I have the number: a = 3.860575156847749e+003; and I would show it in a normal manner. So I write b = sprintf('%0.1f' a);. If I print b I will get: 3860.6. This is perfect. Matter of fact, while a is a double type, b has been converted in char.
What can I do to proper format that number and still have a number as final result?
Best regards
Well, you have to distinguish between both the numerical value (the number stored in your computer's memory) and its decimal representation (the string/char array you see on your screen). You can't really impose a format on a number: a number has a value which can be represented as a string in different ways (e.g. 1234 = 1.234e3 = 12.34e2 = 0.1234e4 = ...).
If you want to store a number with less precision, you can use round, floor, ceil to calculate a number which has less precision than the original number.
E.g. if you have a = 3.860575156847749e+003 and you want a number that only has 5 significant digits, you can do so by using round:
a = 3.860575156847749e+003;
p = 0.1; % absolute precision you want
b = p .* round(a./p)
This will yield a variable b = 3.8606e3 which can be represented in different ways, but should contain zeros (in practice: very small values are sometimes unavoidable) after the fifth digit. I think that is what you actually want, but remember that for a computer this number is equal to 3.86060000 as well (it is just another string representation of the same value), so I want to stress again that the decimal representation is not set by rounding the number but by (implicitly) calling a function that converts the double to a string, which happens either by sprintf, disp or possibly some other functions.
Result of sprintf y a text variable. have you tried to declare a variable as integer (for example) and use this as return value for sprintf instruction?
This can be useful to you: http://blogs.mathworks.com/loren/2006/12/27/displaying-numbers-in-matlab/
Related
I'm trying to implement the 'Sport Scheduling Problem' (with a Round-Robin approach to break symmetries). The actual problem is of no importance. I simply want to declare the value at x[1,1] to be the set {1,2} and base the sets in the same column upon the first set. This is modelled as in the code below. The output is included in a screenshot below it. The problem is that the first set is not printed as a set but rather some sort of range while the values at x[2,1] and x[3,1] are indeed printed as sets and x[4,1] again as a range. Why is this? I assume that in the declaration of x that set of 1..n is treated as an integer but if it is not, how to declare it as integers?
EDIT: ONLY the first column of the output is of importance.
int: n = 8;
int: nw = n-1;
int: np = n div 2;
array[1..np, 1..nw] of var set of 1..n: x;
% BEGIN FIX FIRST WEEK $
constraint(
x[1,1] = {1, 2}
);
constraint(
forall(t in 2..np) (x[t,1] = {t+1, n+2-t} )
);
solve satisfy;
output[
"\(x[p,w])" ++ if w == nw then "\n" else "\t" endif | p in 1..np, w in 1..nw
]
Backend solver: Gecode
(Here's a summarize of my comments above.)
The range syntax is simply a shorthand for contiguous values in a set: 1..8 is a shorthand of the set {1,2,3,4,5,6,7,8}, and 5..6 is a shorthand for the set {5,6}.
The reason for this shorthand is probably since it's often - and arguably - easier to read the shorthand version than the full list, especially if it's a long list of integers, e.g. 1..1024. It also save space in the output of solutions.
For the two set versions, e.g. {1,2}, this explicit enumeration might be clearer to read than 1..2, though I tend to prefer the shorthand version in all cases.
I am well aware that one is able to assign a value to an array or constant in Swift and have those value represented in different formats.
For Integer: One can declare in the formats of decimal, binary, octal or hexadecimal.
For Float or Double: One can declare in the formats of either decimal or hexadecimal and able to make use of the exponent too.
For instance:
var decInt = 17
var binInt = 0b10001
var octInt = 0o21
var hexInt = 0x11
All of the above variables gives the same result which is 17.
But what's the catch? Why bother using those other than decimal?
There are some notations that can be way easier to understand for people even if the result in the end is the same. You can for example think in cases like colour notation (hexadecimal) or file permission notation (octal).
Code is best written in the most meaningful way.
Using the number format that best matches the domain of your program, is just one example. You don't want to obscure domain specific details and want to minimize the mental effort for the reader of your code.
Two other examples:
Do not simplify calculations. For example: To convert a scaled integer value in 1/10000 arc minutes to a floating point in degrees, do not write the conversion factor as 600000.0, but instead write 10000.0 * 60.0.
Chose a code structure that matches the nature of your data. For example: If you have a function with two return values, determine if it's a symmetrical or asymmetrical situation. For a symmetrical situation always write a full if (condition) { return A; } else { return B; }. It's a common mistake to write if (condition) { return A; } return B; (simply because 'it works').
Meaning matters!
I have a vector of cells (say, size of 50x1, called tokens) , each of which is a struct with properties x,f1,f2 which are strings representing numbers. for example, tokens{15} gives:
x: "-1.4343429"
f1: "15.7947111"
f2: "-5.8196158"
and I am trying to put those numbers into 3 vectors (each is also 50x1) whose type is float. So I create 3 vectors:
x = zeros(50,1,'single');
f1 = zeros(50,1,'single');
f2 = zeros(50,1,'single');
and that works fine (why wouldn't it?). But then when I try to populate those vectors: (L is a for loop index)
x(L)=tokens{L}.x;
.. also for the other 2
I get :
The following error occurred converting from string to single:
Conversion to single from string is not possible.
Which I can understand; implicit conversion doesn't work for single. It does work if x, f1 and f2 are of type 50x1 double.
The reason I am doing it with floats is because the data I get is from a C program which writes the some floats into a file to be read by matlab. If I try to convert the values into doubles in the C program I get rounding errors...
So, (after what I hope is a good question,) how might I be able to get the numbers in those strings, at the right precision? (all the strings have the same number of decimal places: 7).
The MCVE:
filedata = fopen('fname1.txt','rt');
%fname1.txt is created by a C program. I am quite sure that the problem isn't there.
scanned = textscan(filedata,'%s','Delimiter','\n');
raw = scanned{1};
stringValues = strings(50,1);
for K=1:length(raw)
stringValues(K)=raw{K};
end
clear K %purely for convenience
regex = 'x=(?<x>[\-\.0-9]*),f1=(?<f1>[\-\.0-9]*),f2=(?<f2>[\-\.0-9]*)';
tokens = regexp(stringValues,regex,'names');
x = zeros(50,1,'single');
f1 = zeros(50,1,'single');
f2 = zeros(50,1,'single');
for L=1:length(tokens)
x(L)=tokens{L}.x;
f1(L)=tokens{L}.f1;
f2(L)=tokens{L}.f2;
end
Use function str2double before assigning into yours arrays (and then cast it to single if you want). Strings (char arrays) must be explicitely converted to numbers before using them as numbers.
I'm creating a program to simulate a random walk and it requires the user to input an integer number of steps to take for the walk.
The prompt for this uses code very similar to this:
**% Ask user for a number.
defaultValue = 45;
titleBar = 'Enter a value';
userPrompt = 'Enter the integer';
caUserInput = inputdlg(userPrompt, titleBar, 1,{num2str(defaultValue)});
if isempty(caUserInput),return,end; % Bail out if they clicked Cancel.
% Round to nearest integer in case they entered a floating point number.
integerValue = round(str2double(cell2mat(caUserInput)));
% Check for a valid integer.
if isnan(integerValue)
% They didn't enter a number.
% They clicked Cancel, or entered a character, symbols, or something else not allowed.
integerValue = defaultValue;
message = sprintf('I said it had to be an integer.\nI will use %d and continue.', integerValue);
uiwait(warndlg(message));
end**
However, I want it to simply display the "Enter a value" prompt again if the user does not enter an integer the first time i.e. 4.4.
Any ideas?
Thanks!
if (mod(integerValue,1) == 0)
will evaluate to true if integerValue is an integer. Simply augment your if statement w/ this logic. You might want to consider changing to using a while loop so the user can enter bad input more than once.
The first answer is totally correct for checking for an integer value, but to address the "show prompt again" issue you can just use a loop conditioning it to get the exact kind of data you want:
caUserInput = nan; %or anything worng for that matter
while isempty(caUserInput) || isnan(caUserInput)
caUserInput = inputdlg(userPrompt, titleBar, 1,{num2str(defaultValue)});
end
if you want you can start it again with different argument lines in a more fancy style:
inputiswrong = 1; %or anything worng for that matter
while inputiswrong
inputiswrong = 0;
caUserInput = inputdlg(userPrompt, titleBar, 1,{num2str(defaultValue)});
if isempty(caUserInput )
userPrompt = 'Try again with an input';
inputiswrong = 1;
end
if isnan(caUserInput )
userPrompt = 'not really a number';
inputiswrong = 1;
end
%and so on
end
In both scenarios you should consider transforming the caUserInput to something you could use, i think inputdlg returns a cell so maybe a cell2mat() around the inputdlg().
Remember that unspecified inputs in MATLAB are double-precision by default. For instance a=3 is not an integer. So you should consider two cases:
Integer type
If you are talking about integer type in MATLAB the easiest way is to use isinteger function by MATLAB:
tf = isinteger(A)
for instance:
isinteger(4.4)
=
0
as I mentioned before, 3 is not an integer:
isinteger(3)
=
0
but this one is integer actually:
isinteger(uint8(3))
=
1
To repeat the input query also easily use the same function in a while loop
while ~isinteger(a)
disp('enter an integer');
....
end
Constant double-precision with no decimal
But if you are considering normal constant inputs to be integers you could convert them to integer and compare the result with the original value:
while a ~= double(int64(a))
disp('enter an integer');
....
end
int64 converts the double type to integer, and double converts it back to double. If in this process the number remains unchanged, then you could consider that it was intended to be an integer.
Recommendation for you specific program
I would use a fix function to get rid of the decimal parts. Usually when you receive a double-precision number including decimal values, the main intention is the the number before the floating point. So in many algorithms it is common practice to use fix to round each element of the given number to the nearest integer toward zero.
Say I have an array that contains the following elements:
1.0e+14 *
1.3325 1.6485 2.0402 1.0485 1.2027 2.0615 1.7432 1.9709 1.4807 0.9012
Now, is there a way to grab 1.0e+14 * (base and exponent) individually?
If I do arr(10), then this will return 9.0120e+13 instead of 0.9012e+14.
Assuming the question is to grab any elements in the array with coefficient less than one. Is there a way to obtain 1.0e+14, so that I could just do arr(i) < 1.0e+14?
I assume you want string output.
Let a denote the input numeric array. You can do it this way, if you don't mind using evalc (a variant of eval, which is considered bad practice):
s = evalc('disp(a)');
s = regexp(s, '[\de+-\.]+', 'match');
This produces a cell array with the desired strings.
Example:
>> a = [1.2e-5 3.4e-6]
a =
1.0e-04 *
0.1200 0.0340
>> s = evalc('disp(a)');
>> s = regexp(s, '[\de+-\.]+', 'match')
s =
'1.0e-04' '0.1200' '0.0340'
Here is the original answer from Alain.
Basic math can tell you that:
floor(log10(N))
The log base 10 of a number tells you approximately how many digits before the decimal are in that number.
For instance, 99987123459823754 is 9.998E+016
log10(99987123459823754) is 16.9999441, the floor of which is 16 - which can basically tell you "the exponent in scientific notation is 16, very close to being 17".
Now you have the exponent of the scientific notation. This should allow you to get to whatever your goal is ;-).
And depending on what you want to do with your exponent and the number, you could also define your own method. An example is described in this thread.