How can I check if string contains only floats separated by commas with no space.
Something like below:
str = "0.0687987167581341,0.120311605902415,89.8399554017928,198.151088713489" #true
str = "0.068798716758f1341,0.120311605902415, 89.8399554017928,198.151088713489" #False because of "f" in the first value.
str = "0.0687987167581341 0.120311605902415" # False because of no space and comma.
Basically, how can I check if a string is in the form below:
str = "<value>,<value>,<value>" # where value may only contains, integers, floats.
How about this regexp:
str.split(',').all? {|val| val =~ /\A-?\d+(\.\d+)?\Z/}
If the following matches, then it means that str does not satisfy the condition.
str =~ /[^\d.,]/
So, taking the negation:
re = /[^\d.,]/
"0.0687987167581341,0.120311605902415,89.8399554017928,198.151088713489" !~ re
# => true
"0.068798716758f1341,0.120311605902415, 89.8399554017928,198.151088713489" !~ re
# => false
"0.0687987167581341 0.120311605902415" !~ re
# => false
Edit: I thought the title of the question was pretty clear, but obviously not, as I now see that the string may also contain integers. Whether negative numbers are permitted is left unsaid, but -2.1 is a float and -3 is an integer. In any event, my solution works only for floats, though it would not be difficult to modify it to permit integers as well, but that would be too messy to bother with.
I prefer #Linuxios' solution, but I'll mention another approach in the interest of diversity. If we are not concerned about the possibility of extra leading or trailing zeroes, this should work:
str = "0.0687987167581341,0.120311605902415,89.8399554017928,198.151088713489"
str.split(',').all? { |s| s.to_f.to_s == s.strip }
#=>true
If there be extra leading or trailing zeros, or extra whitespace, I believe this would work:
str.split(',').all? {|s| s.to_f.to_s ==
s.match(/^\s*(-)?\s*?0*?(0.\d+?|[1-9]\d*\.\d+?)0*\s*$/)[1..-1].join
Related
I'm trying to use the match method with an argument of a regex to select a valid phone number, by definition, any string with nine digits.
For example:
9347584987 is valid,
(456)322-3456 is valid,
(324)5688890 is valid.
But
(340)HelloWorld is NOT valid and
456748 is NOT valid.
So far, I'm able to use \d{9} to select the example string of 9 digit characters in a row, but I'm not sure how to specifically ignore any character, such as '-' or '(' or ')' in the middle of the sequence.
What kind of Regex could I use here?
Given:
nums=['9347584987','(456)322-3456','(324)5688890','(340)HelloWorld', '456748 is NOT valid']
You can split on a NON digit and rejoin to remove non digits:
> nums.map {|s| s.split(/\D/).join}
["9347584987", "4563223456", "3245688890", "340", "456748"]
Then filter on the length:
> nums.map {|s| s.split(/\D/).join}.select {|s| s.length==10}
["9347584987", "4563223456", "3245688890"]
Or, you can grab a group of numbers that look 'phony numbery' by using a regex to grab digits and common delimiters:
> nums.map {|s| s[/[\d\-()]+/]}
["9347584987", "(456)322-3456", "(324)5688890", "(340)", "456748"]
And then process that list as above.
That would delineate:
> '123 is NOT a valid area code for 456-7890'[/[\d\-()]+/]
=> "123" # no match
vs
> '123 is NOT a valid area code for 456-7890'.split(/\D/).join
=> "1234567890" # match
I suggest using one regular expression for each valid pattern rather than constructing a single regex. It would be easier to test and debug, and easier to maintain the code. If, for example, "123-456-7890" or 123-456-7890 x231" were in future deemed valid numbers, one need only add a single, simple regex for each to the array VALID_PATTERS below.
VALID_PATTERS = [/\A\d{10}\z/, /\A\(\d{3}\)\d{3}-\d{4}\z/, /\A\(\d{3}\)\d{7}\z/]
def valid?(str)
VALID_PATTERS.any? { |r| str.match?(r) }
end
ph_nbrs = %w| 9347584987 (456)322-3456 (324)5688890 (340)HelloWorld 456748 |
ph_nbrs.each { |s| puts "#{s.ljust(15)} \#=> #{valid?(s)}" }
9347584987 #=> true
(456)322-3456 #=> true
(324)5688890 #=> true
(340)HelloWorld #=> false
456748 #=> false
String#match? made its debut in Ruby v2.4. There are many alternatives, including str.match(r) and str =~ r.
"9347584987" =~ /(?:\d.*){9}/ #=> 0
"(456)322-3456" =~ /(?:\d.*){9}/ #=> 1
"(324)5688890" =~ /(?:\d.*){9}/ #=> 1
"(340)HelloWorld" =~ /(?:\d.*){9}/ #=> nil
"456748" =~ /(?:\d.*){9}/ #=> nil
Pattern: (Rubular Demo)
^\(?\d{3}\)?\d{3}-?\d{4}$ # this makes the expected symbols optional
This pattern will ensure that an opening ( at the start of the string is followed by 3 numbers the a closing ).
^(\(\d{3}\)|\d{3})\d{3}-?\d{4}$
On principle, though, I agree with melpomene in advising that you remove all non-digital characters, test for 9 character length, then store/handle the phone numbers in a single/reliable/basic format.
I am trying to call the first duplicate character in my string in Ruby.
I have defined an input string using gets.
How do I call the first duplicate character in the string?
This is my code so far.
string = "#{gets}"
print string
How do I call a character from this string?
Edit 1:
This is the code I have now where my output is coming out to me No duplicates 26 times. I think my if statement is wrongly written.
string "abcade"
puts string
for i in ('a'..'z')
if string =~ /(.)\1/
puts string.chars.group_by{|c| c}.find{|el| el[1].size >1}[0]
else
puts "no duplicates"
end
end
My second puts statement works but with the for and if loops, it returns no duplicates 26 times whatever the string is.
The following returns the index of the first duplicate character:
the_string =~ /(.)\1/
Example:
'1234556' =~ /(.)\1/
=> 4
To get the duplicate character itself, use $1:
$1
=> "5"
Example usage in an if statement:
if my_string =~ /(.)\1/
# found duplicate; potentially do something with $1
else
# there is no match
end
s.chars.map { |c| [c, s.count(c)] }.drop_while{|i| i[1] <= 1}.first[0]
With the refined form from Cary Swoveland :
s.each_char.find { |c| s.count(c) > 1 }
Below method might be useful to find the first word in a string
def firstRepeatedWord(string)
h_data = Hash.new(0)
string.split(" ").each{|x| h_data[x] +=1}
h_data.key(h_data.values.max)
end
I believe the question can be interpreted in either of two ways (neither involving the first pair of adjacent characters that are the same) and offer solutions to each.
Find the first character in the string that is preceded by the same character
I don't believe we can use a regex for this (but would love to be proved wrong). I would use the method suggested in a comment by #DaveNewton:
require 'set'
def first_repeat_char(str)
str.each_char.with_object(Set.new) { |c,s| return c unless s.add?(c) }
nil
end
first_repeat_char("abcdebf") #=> b
first_repeat_char("abcdcbe") #=> c
first_repeat_char("abcdefg") #=> nil
Find the first character in the string that appears more than once
r = /
(.) # match any character in capture group #1
.* # match any character zero of more times
? # do the preceding lazily
\K # forget everything matched so far
\1 # match the contents of capture group 1
/x
"abcdebf"[r] #=> b
"abccdeb"[r] #=> b
"abcdefg"[r] #=> nil
This regex is fine, but produces the warning, "regular expression has redundant nested repeat operator '*'". You can disregard the warning or suppress it by doing something clunky, like:
r = /([^#{0.chr}]).*?\K\1/
where ([^#{0.chr}]) means "match any character other than 0.chr in capture group 1".
Note that a positive lookbehind cannot be used here, as they cannot contain variable-length matches (i.e., .*).
You could probably make your string an array and use detect. This should return the first char where the count is > 1.
string.split("").detect {|x| string.count(x) > 1}
I'll use positive lookahead with String#[] method :
"abcccddde"[/(.)(?=\1)/] #=> c
As a variant:
str = "abcdeff"
p str.chars.group_by{|c| c}.find{|el| el[1].size > 1}[0]
prints "f"
I have a string composed by words divided by'#'. For instance 'this#is#an#example' and I need to extract the last word or the last two words according to the second to last word.
If the second to last is 'myword' I need the last two words otherwise just the last one.
'this#is#an#example' => 'example'
'this#is#an#example#using#myword#also' => 'myword#also'
Is there a better way than splitting and checking the second to last? perhaps using regular expression?
Thanks.
You can use the end-of-line anchor $ and make the myword# prefix optional:
str = 'this#is#an#example'
str[/(?:#)((myword#)?[^#]+)$/, 1]
#=> "example"
str = 'this#is#an#example#using#myword#also'
str[/(?:#)((myword#)?[^#]+)$/, 1]
#=> "myword#also"
However, I don't think using a regular expression is "better" in this case. I would use something like Santosh's (deleted) answer: split the line by # and use an if clause.
def foo(str)
*, a, b = str.split('#')
if a == 'myword'
"#{a}##{b}"
else
b
end
end
str = 'this#is#an#example#using#myword#also'
array = str.split('#')
array[-2] == 'myword' ? array[-2..-1].join('#') : array[-1]
With regex:
'this#is#an#example'[/(myword\#)*\w+$/]
# => "example"
'this#is#an#example#using#myword#also'[/(myword\#)*\w+$/]
# => "myword#also"
Let's say I have the following array:
arr = ["", "2121", "8", "myString"]
I want to return false in case the array contains any non-digit symbols.
arr.all? { |s| s =~ /^\d+$/ }
This will check for each element if it consists only of digits (\d) – If any of them does not, false will be returned.
Edit: You didn't completely specify if the empty string is valid or not. If it is, the line has to be rewritten as follows (as per DarkDust):
arr.all? {|s| s =~ /^\d*$/ }
If empty strings are allowed:
def contains_non_digit(array)
!array.select {|s| s =~ /^.*[^0-9].*$/}.empty?
end
Explanation: this filters the array for all strings that match a regular expression. This regex is true for a string that contains at least one non-digit character. If the resulting array is empty, the array contains no non-digit strings. Finally, we need to negate the result, because we want to know the array does contain non-digit strings.
I want to replace the last occurrence of a substring in Ruby. What's the easiest way?
For example, in abc123abc123, I want to replace the last abc to ABC. How do I do that?
How about
new_str = old_str.reverse.sub(pattern.reverse, replacement.reverse).reverse
For instance:
irb(main):001:0> old_str = "abc123abc123"
=> "abc123abc123"
irb(main):002:0> pattern="abc"
=> "abc"
irb(main):003:0> replacement="ABC"
=> "ABC"
irb(main):004:0> new_str = old_str.reverse.sub(pattern.reverse, replacement.reverse).reverse
=> "abc123ABC123"
"abc123abc123".gsub(/(.*(abc.*)*)(abc)(.*)/, '\1ABC\4')
#=> "abc123ABC123"
But probably there is a better way...
Edit:
...which Chris kindly provided in the comment below.
So, as * is a greedy operator, the following is enough:
"abc123abc123".gsub(/(.*)(abc)(.*)/, '\1ABC\3')
#=> "abc123ABC123"
Edit2:
There is also a solution which neatly illustrates parallel array assignment in Ruby:
*a, b = "abc123abc123".split('abc', -1)
a.join('abc')+'ABC'+b
#=> "abc123ABC123"
Since Ruby 2.0 we can use \K which removes any text matched before it from the returned match. Combine with a greedy operator and you get this:
'abc123abc123'.sub(/.*\Kabc/, 'ABC')
#=> "abc123ABC123"
This is about 1.4 times faster than using capturing groups as Hirurg103 suggested, but that speed comes at the cost of lowering readability by using a lesser-known pattern.
more info on \K: https://www.regular-expressions.info/keep.html
Here's another possible solution:
>> s = "abc123abc123"
=> "abc123abc123"
>> s[s.rindex('abc')...(s.rindex('abc') + 'abc'.length)] = "ABC"
=> "ABC"
>> s
=> "abc123ABC123"
When searching in huge streams of data, using reverse will definitively* lead to performance issues. I use string.rpartition*:
sub_or_pattern = "!"
replacement = "?"
string = "hello!hello!hello"
array_of_pieces = string.rpartition sub_or_pattern
( array_of_pieces[(array_of_pieces.find_index sub_or_pattern)] = replacement ) rescue nil
p array_of_pieces.join
# "hello!hello?hello"
The same code must work with a string with no occurrences of sub_or_pattern:
string = "hello_hello_hello"
# ...
# "hello_hello_hello"
*rpartition uses rb_str_subseq() internally. I didn't check if that function returns a copy of the string, but I think it preserves the chunk of memory used by that part of the string. reverse uses rb_enc_cr_str_copy_for_substr(), which suggests that copies are done all the time -- although maybe in the future a smarter String class may be implemented (having a flag reversed set to true, and having all of its functions operating backwards when that is set), as of now, it is inefficient.
Moreover, Regex patterns can't be simply reversed. The question only asks for replacing the last occurrence of a sub-string, so, that's OK, but readers in the need of something more robust won't benefit from the most voted answer (as of this writing)
You can achieve this with String#sub and greedy regexp .* like this:
'abc123abc123'.sub(/(.*)abc/, '\1ABC')
simple and efficient:
s = "abc123abc123abc"
p = "123"
s.slice!(s.rindex(p), p.size)
s == "abc123abcabc"
string = "abc123abc123"
pattern = /abc/
replacement = "ABC"
matches = string.scan(pattern).length
index = 0
string.gsub(pattern) do |match|
index += 1
index == matches ? replacement : match
end
#=> abc123ABC123
I've used this handy helper method quite a bit:
def gsub_last(str, source, target)
return str unless str.include?(source)
top, middle, bottom = str.rpartition(source)
"#{top}#{target}#{bottom}"
end
If you want to make it more Rails-y, extend it on the String class itself:
class String
def gsub_last(source, target)
return self unless self.include?(source)
top, middle, bottom = self.rpartition(source)
"#{top}#{target}#{bottom}"
end
end
Then you can just call it directly on any String instance, eg "fooBAR123BAR".gsub_last("BAR", "FOO") == "fooBAR123FOO"
.gsub /abc(?=[^abc]*$)/, 'ABC'
Matches a "abc" and then asserts ((?=) is positive lookahead) that no other characters up to the end of the string are "abc".