
0.42575 -0.054071 0.26255 -0.097088
0.056052 -0.0071187 -0.097088 0.98722
A(:,:,6,1) =
0.9818 -0.10286 0.085279 0.0034902
-0.10286 0.41855 0.48208 0.01973
0.085279 0.48208 0.60031 -0.016358
0.0034902 0.01973 -0.016358 0.99933
A(:,:,1,2) =
0.99071 -0.091198 0.0020814 -0.029755
-0.091198 0.10503 0.020426 -0.292
0.0020814 0.020426 0.99953 0.0066643
-0.029755 -0.292 0.0066643 0.90473
A(:,:,2,2) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
A(:,:,3,2) =
0.97125 -0.15889 -0.0080537 -0.051131
-0.15889 0.12194 -0.044507 -0.28256
-0.0080537 -0.044507 0.99774 -0.014323
-0.051131 -0.28256 -0.014323 0.90907
A(:,:,4,2) =
0.91488 -0.16388 -0.18495 0.12967
-0.16388 0.6845 -0.35607 0.24964
-0.18495 -0.35607 0.59815 0.28174
0.12967 0.24964 0.28174 0.80247
A(:,:,5,2) =
0.95461 0.16812 0.10326 0.066372
0.16812 0.37733 -0.38244 -0.24582
0.10326 -0.38244 0.76511 -0.15098
0.066372 -0.24582 -0.15098 0.90295
A(:,:,6,2) =
0.99628 0.012018 0.052874 0.027665
0.012018 0.96117 -0.17085 -0.089393
0.052874 -0.17085 0.24833 -0.39329
0.027665 -0.089393 -0.39329 0.79422
A(:,:,1,3) =
0.46769 0.019281 -0.49725 0.036486
0.019281 0.9993 0.018011 -0.0013215
-0.49725 0.018011 0.53551 0.034083
0.036486 -0.0013215 0.034083 0.9975
A(:,:,2,3) =
0.97125 -0.15889 -0.0080537 -0.051131
-0.15889 0.12194 -0.044507 -0.28256
-0.0080537 -0.044507 0.99774 -0.014323
-0.051131 -0.28256 -0.014323 0.90907
A(:,:,3,3) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
A(:,:,4,3) =
0.98622 0.043449 -0.066709 0.085142
0.043449 0.86297 0.21038 -0.26852
-0.066709 0.21038 0.67698 0.41227
0.085142 -0.26852 0.41227 0.47382
A(:,:,5,3) =
0.62859 0.041458 0.47558 0.074661
0.041458 0.99537 -0.053085 -0.0083339
0.47558 -0.053085 0.39105 -0.0956
0.074661 -0.0083339 -0.0956 0.98499
A(:,:,6,3) =
0.95505 -0.16608 0.12371 0.0067153
-0.16608 0.38639 0.45705 0.02481
0.12371 0.45705 0.65956 -0.01848
0.0067153 0.02481 -0.01848 0.999
A(:,:,1,4) =
0.96774 0.063488 -0.10826 0.12438
0.063488 0.87506 0.21304 -0.24477
-0.10826 0.21304 0.63673 0.41737
0.12438 -0.24477 0.41737 0.52047
A(:,:,2,4) =
0.91488 -0.16388 -0.18495 0.12967
-0.16388 0.6845 -0.35607 0.24964
-0.18495 -0.35607 0.59815 0.28174
0.12967 0.24964 0.28174 0.80247
A(:,:,3,4) =
0.98622 0.043449 -0.066709 0.085142
0.043449 0.86297 0.21038 -0.26852
-0.066709 0.21038 0.67698 0.41227
0.085142 -0.26852 0.41227 0.47382
A(:,:,4,4) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
A(:,:,5,4) =
0.73864 0.20112 -0.011394 0.39048
0.20112 0.84524 0.0087678 -0.30047
-0.011394 0.0087678 0.9995 0.017023
0.39048 -0.30047 0.017023 0.41662
A(:,:,6,4) =
0.87322 -0.15647 0.0029936 0.29363
-0.15647 0.80689 0.0036946 0.36238
0.0029936 0.0036946 0.99993 -0.0069332
0.29363 0.36238 -0.0069332 0.31996
A(:,:,1,5) =
0.7542 0.031217 0.42575 0.056052
0.031217 0.99604 -0.054071 -0.0071187
0.42575 -0.054071 0.26255 -0.097088
0.056052 -0.0071187 -0.097088 0.98722
A(:,:,2,5) =
0.95461 0.16812 0.10326 0.066372
0.16812 0.37733 -0.38244 -0.24582
0.10326 -0.38244 0.76511 -0.15098
0.066372 -0.24582 -0.15098 0.90295
A(:,:,3,5) =
0.62859 0.041458 0.47558 0.074661
0.041458 0.99537 -0.053085 -0.0083339
0.47558 -0.053085 0.39105 -0.0956
0.074661 -0.0083339 -0.0956 0.98499
A(:,:,4,5) =
0.73864 0.20112 -0.011394 0.39048
0.20112 0.84524 0.0087678 -0.30047
-0.011394 0.0087678 0.9995 0.017023
0.39048 -0.30047 0.017023 0.41662
A(:,:,5,5) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
A(:,:,6,5) =
0.93556 0.24481 -0.0093576 0.016177
0.24481 0.069855 0.035553 -0.061461
-0.0093576 0.035553 0.99864 0.0023492
0.016177 -0.061461 0.0023492 0.99594
A(:,:,1,6) =
0.9818 -0.10286 0.085279 0.0034902
-0.10286 0.41855 0.48208 0.01973
0.085279 0.48208 0.60031 -0.016358
0.0034902 0.01973 -0.016358 0.99933
A(:,:,2,6) =
0.99628 0.012018 0.052874 0.027665
0.012018 0.96117 -0.17085 -0.089393
0.052874 -0.17085 0.24833 -0.39329
0.027665 -0.089393 -0.39329 0.79422
A(:,:,3,6) =
0.95505 -0.16608 0.12371 0.0067153
-0.16608 0.38639 0.45705 0.02481
0.12371 0.45705 0.65956 -0.01848
0.0067153 0.02481 -0.01848 0.999
A(:,:,4,6) =
0.87322 -0.15647 0.0029936 0.29363
-0.15647 0.80689 0.0036946 0.36238
0.0029936 0.0036946 0.99993 -0.0069332
0.29363 0.36238 -0.0069332 0.31996
A(:,:,5,6) =
0.93556 0.24481 -0.0093576 0.016177
0.24481 0.069855 0.035553 -0.061461
-0.0093576 0.035553 0.99864 0.0023492
0.016177 -0.061461 0.0023492 0.99594
A(:,:,6,6) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
Edit 2:
Added relevant code. I've been pausing the code and getting the errors inside the for loops at the end. (I believe it's also giving errors in S, but I've been focusing on D trying to figure it out.)
mtimesx is from here.
n = 4;
M = 6;
P = Normalize(2*rand(n,1,M)-1);
%differences between p_i and p_j
%sum of p_i and p_j
d = Normalize(repmat(permute(P,[1,3,2]),[1,1,M]) - repmat(P,[1,M,1]));
s = Normalize(repmat(permute(P,[1,3,2]),[1,1,M]) + repmat(P,[1,M,1]));
d(isnan(d)) = 0;
%orthogonal projection onto d(:,i,j), i.e. outer product of differences
%orthogonal projection onto s(:,i,j), i.e. outer product of sums
D = mtimesx(permute(d,[1,4,2,3]), permute(d,[4,1,2,3]));
S = mtimesx(permute(s,[1,4,2,3]), permute(s,[4,1,2,3]));
D2 = D;
S2 = S;
%projection onto the complement of d(:,i,j)
%projection onto the complement of s(:,i,j)
D = repmat(eye(n),[1,1,M,M]) - D;
S = repmat(eye(n),[1,1,M,M]) - S;
%total distance to the nearest subspace
PDist = zeros([1,M]);
PDist2 = PDist;
for j = 1:M
for k = 1:M-1
for l = k:M
if j~=k && j~=l
PDist(j) = PDist(j) + min(norm(P(:,1,j) - mtimes(D(:,:,k,l),P(:,1,j))), norm(P(:,1,j) - mtimes(S(:,:,k,l),P(:,1,j))));
PDist2(j) = PDist2(j) + min(norm(D2(:,:,k,1)*P(:,1,j)),norm(S2(:,:,k,1)*P(:,1,j)));
end
end
end
end
PDist-PDist2
Normalize.m
%Normalize
%Accepts an array (of column vectors) and normalizes the columns
function B = Normalize(A)
B = A./repmat(sqrt(sum(A.*A)),size(A,1),1);
end

The problem is that you indexed the matrices using the constant 1 instead of the variable l (lowercase L), both in the first example and in the code for computing PDist2.
In general it is good to avoid using variable names that look similar to each other and/or similar to numbers.
This can be avoided by using an editor that highlights uses different colors for variables and constants (I don't know if this is possible in MATLAB). In fact, this is how I found the error in your code. As you can see, when indexing D2 for the computation of PDist2 the number 1 is colored red.

Lua: Search key in table, increment index if not and retry

Situation:
table = { this, that, something else, x_coord, y_coord }
table.x_coord = { 1,2,3,4,7,8,n}
table.y_coord = { 2,4,5,9,n} -- numbers aren't fix
table.check_boxes = { [12] = a function,
[14] = a function,
[15] = a function,
[24] = a function,
[29] = a function,
....(n) }
As you can see, the x/y_coords forming check_boxes. For example:
table.x_coord[1]..table.y_coord[1] ~ table.check_boxes[1]
I use this to move the cursor in the Terminal between the check_boxes.
The problem now is in my cursormovement.
Currently I got a function that's searching for the next x/y_coord to the left/right/up/down depending on the given input (arrow-keys).
With return/space I call the function behind the checkbox.
Now, that could set the Cursor on positions where no check_boxes are given. Actually that's not a big deal, because when input == space/return, an inputhandler calls the function at
table.check_boxes[table.x_coorx[x_index]..table.y_coords[y_index]]
So if the cursor doesn't point on a function, just nothing happens.
But now I want the cursor to be forced to the next check_box. What can I do?
My Idea so far:
following function either for x or y, depending on input left/right up/down:
while true do
for k, v in pairs(table.check_boxes) do
if(table.x_coord[x_index] .. table.y_coord[y_index] == k then break end
end -- break -> okay, coord is at a checkbox
x_index = x_index + 1 -- or -1
if table.x_coord[x_index] == nil then
x_index = 1
end
end
The Problem now is that the last if will not allow cases like x_coord = {1,3} because it will set x_index to 1 if 2 is reached.
Any tips?
Edit:
Now I got that one going:
function cursorTioNextBoxRight()
searc_index = x_index
search = true
while search do
search_index = search_index + 1
if search_index > #table.x_coord then
search_index = 1
end
for k, v in pairs(table.check_boxes) do
if tonumber(table.x_coord[search_index..table.y_coord[y_index] == k then
x_index = search_index -- YAAAY
search = false
break
end
end
end
I'ts damn slow.

local x_newIndex = x_index + 1 --[[ or -1 --]]
x_index = table.x_coord[x_newIndex] and x_newIndex or x_index
x_index becomes x_newIndex when x_newIndex exists in the table otherwise it stays the old x_index

function cursorTioNextBoxRight()
searc_index = x_index
search = true
while search do
search_index = search_index + 1
if search_index > #table.x_coord then
search_index = 1
end
for k, v in pairs(table.check_boxes) do
if tonumber(table.x_coord[search_index..table.y_coord[y_index] == k then
x_index = search_index -- YAAAY
search = false
break
end
end
end

Evaluating equality in perl using elements taken from an array ref

I have a small perl script that needs to evaluate the equality of two parameters and a small return from the database.
my ($firstId, $secondId, $firstReturnedId, $secondReturnedId, $picCount);
my $pics = $dbh->prepare(qq[select id from pictures limit 10]);
$firstId = q->param('firstId');
$secondId = q->param('secondId');
$pics->execute or die;
my $picids = $pics->fetchall_arrayref;
$picCount = scalar(#{$picids});
$firstReturnedId = $picCount > 0 ? shift(#{$picids}) : 0;
$secondReturnedId = $picCount > 1 ? pop(#{$picids}) : $firstReturnedId;
Here, a quick look at my debugger shows that $picCount = 1 and $firstReturnedId = 9020 and $secondReturnedId = 9020. However, they are both denoted as
ARRAY(0x9e79184)
0 9020
in the debugger so when I perform the final check
my $result = (($firstId == $firstReturnedId) && ($secondId == $secondReturnedId)) ? 1 : 0;
I get $result = 0, which is not what I want.
What am I doing wrong?

DBI::fetchall_arrayref returns a reference to a list of "row results". But since there could be more than one value in a row result (e.g., your query could have been select id,other_field from pictures), each row result is also a reference to a list. This means you have one more dereferencing to do in order to get the result you want. Try:
$picCount = scalar(#{$picids});
if ($picCount > 0) {
my $result = shift #{$picids};
$firstReturnedId = $result->[0];
} else {
$firstReturnedId = 0;
}
if ($picCount > 1) {
my $result = pop #{$picids};
$secondReturnedId = $result->[0];
} else {
$secondReturnedId = $firstReturnedId;
}
or if you still want to use a concise style:
$firstReturnedId = $picCount > 0 ? shift(#{$picids})->[0] : 0;
$secondReturnedId = $picCount > 1 ? pop(#{$picids})->[0] : $firstReturnedId;

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