I'm trying to use a regular expression to solve a reverse polish calculator problem, but I'm having issues with converting the mathematical expressions into conventional form.
I wrote:
puts '35 29 1 - 5 + *'.gsub(/(\d*) (\d*) (\W)/, '(\1\3\2)')
which prints:
35 (29-1)(+5) *
expected
(35*((29-1)+5))
but I'm getting a different result. What am I doing wrong?
I'm assuming you meant you tried
puts '35 29 1 - 5 + *'.gsub(/(\d*) (\d*) (\W)/, '(\1\3\2)')
^ ^
Anyway, you have to use the quantifier + instead of *, since otherwise you will match an empty string for \d* as one of your captures, hence the (+5):
/(\d+) (\d+) (\W)/
I would further extend/constrain the expression to something like:
/([\d+*\/()-]+)\s+([\d+*\/()-]+)\s+([+*\/-])/
| | | | |
| | | | Valid operators, +, -, *, and /.
| | | |
| | | Whitespace.
| | |
| | Arbitrary atom, e.g. "35", "(29-1)", "((29-1)+5)".
| |
| Whitepsace.
|
Arbitrary atom, e.g. "35", "(29-1)", "((29-1)+5)".
...and instead of using gsub, use sub in a while loop that quits when it detects that no more substitutions can be made. This is very important because otherwise, you will violate the order of operations. For example, take a look at this Rubular demo. You can see that by using gsub, you might potentially replace the second triad of atoms, "5 + *", when really a second iteration should substitute an "earlier" triad after substituting the first triad!
WARNING: The - (minus) character must appear first or last in a character class, since otherwise it will specify a range! (Thanks to #JoshuaCheek.)
Related
There are some lesser known bash variable expansions:
+----------------------------------------------------------+----------------+
| description | expression |
+----------------------------------------------------------+----------------+
| Remove everything **after** the **last** '7' | ${var%7*} |
| Remove everything **after** the **first** '7' | ${var%%7*} |
| Remove everything **before** the **first** '7' | ${var#*7} |
| Remove everything **before** the **last** '7' | ${var##*7} |
| First char upper case | ${var^} |
| All upper case | ${var^^} |
| First char lower case | ${var,} |
| All lower case | ${var,,} |
| Show how variable was set | ${var#A} |
| ?? something cool ?? | ${var#E} |
| Print variable as though it were the prompt variable PS1 | ${var#P} |
| ?? something cool ?? | ${var#Q} |
+----------------------------------------------------------+----------------+
I have been struggling to find a source that documents all of these tricks. So far the best one I have found is this cheat sheet. But even that page is missing some of these expansion rules. For the purposes of writing good bash code, and making that code portable I am looking for several things:
What are all of the bash variable expansion tricks?
Where is there a document that shows all of them (with examples ideally)?
What versions of bash do which tricks work with?
Some good pointers on parameter expansions:
https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
http://mywiki.wooledge.org/BashFAQ/073
https://wiki.bash-hackers.org/syntax/pe
You missed many, like
single substitution a -> b : ${x/a/b}
multiple substitutions a -> b : ${x//a/b}
offset manipulation: ${x:1:3}
${var-word} if var is defined, use var; otherwise, "word"
${var+word} if var is defined, use "word"; otherwise, nothing
${var=word} if var is defined, use var; otherwise, use "word" AND also assign "word" to var
${var?error} if var is defined, use var; otherwise print "error" and exit
array slice ${files[#]: -4}
Note that most of PE works with array too
How to count number of integers in a file using egrep?
I tried to solve it as a pattern finding problem. Actually, I am facing problem of how to represent range of characters [0-9] continuously which include "space" before the beginning and "space or dot" after the end. I think the latter can be solved by using \< and \> respectively. Also, It should not include dot in between otherwise it will not be an integer. I am unable to convert this logic into regular expression using available tools and techniques.
My name is 2322.
33 is my sister.
I am blessed with a son named 55.
Why are you so 69. Is everything 33.
66.88 is not an integer
55whereareyou?
The right answer should be 5 i.e. for 2322, 33, 55, 69 and 33.
grep -Eo '(^| )([0-9]+[\.\?\=\:]?( |$))+' | wc -w
^^ ^ ^ ^ ^ ^ ^
|| | | | | | |
E = extended regex--------+| | | | | | |
o = extract what found-----+ | | | | | |
starts with new line or space---+ | | | | |
digits--------------------------------+ | | | |
optional dot, question mark, etc.-------------+ | | |
ends with end line or space----------------------------+ | |
repeat 1 time or more (to detect integers like "123 456")--+ |
count words------------------------------------------------------+
Note: 123. 123? 123: are also counted as integer
Test:
#!/bin/bash
exec 3<<EOF
My name is 2322.
33 is my sister.
I am blessed with a son named 55.
Why are you so 69. Is everything 33.
66.88 is not an integer
55whereareyou?
two integers 123 456.
how many tables in room 400? 50.
50? oh I thought it was 40.
23: It's late, 23:00 already
EOF
grep -Eo '(^| )([0-9]+[\.\?\=\:]?( |$))+' <&3 | \
tee >(sleep 0.5; echo -n "integer counted: "; wc -w; )
Outputs:
2322.
33
55.
69.
33.
123 456.
400? 50.
50?
40.
23:
integer counted: 12
Based on the observation that you want 66.88 excluded, I'm guessing
grep -Ec '[0-9]\.?( |$)' file
which finds a digit, optionally followed by a dot, followed by either a space or end of line.
The -c option says to report the number of lines which contain a match (so not strictly the number of matches, if there are lines which contain multiple matches) and the -E option enables extended regular expression syntax, i.e. what was traditionally calned egrep (though the command name is now obsolescent).
If you need to count matches, the -o option prints each match on a separate line, which you can then pass to wc -l (or in lucky cases combine with grep -c, but check first; this doesn't work e.g. with GNU grep currently).
On my ubuntu this code working fine
grep -P '((^)|(\s+))[-+]?\d+\.?((\s+)|($))' test
I need to tell Ruby in regex to split before and after the + - * / symbols in my program.
Examples:
I need to turn "1+12" into [1.0, "+", 12.0]
and "6/0.25" into [6.0, "/", 0.25]
There could be cases like "3/0.125" but highly unlikely. If first two I listed above are satisfied it should be good.
On the Ruby docs, "hi mom".split(%r{\s*}) #=> ["h", "i", "m", "o", "m"]
I looked up a cheat-sheet to try to understand %r{\s*}, and I know that the stuff inside %r{} such as \s are skipped and \s means white space in regex.
'1.0+23.7'.scan(/(((\d\.?)+)|[\+\-\*\/])/)
instead of splitting, match with capture groups to parse your inputs:
(?<operand1>(?:\d+(?:\.\d+)?)|(?:\.\d+))\s*(?<operator>[+\/*-])\s*(?<operand2>(?:\d+(?:\.\d+)?)|(?:\.\d+))
explanation:
I've used named groups (?<groupName>regex) but they aren't necessary and could just be ()'s - either way, the sub-captures will still be available as 1,2,and 3. Also note the (?:regex) constructs that are for grouping only and do not "remember" anything, and won't mess up your captures)
(?:\d+(?:\.\d+)?)|(?:\.\d+)) first number: either leading digit(s) followed optionally by a decimal point and digit(s), OR a leading decimal point followed by digit(s)
\s* zero or more spaces in between
[+\/*-] operator: character class meaning a plus, division sign, minus, or multiply.
\s* zero or more spaces in between
(?:\d+(?:\.\d+)?)|(?:\.\d+) second number: same pattern as first number.
regex demo output:
I arrived a little late to this party, and found that many of the good answers had already been taken. So, I set out to expand on the theme slightly and compare the performance and robustness of each of the solutions. It seemed like a fun way to entertain myself this morning.
In addition to the 3 examples given in the question, I added test cases for each of the four operators, as well as for some new edge cases. These edge cases included handling of negative numbers and arbitrary spaces between operands, as well as how each of the algorithms handled expected failures.
The answers revolved around 3 methods: split, scan, and match. I also wrote new solutions using each of these 3 methods, specifically respecting the additional edge cases that I added to here. I ran all of the algorithms against this full set of test cases, and ended up with a table of pass/fail results.
Next, I created a benchmark that created 1,000,000 test strings that each of the solutions would be able to parse properly, and ran each solution against that sample set.
On first benchmarking, Cary Swoveland's solution performed far better than the others, but didn't pass the added test cases. I made very minor changes to his solution to produce a solution that supported both negative numbers and arbitrary spaces, and included that test as Swoveland+.
The final results printed from to the console are here (note: horizontal scroll to see all results):
| Test Case | match | match | scan | scan |partition| split | split | split | split |
| | Gaskill | sweaver | Gaskill | techbio |Swoveland| Gaskill |Swoveland|Swoveland+| Lilue |
|------------------------------------------------------------------------------------------------------|
| "1+12" | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass |
| "6/0.25" | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass |
| "3/0.125" | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass |
| "30-6" | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass |
| "3*8" | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass |
| "20--4" | Pass | -- | Pass | -- | Pass | Pass | -- | Pass | Pass |
| "33+-9" | Pass | -- | Pass | -- | Pass | Pass | -- | Pass | Pass |
| "-12*-2" | Pass | -- | Pass | -- | Pass | Pass | -- | Pass | Pass |
| "-72/-3" | Pass | -- | Pass | -- | Pass | Pass | -- | Pass | Pass |
| "34 - 10" | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass |
| " 15+ 9" | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass |
| "4*6 " | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass | Pass |
| "b+0.5" | Pass | Pass | Pass | -- | -- | -- | -- | -- | -- |
| "8---0.5" | Pass | Pass | Pass | -- | -- | -- | -- | -- | -- |
| "8+6+10" | Pass | -- | Pass | -- | -- | -- | -- | -- | -- |
| "15*x" | Pass | Pass | Pass | -- | -- | -- | -- | -- | -- |
| "1.A^ff" | Pass | Pass | Pass | -- | -- | -- | -- | -- | -- |
ruby 2.2.5p319 (2016-04-26 revision 54774) [x86_64-darwin14]
============================================================
user system total real
match (Gaskill): 4.770000 0.090000 4.860000 ( 5.214996)
match (sweaver2112): 4.640000 0.040000 4.680000 ( 4.911849)
scan (Gaskill): 7.360000 0.080000 7.440000 ( 7.719646)
scan (techbio): 12.930000 0.140000 13.070000 ( 13.791613)
partition (Swoveland): 5.390000 0.050000 5.440000 ( 5.648762)
split (Gaskill): 5.150000 0.100000 5.250000 ( 5.455094)
split (Swoveland): 3.860000 0.060000 3.920000 ( 4.040774)
split (Swoveland+): 4.240000 0.040000 4.280000 ( 4.537570)
split (Lilue): 7.540000 0.090000 7.630000 ( 8.022252)
In order to keep this post from being far too long, I've included the complete code for this test at https://gist.github.com/mgaskill/96f04e7e1f72a86446f4939ac690759a
The robustness test cases can be found in the first table above. The Swoveland+ solution is:
f,op,l = formula.split(/\b\s*([+\/*-])\s*/)
return [f.to_f, op, l.to_f]
This includes a \b metacharacter prior to splitting on an operator ensures that the previous character is a word character, giving support for negative numbers in the second operand. The \s* metacharacter expressions support arbitrary spaces between operands and operator. These changes incur less than 10% performance overhead for the additional robustness.
The solutions that I provided are here:
def match_gaskill(formula)
return [] unless (match = formula.match(/^\s*(-?\d+(?:\.\d+)?)\s*([+\/*-])\s*(-?\d+(?:\.\d+)?)\s*$/))
return [match[1].to_f, match[2], match[3].to_f]
end
def scan_gaskill(formula)
return [] unless (match = formula.scan(/^\s*(-?\d+(?:\.\d+)?)\s*([+*\/-])\s*(-?\d+(?:\.\d+)?)\s*$/))[0]
return [match[0][0].to_f, match[0][1], match[0][2].to_f]
end
def split_gaskill(formula)
match = formula.split(/(-?\d+(?:\.\d+)?)\s*([+\/*-])\s*(-?\d+(?:\.\d+)?)/)
return [match[1].to_f, match[2], match[3].to_f]
end
The match and scan solutions are very similar, but perform significantly differently, which is very interesting, because they use the exact same regex to do the work. The split solution is slightly simpler, and only splits on the entire expression, capturing each operand and the operator, separately.
Note that none of the split solutions was able to properly identify failures. Adding this support requires additional parsing of the operands, which significantly increases the overhead of the solution, typically running about 3 times slower.
For both performance and robustness, match is the clear winner. If robustness isn't a concern, but performance is, use split. On the other hand, scan provided complete robustness, but was more than 50% slower than the equivalent match solution.
Also note that using an efficient way to extract the results from the solution into the result array is as important to performance as is the algorithm chosen. The technique of capturing the results array into multiple variables (used in Woveland) outperformed the map solutions dramatically. Early testing showed that the map extraction solution more than doubled the runtimes of even the highest-performing solutions, hence the exceptionally high runtime numbers for Lilue.
I think this could be useful:
"1.2+3.453".split('+').flat_map{|elem| [elem, "+"]}[0...-1]
# => ["1.2", "+", "3.453"]
"1.2+3.453".split('+').flat_map{|elem| [elem.to_f, "+"]}[0...-1]
# => [1.2, "+", 3.453]
Obviously this work only for +. But you can change the split character.
EDIT:
This version work for every operator
"1.2+3.453".split(%r{(\+|\-|\/|\*)}).map do |x|
unless x =~ /(\+|\-|\/|\*)/ then x.to_f else x end
end
# => [1.2, "+", 3.453]
R = /
(?<=\d) # match a digit in a positive lookbehind
[^\d\.] # match any character other than a digit or period
/x # free-spacing regex definition mode
def split_it(str)
f,op,l = str.delete(' ').partition(R)
[convert(f), op, convert(l)]
end
def convert(str)
(str =~ /\./) ? str.to_f : str.to_i
end
split_it "1+12"
#=> [1, "+", 12]
split_it "3/ 5.2"
#=> [3, "/", 5.2]
split_it "-4.1 * 6"
#=> [-4.1, "*", 6]
split_it "-8/-2"
#=> [-8, "/", -2]
The regex can of course be written in the conventional way:
R = /(?<=\d)[^\d\.]/
I'm currently trying to test a service that should be properly replacing certain special characters within a certain Unicode range, including emojis, transportation icons, emoticons and dingbats. I have been using Cucumber and Ruby to do the testing, and my latest scenario outline won't work. I've tried looking up other ways of getting the character from the examples table, however I can't seem to get it working, and the cucumber printout just complains that the Given step isn't defined.
Here is my feature scenario:
Scenario Outline: I update a coupon with a name including emojis/emoticons/dingbats/symbols
Given I have a name variable with a <character> included
When I patch my coupon with this variable
Then the patch should succeed
And the name should include replacement characters
Examples:
| character |
| π³ |
| π
|
| π₯ |
| π |
| β |
| β |
| β |
| β |
| β¨ |
| π |
| π¦Ώ° |
And Here is my step definition for the Given (which is the step that is complaining that it isn't defined)
Given(/^I have a name variable with a (\w+) included$/) do |char|
#name = 'min length ' + char
#json = { 'name' => #name }.to_json
end
I've tried using some regex's to capture the character, and a (\w+) and (\d+), although I can't find information on how to capture the special character. It's possible for me to write 11 different step definitions, but that would be such poor practice it would drive me nuts.
Unless you have spaces in your specials, itβs safe to use non-space \S:
Given(/^I have a name variable with a (\S+) included$/) do |char|
...
\w would not give you the desired result, since \w is resolved to [a-zA-Z0-9_].
I am using a scenario table (multiline step arguments) to check some data from a screen using cucumber, using the in built .diff! method on the Cucumber AST table.
I would like to check the content matches against regular expressions.
Scenario: One
Then the table appears as:
| One | Two | Three |
| /\d+/ | /\d+/ | /\d+/ |
The actual table could look something like
| One | Two | Three |
| 123 | 456 | 789 |
which this scenario is translated to "as long as there are some digits, I don't care"
An example step implementation that fails:
Then /^the table appears as:$/ do |expected_table|
actual_table = [['One','Two', 'Three'],['123', '456', '789']]
expected_table.diff! actual_table
end
Error:
Then the table appears as: # features/step_definitions/my_steps.rb:230
| One | Two | Three |
| /\\d+/ | /\\d+/ | /\\d+/ |
| 123 | 456 | 789 |
Tables were not identical (Cucumber::Ast::Table::Different)
I have tried using step transforms to transform the cells into regular expressions, but they still aren't identical.
Transform code:
expected_table.raw[0].each do |column|
expected_table.map_column! column do |cell|
if cell.respond_to? :start_with?
if cell.start_with? "/"
cell.to_regexp
else
cell
end
else
cell
end
end
end
which provides the eror:
Then the table appears as: # features/step_definitions/my_steps.rb:228
| One | Two | Three |
| (?-mix:\\d+) | (?-mix:\\d+) | (?-mix:\\d+) |
| 123 | 456 | 789 |
Tables were not identical (Cucumber::Ast::Table::Different)
Any ideas? I am stuck.
Using regular expressions in a scenario is almost certainly the wrong approach. Cucumber features are intended to be read and understood by business-focussed stakeholders.
How about writing the step at a higher level, such as as:
Then the first three columns of the table should contain a digit
There is no way to do it without writing your own implementation of diff! method from Ast::Table. Take a look into cucumber/lib/ast/table.rb. Internally it uses diff-lcs library to do an actual comparison which doesn't support regex match.
It seems that you want to write this in a way that provides the cool diff output. Otherwise, I'd look at writing this such that you simply check the rows. It won't be as pretty, and it won't get you the diff of the entire table, but it's something.
Then /^the table appears as:$/ do |expected_table|
actual_table = [['One','Two', 'Three'],['123', '456', '789']]
expected_table.raw.each_with_index { |row, y|
row.each_with_index { |cell, x|
actual_table[x][y].should == cell
}
}
end