Is there a simple way to print a number with either one decimal place or none?
I've searched the net for a method to do that but all of them try to always have a zero after the decimal point..
I want 3.0 to be printed as just 3, and 3.5 to be printed as 3.5.
I tried print('{:.1f}'.format(num)) but this prints 3.0
You have not specified the programming language, so I will provide an answer in pseudocode.
Using if-else
printWithOneOrNoDecimals(n)
if (isNumberInteger(n))
printWithoutDecimals(n)
else
printNumberWithOneDecimal(n)
isNumberInteger(n)
return round(n) == n
The method round(n) should round the number, for example 2.4 to 2. Because 2.4 != 2, isNumberInteger(2.4) would return false and the else statement is exectuted.
Now you can define different formats for printing numbers with or without decimal in printNumberWithOneDecimal and printWithoutDecimals.
Using “right trim” or “right strip”
Another way to achieve the result is to first make the number a string (maybe formatting it to have one decimal) and then “trimming” or “stripping” it from the right, first “all” zeros (in your case at most one) and then the decimal point (if any on the right). Note: trim or strip methods do not give an error if there’s nothing to trim/strip.
I guess that in Python it would be:
'{:.1f}'.format(num).rstrip('0').rstrip('.')
Happy coding, good luck!
Related
I'm making a list for buying groceries in Google Sheets and have the following value in cell B4.
0.95 - Lemon Juice
2.49 - Pringle Chips
1.29 - Baby Carrots
9.50 - Chicken Kebab
What I'm trying to do is split using the dash character and combine the costs (0.95+2.49+1.29+9.50).
I've tried to use Index(SPLIT(B22,"-"), 7) and SPLIT(B22,"-") but I don't know how to use only numbers from the split string.
Does someone know how to do this? Here's a sample sheet.
Answer
The following formula should produce the result you desire:
=SUM(ARRAYFORMULA(VALUE(REGEXEXTRACT(SPLIT(B4,CHAR(10)),"(.*)-"))))
Explanation
The first thing to do is to split the entry in B4 into its component parts. This is done by using the =SPLIT function, which takes the text in B4 and returns a separate result every time it encounters a specific delimiter. In this case, that is =CHAR(10), the newline character.
Next, all non-number information needs to be removed. This is relatively easy in your sample data because the numbers always appear to the left of a dash. =REGEXEXTRACT uses a regular expression to only return the text to the left of the dash.
Before the numbers can be added together, however, they must be converted to be in a number format. The =VALUE function is used to convert each result from a text string containing a number to an actual number.
All of this is wrapped in an =ARRAYFORMULA so that =VALUE and =REGEXEXTRACT parse each returned value from =SPLIT, rather than just the first.
Finally, all results are added together using =SUM.
Functions used:
=CHAR
=SPLIT
=REGEXEXTRACT
=VALUE
=ARRAYFORMULA
=SUM
Firstly you can add , symbols start and ends of numbers with below code:
REGEXREPLACE(B4,"([0-9\.]+)",",$1,")
Then split it based of , sign.
SPLIT(A8, ",")
Try below formula (see your sheet)-
=SUM(ArrayFormula(--REGEXEXTRACT(SPLIT(B4,CHAR(10)),"-*\d*\.?\d+")))
I've been looking for a good way to see if a string of items are all numbers, and thought there might be a way of specifying a range from 0 to 9 and seeing if they're included in the string, but all that I've looked up online has really confused me.
def validate_pin(pin)
(pin.length == 4 || pin.length == 6) && pin.count("0-9") == pin.length
end
The code above is someone else's work and I've been trying to identify how it works. It's a pin checker - takes in a set of characters and ensures the string is either 4 or 6 digits and all numbers - but how does the range work?
When I did this problem I tried to use to_a? Integer and a bunch of other things including ranges such as (0..9) and ("0..9) and ("0".."9") to validate a character is an integer. When I saw ("0-9) it confused the heck out of me, and half an hour of googling and youtube has only left me with regex tutorials (which I'm interested in, but currently just trying to get the basics down)
So to sum this up, my goal is to understand a more semantic/concise way to identify if a character is an integer. Whatever is the simplest way. All and any feedback is welcome. I am a new rubyist and trying to get down my fundamentals. Thank You.
Regex really is the right way to do this. It's specifically for testing patterns in strings. This is how you'd test "do all characters in this string fall in the range of characters 0-9?":
pin.match(/\A[0-9]+\z/)
This regex says "Does this string start and end with at least one of the characters 0-9, with nothing else in between?" - the \A and \z are start-of-string and end-of-string matchers, and the [0-9]+ matches any one or more of any character in that range.
You could even do your entire check in one line of regex:
pin.match(/\A([0-9]{4}|[0-9]{6})\z/)
Which says "Does this string consist of the characters 0-9 repeated exactly 4 times, or the characters 0-9, repeated exactly 6 times?"
Ruby's String#count method does something similar to this, though it just counts the number of occurrences of the characters passed, and it uses something similar to regex ranges to allow you to specify character ranges.
The sequence c1-c2 means all characters between c1 and c2.
Thus, it expands the parameter "0-9" into the list of characters "0123456789", and then it tests how many of the characters in the string match that list of characters.
This will work to verify that a certain number of numbers exist in the string, and the length checks let you implicitly test that no other characters exist in the string. However, regexes let you assert that directly, by ensuring that the whole string matches a given pattern, including length constraints.
Count everything non-digit in pin and check if this count is zero:
pin.count("^0-9").zero?
Since you seem to be looking for answers outside regex and since Chris already spelled out how the count method was being implemented in the example above, I'll try to add one more idea for testing whether a string is an Integer or not:
pin.to_i.to_s == pin
What we're doing is converting the string to an integer, converting that result back to a string, and then testing to see if anything changed during the process. If the result is =>true, then you know nothing changed during the conversion to an integer and therefore the string is only an Integer.
EDIT:
The example above only works if the entire string is an Integer and won’t properly deal with leading zeros. If you want to check to make sure each and every character is an Integer then do something like this instead:
pin.prepend(“1”).to_i.to_s(1..-1) == pin
Part of the question seems to be exactly HOW the following portion of code is doing its job:
pin.count("0-9")
This piece of the code is simply returning a count of how many instances of the numbers 0 through 9 exist in the string. That's only one piece of the relevant section of code though. You need to look at the rest of the line to make sense of it:
pin.count("0-9") == pin.length
The first part counts how many instances then the second part compares that to the length of the string. If they are equal (==) then that means every character in the string is an Integer.
Sometimes negation can be used to advantage:
!pin.match?(/\D/) && [4,6].include?(pin.length)
pin.match?(/\D/) returns true if the string contains a character other than a digit (matching /\D/), in which case it it would be negated to false.
One advantage of using negation here is that if the string contains a character other than a digit pin.match?(/\D/) would return true as soon as a non-digit is found, as opposed to methods that examine all the characters in the string.
I couldn't really clarify what I'm asking in the title. I an integer for a day and a month. I have to print the month with a 0 in front of it if it's one digit only.
For example 04 if month = 4 and so on.
This is how it's supposed to look like in C#:
Console.WriteLine("{0}.{1:00}", day, month);
Thank you.
int month = 4;
DecimalFormat formater = new DecimalFormat("00");
String month_formated = formater.format(month);
Besides the answer Fernando Lahoz provided (which is pretty specific to your case: decimal formating) you can also use System.out.format in Java which allows you to specify a format-string while printing to System.out (the format function is applicable to any PrintStream though). In your case
System.out.format("%2d %2d", day, month)
should do the trick. The %dis used for decimal integers and you can then specify any width you want just before the 'd' (2 in your case).
If you want to access the string formed for later use and not (only) print it you can use String.format. It uses the same format as System.out.format but returns the String that is formed.
A complete syntax for all formats(string, decimal, floating point, calendar, date/time, ...) can be found here.
If you'd like a quick tuto on number-formatting you can check this link or this link instead.
Good luck!
Trying to work out how to parse out phone numbers that are left in a string.
e.g.
"Hi Han, this is Chewie, Could you give me a call on 02031234567"
"Hi Han, this is Chewie, Could you give me a call on +442031234567"
"Hi Han, this is Chewie, Could you give me a call on +44 (0) 203 123 4567"
"Hi Han, this is Chewie, Could you give me a call on 0207-123-4567"
"Hi Han, this is Chewie, Could you give me a call on 02031234567 OR +44207-1234567"
And be able to consistently replace any one of them with some other item (e.g. some text, or a link).
Am assuming it's a regex type approach (I'm already doing something similar with email which works well).
I've got to
text.scan(/([^A-Z|^"]{6,})/i)
Which leaves me a leading space I can't work out how to drop (would appreciate the help there).
Is there a standard way of doing this that people use?
It also drops things into arrays, which isn't particularly helpful
i.e. if there were multiple numbers.
[["02031234567"]["+44207-1234567"]]
as opposed to
["02031234567","+44207-1234567"]
Adding in the third use-case with spaces is difficult. I think the only way to successfully meet that acceptance criteria would be to chain a #gsub call on to your #scan.
Thus:
text.gsub(/\s+/, "").scan(/([^A-Z|^"|^\s]{6,})/i)
The following code will extract all the numbers for you:
text.scan(/(?<=[ ])[\d \-+()]+$|(?<=[ ])[\d \-+()]+(?=[ ]\w)/)
For the examples you supplied this results in:
["02031234567"]
["+442031234567"]
["+44 (0) 203 123 4567"]
["0207-123-4567"]
["02031234567", "+44207-1234567"]
To understand this regex, what we are matching is:
[\d \-+()]+ which is a sequence of one or more digits, spaces, minus, plus, opening or closing brackets (in any order - NB regex is greedy by default, so it will match as many of these characters next to each other as possible)
that must be preceded by a space (?<=[ ]) - NB the space in the positive look-behind is not captured, and therefore this makes sure that there are no leading spaces in the results
and is either at the end of the string $, or | is followed by a space then a word character (?=[ ]\w) (NB this lookahead is not captured)
This pattern will get rid of the space but not match your third case with spaces:
/([^A-Z|^"|^\s]{6,})/i
This is what I came to in the end in case it helps somebody
numbers = text.scan(/([^A-Z|^"]{6,})/i).collect{|x| x[0].strip }
That gives me an array of
["+442031234567", "02031234567"]
I'm sure there is a more elegant way of doing this and possibly you'd want to check the numbers for likelihood of being phonelike - e.g. using the brilliant Phony gem.
numbers = text.scan(/([^A-Z|^"]{6,})/i).collect{|x| x[0].strip }
real_numbers = numbers.keep_if{|n| Phony.plausible? PhonyRails.normalize_number(n, default_country_code: "GB")}
Which should help exclude serial numbers or the like from being identified as numbers. You'll obviously want to change the country code to something relevant for you.
How do I check a value starting with a decimal
is_a_number(value) .... works for 12, 12.0, 12.2, 0.23 but not .23
Basically I'm doing a validation in a form, and I want to allow values starting with . i.e .23
but obviously pop up a flag (false) when its not a number
".23" isn't really a number, in my book.
If you want to treat it like one, check if the first character is a decimal point, if it is, prepend a "0" and try again.
Actually, you could probably prepend a zero regardless. It shouldn't affect the value of any "legitimate" number. (EDIT: As long as you can explicitly specify base 10 when actually converting to a number)
Read your input into a string and dynamically add the zero if needed. For example:
if (inputvar[0] == '.')
inputvar = "0#{inputvar}"
end
The resulting value can be converted into a number by .to_i, .to_f, etc.