I have figures representing monetary amounts coming in -- 45.10, 24.35, 17.99, and so on.
I want to split these into the dollar and cent portions, do something to the dollars, and then output a string of dollars + '.' + cents.
PROBLEM: The figure .10 obviously becomes 1, and I don't want to output $84.1. I want $84.10. So the format string should specify "two-digit integer, with a trailing zero if there's only one digit."
Any search I do just turns up results for leading zeroes. Is this possible?
You need sprintf:
sprintf("%d.%02d", dollars, cents) # must be numbers
Related
I have a long string which contains only decimal numbers with two signs after comma
str = "123,457568,22321,5484123,77"
The numbers in string only decimals with two signs after comma. How I can separate them in different numbers like that
arr = ["123,45" , "7568,22" , "321,54" , "84123,77"]
You could try a regex split here:
str = "123,457568,22321,5484123,77"
nums = str.split(/(?<=,\d{2})/)
print nums
This prints:
123,45
7568,22
321,54
84123,77
The logic above says to split at every point where a comma followed by two digits precedes.
Scan String for Commas Followed by Two Digits
This is a case where you really need to know your data. If you always have floats with two decimal places, and commas are decimals in your locale, then you can use String#scan as follows:
str.scan /\d+,\d{2}/
#=> ["123,45", "7568,22", "321,54", "84123,77"]
Since your input data isn't consistent (which can be assumed by the lack of a reliable separator between items), you may not be able to guarantee that each item has a fractional component at all, or that the component has exactly two digits. If that's the case, you'll need to find a common pattern that is reliable for your given inputs or make changes to the way you assign data from your data source into str.
For no particular reason, I am trying to add a #reverse method to the Integer class:
class Integer
def reverse
self.to_s.reverse.to_i
end
end
puts 1337.reverse # => 7331
puts 1000.reverse # => 1
This works fine except for numbers ending in a 0, as shown when 1000.reverse returns 1 rather than 0001. Is there any way to keep leading zeroes when converting a string into an integer?
Short answer: no, you cant.
2.1.5 :001 > 0001
=> 1
0001 doesn't make sense at all as Integer. In the Integer world, 0001 is exactly as 1.
Moreover, the number of leading integer is generally irrelevant, unless you need to pad some integer for displaying, but in this case you are probably converting it into another kind of object (e.g a String).
If you want to keep the integer as Fixnum you will not be able to add leading zeros.
The real question is: why do you want/need leading zeros? You didn't provide such information in the question. There are probably better ways to achieve your result (such as wrapping the value into a decorator object if the goal is to properly format a result for display).
Does rjust work for you?
1000.to_s.reverse.to_i.to_s.rjust(1000.to_s.size,'0') #=> "0001"
self.to_s.to_i does convert the integer to a string and this string "0001" to an integer value. Since leading zeros are not required for regular numbers they are dropped. In other words: Keeping leading zeros does not make sense for calculations, so they are dropped. Just ask yourself how the integer 1 would look like if leading zeros would be preserved, since it represents a 32 bit number. If you need the leading zeros, there is no way around a string.
BUT 10 + "0001".to_i returns 11, so you probably need to override the + method of the String class.
I am trying to display a double so it will always have 4 digits after the decimal point. For example, one double would be 0.0182, and another 0.0180. However, my problem occurs when the double ends in 0, which results in truncating the zero and leaving 0.018. I'm trying to add on a zero at the end of my number to fix this, but receiving a syntax error.
Dim minFeature As Double
...
minFeature = Round(minFeature, 4) ' keep only 4 digits of precision. works.
minFeature = CDbl(CStr(minFeature).PadRight(1, "0")) ' add on an extra 0. does not work.
This will return the number as a string and preserve four decimal places.
Format(minFeature, "0.0000")
For more information, see the docs on Format().
I have the same problem as is found here for python, but for ruby.
I need to output a small number like this: 0.00001, not 1e-5.
For more information about my particular problem, I am outputting to a file using f.write("My number: " + small_number.to_s + "\n")
For my problem, accuracy isn't that big of an issue, so just doing an if statement to check if small_number < 1e-5 and then printing 0 is okay, it just doesn't seem as elegant as it should be.
So what is the more general way to do this?
f.printf "My number: %.5f\n", small_number
You can replace .5 (5 digits to the right of the decimal) with any particular formatting size you like, e.g., %8.3f would be total of 8 digits with three to the right of the decimal, much like C/C++ printf formatting strings.
If you always want 5 decimal places, you could use:
"%.5f" % small_number
I would do something like this so you can strip off trailing zero's:
puts ("%.15f" % small_number).sub(/0*$/,"")
Don't go too far past 15, or you will suffer from the imprecision of floating point numbers.
puts ("%.25f" % 0.01).sub(/0*$/,"")
0.0100000000000000002081668
This works also on integers, trim excess zeros, and always returns numbers as a valid floating point number. For clarity, this uses the sprintf instead of the more cryptic % operator.
def format_float(number)
sprintf('%.15f', number).sub(/0+$/, '').sub(/\.$/, '.0')
end
Examples:
format_float(1) => "1.0"
format_float(0.00000001) => "0.00000001"
Dim str as String
str = "30 40 50 60"
I want to count the number of substrings.
Expected Output: 4
(because there are 4 total values: 30, 40, 50, 60)
How can I accomplish this in VB6?
You could try this:
arrStr = Split(str, " ")
strCnt = UBound(arrStr) + 1
msgBox strCnt
Of course, if you've got Option Explicit set (which you should..) then declare the variables above first..
Your request doesn't make any sense. A string is a sequence of text. The fact that that sequence of text contains numbers separated by spaces is quite irrelevant. Your string looks like this:
30 40 50 60
There are not 4 separate values, there is only one value, shown above—a single string.
You could also view the string as containing 11 individual characters, so it could be argued that the "count" of the string would be 11, but this doesn't get you any further towards your goal.
In order to get the result that you expect, you need to split the string into multiple strings at each space, producing 4 separate strings, each containing a 2-digit numeric value.
Of course, the real question is why you're storing this value in a string in the first place. If they're numeric values, you should store them in an array (for example, an array of Integers). Then you can easily obtain the number of elements in the array using the LBound() and UBound() functions.
I agree with everything Cody stated.
If you really wanted to you could loop through the string character by character and count the number of times you find your delimiter. In your example, it is space delimited, so you would simply count the number of spaces and add 1, but as Cody stated, those are not separate values..
Are you trying to parse text here or what? Regardless, I think what you really need to do is store your data into an array. Make your life easier, not more difficult.