I want to find a specific character in a given string of number for example if my input is:
1 4 5 7 9 12
Then for 4 the answer should be 1. My code is as follows:
secarr = second.split(" ")
answer = secarr.index(number) #here number is a variable which gets the character
puts answer
The above method works if I write "4" instead of number or any other specific character but does not work if I write a variable. Is there a method in ruby to do the same?
This is probably your variable number is an Integer, and secarr is an Array of Strings. Try to cast the number to string:
answer = secarr.index(number.to_s)
Related
I have a text file which contains instructions. I'm reading it using File.readlines(filename). I want to check that the file is formatted as follows:
Has 3 lines
Line 1: two integers (including negatives) separated by a space
Line 2: two integers (including negatives) separated by a space and 1 capitalised letter of the alphabet also separated by a space.
Line 3: capitalised letters of the alphabet without any spaces (or punctuation).
This is what the file should look like:
8 10
1 2 E
MMLMRMMRRMML
So far I have calculated the number of lines using File.readlines(filename).length. How do I check the format of each line, do I need to loop through the file?
EDIT:
I solved the problem by creating three methods containing regular expressions, then I passed each line into it's function and created a conditional statement to check if the out put was true.
Suppose IO::read is used to return the following string str.
str = <<~END
8 10
1 2 E
MMLMRMMRRMML
END
#=> "8 10\n1 2 E\nMMLMRMMRRMML\n"
You can then test the string with a single regular expression:
r = /\A(-?\d+) \g<1>\n\g<1> \g<1> [A-Z]\n[A-Z]+\n\z/
str.match?(r)
#=> true
I could have written
r = /\A-?\d+ -?\d+\n-?\d+ -?\d+ [A-Z]\n[A-Z]+\n\z/
but matching an integer (-?\d+) is done three times. It's slightly shorter, and reduces the chance of error, to put the first of the three in capture group 1, and then treat that as a subexpression by calling it with \g<1> (not to be confused with a back-reference, which is written \k<1>). Alternatively, I could have use named capture groups:
r = /\A(?<int>-?\d+) \g<int>\n\g<int> \g<int> (?<cap>[A-Z])\n\g<cap>+\n\z/
This question already has answers here:
How Do I Use VBScript to Strip the First n Characters of a String?
(4 answers)
Closed 6 years ago.
I have a program that prints a string to a notepad file, the output being something random:
#'f7ruhigbergbn
I want to however remove the first 3 characters from the pasted result, how can I do this?
myString = "#'f"
result = workings - myString
This does not work, bare in mind the first 3 characters are always going to be #'f
Any thoughts? thanks
You can use:
result = Mid(workings, 4)
You can use Right function for get the X characters of the right side of string. With Len function you can get the length of the string.
Right(myString,Len(myAtring) - 3)
With this, you get a new string whitout the three first characters, now you can assign to the same string:
myString = Right(myString,Len(myAtring) - 3)
Try this:
mystring = "#'f7ruhigbergbn"
result = Mid(mystring, 3, mystring.length)
I have a timestamp
timestamp = 1466754627
And I want to assert that the first 4 characters are 1466 within a test. If i was to use
timestamp.includes('1466')
This will pass, but 1466 could appear anywhere and not necessarily the first 4 characters. How could i ensure that 1466 are the first 4 characters ?
Thanks
There's a start_with matcher:
expect('1466754627').to start_with('1466')
You might have to convert timestamp to a string via to_s if it is indeed an integer:
timestamp = 1466754627
expect(timestamp.to_s).to start_with('1466')
I am wondering how to make something where if X=5 and Y=2, then have it output something like
Hello 2 World 5.
In Java I would do
String a = "Hello " + Y + " World " + X;
System.out.println(a);
So how would I do that in TI-BASIC?
You have two issues to work out, concatenating strings and converting integers to a string representation.
String concatenation is very straightforward and utilizes the + operator. In your example:
"Hello " + "World"
Will yield the string "Hello World'.
Converting numbers to strings is not as easy in TI-BASIC, but a method for doing so compatible with the TI-83+/84+ series is available here. The following code and explanation are quoted from the linked page:
:"?
:For(X,1,1+log(N
:sub("0123456789",ipart(10fpart(N10^(-X)))+1,1)+Ans
:End
:sub(Ans,1,length(Ans)-1?Str1
With our number stored in N, we loop through each digit of N and store
the numeric character to our string that is at the matching position
in our substring. You access the individual digit in the number by
using iPart(10fPart(A/10^(X, and then locate where it is in the string
"0123456789". The reason you need to add 1 is so that it works with
the 0 digit.
In order to construct a string with all of the digits of the number, we first create a dummy string. This is what the "? is used
for. Each time through the For( loop, we concatenate the string from
before (which is still stored in the Ans variable) to the next numeric
character that is found in N. Using Ans allows us to not have to use
another string variable, since Ans can act like a string and it gets
updated accordingly, and Ans is also faster than a string variable.
By the time we are done with the For( loop, all of our numeric characters are put together in Ans. However, because we stored a dummy
character to the string initially, we now need to remove it, which we
do by getting the substring from the first character to the second to
last character of the string. Finally, we store the string to a more
permanent variable (in this case, Str1) for future use.
Once converted to a string, you can simply use the + operator to concatenate your string literals with the converted number strings.
You should also take a look at a similar Stack Overflow question which addresses a similar issue.
For this issue you can use the toString( function which was introduced in version 5.2.0. This function translates a number to a string which you can use to display numbers and strings together easily. It would end up like this:
Disp "Hello "+toString(Y)+" World "+toString(X)
If you know the length of "Hello" and "World," then you can simply use Output() because Disp creates a new line after every statement.
I have a string that
contains at least one number
can contain multiple numbers
Some examples are:
https://www.facebook.com/permalink.php?story_fbid=53199604568&id=218700384
https://www.facebook.com/username_13/posts/101505775425651120
https://www.facebook.com/username/posts/101505775425699820
I need a way to extract the longest number from the string. So for the 3 strings above, it would extract
53199604568
101505775425651120
101505775425699820
How can I do this?
#get the lines first
text = <<ENDTEXT
https://www.facebook.com/permalink.php?story_fbid=53199604568&id=218700384
https://www.facebook.com/username_13/posts/101505775425651120
https://www.facebook.com/username/posts/101505775425699820
ENDTEXT
lines = text.split("\n")
#this bit is the actual answer to your question
lines.collect{|line| line.scan(/\d+/).sort_by(&:length).last}
Note that i'm returning the numbers as strings here. You could convert them to numbers with to_i
parse the list (to get an int array), then use the Max function. array.Max for syntax.
s = "https://www.facebook.com/permalink.php?story_fbid=53199604568&id=218700384"
s.scan(/\d+/).max{|a,b| a.length <=> b.length}.to_i