I want to write a string of variable values in a formatted way, according to the following:
Maximum decimal points is 3.
If there are less than 3 significant points than less are written.
For example:
the number 1.53848 will be written as 1.538
the number 1.0 will be written as 1 (rather than 1.000).
val variable1 = 1.
val variable2 = 1.53848
language = "%s average value is %.3f and %.3f".format(variable1, variable2)
This should do the trick:
def format(d: Double) =
BigDecimal(d).scale match {
case x if x > 2 => "%.3f".format(d)
case _ => d.toInt.toString
}
How about just removing the zeroes (and possibly the comma/separator character)?
def formatted(d: Double) = "%.3f".format(d).replaceAll(",?0+$", "")
Related
I am a newbie at matlab. I have deployed a machine learning model(developed using python) using flask. From matlab, I have called the API and received a string response. The response is like: '[0.8]'. but matlab is showing the size of the string is 1. I need only the value 0.8. My code:
import matlab.net.http.*
import matlab.net.http.field.*
request = RequestMessage( 'POST', ...
[ContentTypeField( 'application/vnd.api+json' ), AcceptField('application/vnd.api+json')], ...
'{"meta": {"Speed_RPM_PU": 0.2}}' );
response = request.send( 'http://127.0.0.1:5000/predict' );
ans=response.Body.Data
length(ans) % equals to 1
% for i = 1:length(ans)
%
% fprintf('%c ',ans(i))
%
% %disp(String(i))
%
% end
Here, ans='[0.8]'
If ans='[0.8]' then length(ans) would be equal to 5 (because there are 5 characters in this char array). I suspect that you actually have ans={'[0.8]'}, i.e. a cell array of length 1 which contains a single char array. Otherwise you are using single quotation marks '_' which indicate a char where MATLAB is actually showing double quotation marks "_" which indicate a string, since the length would be 1 if it was a string too. Chars and strings are not equivalent.
Calling class(ans) instead of length(ans) would tell you which of these is correct.
You can use str2num(ans) in any case, but if it's a cell you'll need str2num(ans{1})
str2num('[0.8]') % = 0.8 (double from char)
str2num("[0.8]") % = 0.8 (double from string)
str2num({'0.8'}) % -> error
a = {'0.8'};
str2num(a{1}) % = 0.8 (double from char element of cell)
str2double is often preferred over str2num, but you would need to remove the square brackets from your char/string or it would give NaN, so something like
str2double( erase( '[0.8]', {'[',']'} ) ) % = 0.8 (double)
Note that ans is the default variable name for unassigned function outputs, consider using a different variable name to avoid unexpected issues.
tl;dr
I'm just looking for two functions, f from double to string and g from string to double, such that g(f(d)) == d for any double d (scalar and real double).
Original question
How do I convert a double to a string or char array in a reversible way? I mean, in such a way that afterward I can convert that string/char array back to double retrieving the original result.
I've found formattedDisplayText, and in some situations it works:
>> x = eps
x =
2.220446049250313e-16
>> double(formattedDisplayText(x, 'NumericFormat', 'long')) - x
ans =
0
But in others it doesn't
x = rand(1)
x =
0.546881519204984
>> double(formattedDisplayText(x, 'NumericFormat', 'long')) - x
ans =
1.110223024625157e-16
As regards this and other tools like num2str, mat2str, at the end they all require me to decide a precision, whereas I would like to express the idea of "use whatever precision is needed for you (MATLAB) to be able to read back your own number".
Here are two simpler solutions to convert a single double value to a string and back without loss.
I want the string to be a human-readable representation of the number
Use num2str to obtain 17 decimal digits in string form, and str2double to convert back:
>> s = mat2str(x,17)
s =
'2.2204460492503131e-16'
>> y = str2double(s);
>> y==x
ans =
logical
1
Note that 17 digits are always enough to represent any IEEE double-precision floating-point number.
I want a more compact string representation of the number
Use matlab.net.base64encode to encode the 8 bytes of the number. Unfortunately you can only encode strings and integer arrays, so we type cast to some integer array (we use uint8 here, but uint64 would work too). We reverse the process to get the same double value back:
>> s = matlab.net.base64encode(typecast(x,'uint8'))
s =
'AAAAAAAAsDw='
>> y = typecast(matlab.net.base64decode(s),'double');
>> x==y
ans =
logical
1
Base64 encodes every 3 bytes in 4 characters, this is the most compact representation you can easily create. A more complex algorithm could likely convert into a smaller UTF-8-encoded string (which uses more than 6 bytes per displayable character).
Function f: from double real-valued scalar x to char vector str
str = num2str(typecast(x, 'uint8'));
str is built as a string containing 8 numbers, which correspond to the bytes in the internal representation of x. The function typecast extracts the bytes as a numerical vector, and num2str converts to a char vector with numbers separated by spaces.
Function g: from char vector str to double real-valued scalar y
y = typecast(uint8(str2double(strsplit(str))), 'double');
The char vector is split at spaces using strsplit. The result is a cell array of char vectors, each of which is then interpreted as a number by str2double, which produces a numerical vector. The numbers are cast to uint8 and then typecast interprets them as the internal representation of a double real-valued scalar.
Note that str2double(strsplit(str)) is preferred over the simpler str2num(str), because str2num internally calls eval, which is considered evil bad practice.
Example
>> format long
>> x = sqrt(pi)
x =
1.772453850905516
>> str = num2str(typecast(x, 'uint8'))
str =
'106 239 180 145 248 91 252 63'
>> y = typecast(uint8(str2double(strsplit(str))), 'double')
y =
1.772453850905516
>> x==y
ans =
logical
1
I am trying to split R15.49 to (R, 15.49) or
ZAR15.49 to (ZAR, 15.49)
I have tried one of the solutions here and implememted the function below:
def splitCurrency(string):
match = re.search(r'([\D]+)([\d,]+)', string)
output = (match.group(1), match.group(2).replace(',',''))
return output
But I am getting (R, 15) or (ZAR, 15). and its ignoring the digits after the decimal place
If you want to fish out these values from a larger text, then use re.findall:
inp = "R15.49 or ZAR15.49"
matches = re.findall(r'\b([A-Z]+)(\d+(?:\.\d+)?)\b', inp)
print(matches)
This prints:
[('R', '15.49'), ('ZAR', '15.49')]
I am trying to convert a string of varchar to ascii. Then i'm trying to make it so any number that's not 3 digits has a 0 in front of it. then i'm trying to add a 1 to the very beginning of the string and then i'm trying to make it a large number that I can apply math to it.
I've tried a lot of different coding techniques. The closest I've gotten is below:
s = 'Ak'
for c in s:
mgk = (''.join(str(ord(c)) for c in s))
num = [mgk]
var = 1
num.insert(0, var)
mgc = lambda num: int(''.join(str(i) for i in num))
num = mgc(num)
print(num)
With this code I get the output: 165107
It's almost doing exactly what I need to do but it's taking out the 0 from the ord(A) which is 65. I want it to be 165. everything else seems to be working great. I'm using '%03d'% to insert the 0.
How I want it to work is:
Get the ord() value from a string of numbers and letters.
if the ord() value is less than 100 (ex: A = 65, add a 0 to make it a 3 digit number)
take the ord() values and combine them into 1 number. 0 needs to stay in from of 65. then add a one to the list. so basically the output will look like:
1065107
I want to make sure I can take that number and apply math to it.
I have this code too:
s = 'Ak'
for c in s:
s = ord(c)
s = '%03d'%s
mgk = (''.join(str(s)))
s = [mgk]
var = 1
s.insert(0, var)
mgc = lambda s: int(''.join(str(i) for i in s))
s = mgc(s)
print(s)
but then it counts each letter as its own element and it will not combine them and I only want the one in front of the very first number.
When the number is converted to an integer, it
Is this what you want? I am kinda confused:
a = 'Ak'
result = '1' + ''.join(str(f'{ord(char):03d}') for char in a)
print(result) # 1065107
# to make it a number just do:
my_int = int(result)
how to convert decimal to binary by using repeated division in python?
i know i have to use a while loop, and use modulus sign and others {%} and {//} to do this...but i need some kind of example for me to understand how its done so i can understand completely.
CORRECT ME, if I'm wrong:
number = int(input("Enter a numberto convert into binary: "))
result = ""
while number != 0:
remainder = number % 2 # gives the exact remainder
times = number // 2
result = str(remainder) + result
print("The binary representation is", result)
break
Thank You
Making a "break" without any condition, makes the loop useless, so the code only executes once no matter what.
-
If you don't need to keep the original number, you can change "number" as you go.
If you do need to keep the original number, you can make a different variable like "times".
You seem to have mixed these two scenarios together.
-
If you want to print all the steps, the print will be inside the loop so it prints multiple times.
If you only want to print the final result, then the print goes outside the loop.
while number != 0:
remainder = number % 2 # gives the exact remainder
number = number // 2
result = str(remainder) + result
print("The binary representation is", result)
-
The concatenation line:
Putting the print inside the loop might help you see how it works.
we can make an example:
the value in result might be "11010" (a string, with quotes)
the value in remainder might be 0 (an integer, no quotes)
str(remainder) turns the remainder into a string = "0" instead of 0
So when we look at the assignment statement:
result = str(remainder) + result
The right side of the assignment operator = is evaulated first.
The right side of the = is
str(remainder) + result
which, as we went over above has the values:
"0" + "11010"
This is string concatenation. It just puts one string on the end of the other one. The result is:
"0 11010"
"011010"
That is the value evaluated on the right side of the assignment statement.
result = "011010"
Now that is the value of result.
B_Number = 0
cnt = 0
while (N != 0):
rem = N % 2
c = pow(10, cnt)
B_Number += rem * c
N //= 2
# Count used to store exponent value
cnt += 1
return B_Number