Expand Comparison Chains

Question

Unlike most languages, Python evaluates a<b<c as it would be done in mathematics, actually comparing the three numbers, as opposed to comparing the boolean a<b to c. The correct way to write this in C (and many others) would be a<b && b<c.

In this challenge, your task is to expand such comparison chains of arbitrary lengths from the Python/intuitive representation, to how it would be written in other languages.

Specifications

Your program will have to handle the operators: ==, !=, <, >, <=, >=.
The input will have comparison chains using only integers.
Don't worry about the trueness of any of the comparisons along the way, this is purely a parsing/syntactical challenge.
The input won't have any whitespace to prevent answers that trivialize parsing by splitting on spaces.
However, your output may have a single space surrounding either just the &&'s, or both the comparison operators and the &&'s, or neither, but be consistent.

Test Cases

Input Output --------------------------------------------------------------- 3<4<5 3<4 && 4<5 3<4<5<6<7<8<9 3<4 && 4<5 && 5<6 && 6<7 && 7<8 && 8<9 3<5==6<19 3<5 && 5==6 && 6<19 10>=5<7!=20 10>=5 && 5<7 && 7!=20 15==15==15==15==15 15==15 && 15==15 && 15==15 && 15==15

This is code-golf, so shortest code in bytes wins!

\$\begingroup\$ related \$\endgroup\$

Maltysen
– Maltysen

2017-10-09 03:50:57 +00:00
Commented Oct 9, 2017 at 3:50 — Maltysen
– Maltysen, Commented Oct 9, 2017 at 3:50
\$\begingroup\$ Can I have two spaces either side of the &&? \$\endgroup\$

H.PWiz
– H.PWiz

2017-10-09 04:34:40 +00:00
Commented Oct 9, 2017 at 4:34 — H.PWiz
– H.PWiz, Commented Oct 9, 2017 at 4:34
\$\begingroup\$ @H.PWiz nope, srry. \$\endgroup\$

Maltysen
– Maltysen

2017-10-09 04:35:28 +00:00
Commented Oct 9, 2017 at 4:35 — Maltysen
– Maltysen, Commented Oct 9, 2017 at 4:35

ovs · Accepted Answer · 2017-10-09 07:05:48Z

6

Retina, 33 22 17 bytes

-5 bytes thanks to @MartinEnder

-2`\D(\d+) $&&&$1

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edited Oct 9, 2017 at 7:05

answered Oct 9, 2017 at 6:35

ovs

61.2k3 gold badges49 silver badges164 bronze badges

\$\begingroup\$ @MartinEnder A pity 1>-2 doesn't work to limit from both ends at once... \$\endgroup\$

Neil
– Neil

2017-10-09 07:57:07 +00:00
Commented Oct 9, 2017 at 7:57
\$\begingroup\$ @Neil yeah, Retina 1.0 will get a new limit syntax where that's possible. \$\endgroup\$

Martin Ender
– Martin Ender

2017-10-09 07:57:54 +00:00
Commented Oct 9, 2017 at 7:57
\$\begingroup\$ Could you post an explanation? \$\endgroup\$

DJMcMayhem
– DJMcMayhem

2017-10-09 19:17:38 +00:00
Commented Oct 9, 2017 at 19:17

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

Husk, 16 14 bytes

Prints a space around each operator.

wJ"&&"m←C2X3ġ±

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Explanation:

 Implicit input, e.g "3<4<5<6" ġ± Group digits ["3","<","4","<","5","<","6"] X3 Sublists of length 3 [["3","<","4"],["<","4","<"],["4","<","5"],["<","5","<"],["5","<","6"]] C2 Cut into lists of length 2 [[["3","<","4"],["<","4","<"]],[["4","<","5"],["<","5","<"]],[["5","<","6"]]] m← Take the first element of each [["3","<","4"],["4","<","5"],["5","<","6"]] J"&&" Join with "&&" ["3","<","4","&&","4","<","5","&&","5","<","6"] w Print, separates by spaces

Nice one. You could use w instead of ; for a more direct approach to joining strings with spaces — Leo
– Leo, Commented Oct 9, 2017 at 4:53

ATaco · Accepted Answer · 2017-10-09 22:12:14Z

3

Retina, 42 47 22 bytes

Massively golfed thanks to Kevin Cruijssen

(\D(\d+))(?=\D) $1&&$2

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edited Oct 9, 2017 at 22:12

answered Oct 9, 2017 at 4:12

ATaco

11.7k30 silver badges82 bronze badges

\$\begingroup\$ (==|!=|<=?|>=?) can be \D+ \$\endgroup\$

ovs
– ovs

2017-10-09 06:37:57 +00:00
Commented Oct 9, 2017 at 6:37
\$\begingroup\$ Similarly (?<!^|\d) can be (?<=\D). Also the (?=\d+) is unnecessary, the operator will always be followed by an operand, at which point you can drop the + after the \D. There's also $& instead of $1$2, and then a further byte can be saved by capturing behind and looking ahead instead of capturing ahead and looking behind. \$\endgroup\$

Neil
– Neil

2017-10-09 08:02:57 +00:00
Commented Oct 9, 2017 at 8:02
\$\begingroup\$ (\D(\d+))(?=\D) on line 1, and $1&&$2 on line two is enough (22 bytes). Try it here. \$\endgroup\$

Kevin Cruijssen
– Kevin Cruijssen

2017-10-09 14:28:21 +00:00
Commented Oct 9, 2017 at 14:28

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DJMcMayhem · Accepted Answer · 2017-10-09 04:31:18Z

V, 37 bytes

òÍ¨ä«„0-9& ]«©¨ä«©¨„0-9& ]«©/±² ¦¦ ²³

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Hexdump:

00000000: f2cd a8e4 ab84 302d 3926 205d aba9 a8e4 ......0-9& ].... 00000010: aba9 a884 302d 3926 205d aba9 2fb1 b220 ....0-9& ]../.. 00000020: a6a6 20b2 b3 .. ..

NikoNyrh · Accepted Answer · 2017-10-09 11:32:40Z

Clojure, 88 bytes

Update: subs instead of clojure.string/join.

#(subs(apply str(for[p(partition 3 2(re-seq #"(?:\d+|[^\d]+)" %))](apply str" && "p)))4)

Community · Accepted Answer · 2020-06-17 09:04:33Z

J, 59 46 bytes

4(}.;)_2{.\3' && '&;\]</.~0+/\@,2~:/\e.&'!<=>'

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How it used to work

 (0 , }. ~:&(e.&'!<=>') }:)

We’re looking for operator boundaries. “Beheaded” and “curtailed” strings are turned into zeroes and ones where 0s are digits, then xored together. Prepend zero to match the length.

 +/\ </. ] Split on boundaries. ┌──┬──┬─┬─┬─┬──┬──┐ │10│>=│5│<│7│!=│20│ └──┴──┴─┴─┴─┴──┴──┘ 3' && '&;\ Add && to infixes of 3. ┌────┬──┬──┬──┐ │ && │10│>=│5 │ ├────┼──┼──┼──┤ │ && │>=│5 │< │ ├────┼──┼──┼──┤ │ && │5 │< │7 │ ├────┼──┼──┼──┤ │ && │< │7 │!=│ ├────┼──┼──┼──┤ │ && │7 │!=│20│ └────┴──┴──┴──┘ _2{.\ Take even numbered rows. ;}., Concatenate after dropping the first box.

Neil · Accepted Answer · 2017-10-09 09:21:23Z

Charcoal, 29 bytes

≔ ηＦθ«¿›ι9«Ｆ›η!⁺&&η≔ωη»≔⁺ηιηι ≔ ηＦθ«Ｆ∧›ι9›η!⁺&&η≔⎇›ι9ω⁺ηιηι

Two slightly different formulations of the same basic algorithm. The input string is iterated by character. As digits are found they are collected in a variable. When a boundary between a number and operator is found, an extra "&&" plus the variable is printed and the variable is cleared. The variable is initially initialised to a space so that the first boundary doesn't trigger the extra "&&".

Erik the Outgolfer · Accepted Answer · 2017-10-09 13:26:51Z

Jelly, 16 bytes

e€ØDŒg⁸ṁṡ3m2j⁾&&

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Explanation:

e€ØDŒg⁸ṁṡ3m2j⁾&& Uses Jelly stringification, thus full program only e€ØD For each each char, 1 if it's in '0123456789', otherwise 0 Œg Split into longest runs of equal elements ⁸ṁ Reshape original input like the list we just made Reshaping will separate numbers from operators ṡ3 Get contiguous sublists of length 3 m2 Keep every second item, starting from the first This will keep every (number operator number) and remove every (operator number operator) substring j⁾&& Join with '&&'

Community · Accepted Answer · 2020-06-17 09:04:33Z

Java 8, 46 bytes

s->s.replaceAll("(\\D(\\d+))(?=\\D)","$1&&$2")

Explanation:

Try it here.

s-> // Method with String as both parameter and return-type s.replaceAll("(\\D(\\d+))(?=\\D)", "$1&&$2") // Replace the match of the regex // with the second String // End of method (implicit / single-line return-statement)

Regex explanation:

(\D(\d+))(?=\D) # 1) For all matches of this regex (\d+) # Capture group 2: a number (one or more digits) (\D \d+ ) # Capture group 1: anything but a number + the number # (`\D` will match the `=`, `!`, `<`, or `>`) (?=\D) # Look-ahead so everything after the match remains as is $1&&$2 # 2) Replace it with $1 # Result of capture group 1 (trailing part of the equation + the number) && # Literal "&&" $2 # Result of capture group 2 (the number)

Step by step example of the replacements:

Initial: 10>=5<7!=20 Match of first replacement: =5 Replace with: =5&&5 After first replacement: 10>=5&&5<7!=20 Match of second replacement: <7 Replace with: <7&&7 After second replacement (result): 10>=5&&5<7&&7!=20

Xcali · Accepted Answer · 2017-10-09 04:31:04Z

0

Perl 5, 47 + 1 (-p) = 48 bytes

$\.=$1." && "x(!!$')while s/(\d+\D+(\d+))/$2/}{

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answered Oct 9, 2017 at 4:31

Xcali

17k2 gold badges17 silver badges42 bronze badges

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TFeld · Accepted Answer · 2017-10-09 08:07:11Z

0

Python 2, 60 59 58 57 bytes

lambda s:re.sub(r'(\D(\d+))(?=\D)',r'\1&&\2',s) import re

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edited Oct 9, 2017 at 8:07

answered Oct 9, 2017 at 6:58

TFeld

19.9k3 gold badges21 silver badges63 bronze badges

\$\begingroup\$ re.sub(r'\D(\d+)(?=\D)',r'\g<0>&&\1',s) saves 1 byte. \$\endgroup\$

Neil
– Neil

2017-10-09 07:59:46 +00:00
Commented Oct 9, 2017 at 7:59

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G B · Accepted Answer · 2017-10-09 10:02:36Z

0

Ruby, 37 bytes

->s{s.gsub(/\D(\d+)(?=\D)/,'\&&&\1')}

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answered Oct 9, 2017 at 10:02

G B

23.4k1 gold badge24 silver badges55 bronze badges

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Ephellon Grey · Accepted Answer · 2017-10-09 23:22:37Z

0

JavaScript (SpiderMonkey), 43 bytes

f=s=>s.replace(/(\W+(\d+))(?!$)/g,"$1&&$2")

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answered Oct 9, 2017 at 23:22

Ephellon Grey

2131 silver badge4 bronze badges

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caird coinheringaahing · Accepted Answer · 2021-01-20 18:33:13Z

Jelly, 15 bytes

e€ØDnƝkṡ3m2j⁾&&

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Outputs with no spaces around the &&, but the Footer in the TIO link adds the spaces to make the output easier to read

How it works

e€ØDnƝkṡ3m2j⁾&& - Main link. Takes a string S on the left e€ØD - Replace digits with 1, otherwise 0 Ɲ - To overlapping pairs [a,b]: n - a ≠ b k - Split S on truthy elements, grouping digits together ṡ3 - Split into overlapping lists of length 3 m2 - Take every other element j⁾&& - Join by "&&"

Stack Exchange Network

Expand Comparison Chains

Specifications

Test Cases

14 Answers 14

Retina, 33 22 17 bytes

Husk, 16 14 bytes

Explanation:

Retina, 42 47 22 bytes

V, 37 bytes

Clojure, 88 bytes

J, 59 46 bytes

How it used to work

Charcoal, 29 bytes

Jelly, 16 bytes

Java 8, 46 bytes

Perl 5, 47 + 1 (-p) = 48 bytes

Python 2, 60 59 58 57 bytes

Ruby, 37 bytes

JavaScript (SpiderMonkey), 43 bytes

Jelly, 15 bytes

How it works

Linked

Hot Network Questions

Expand Comparison Chains

Specifications

Test Cases

14 Answers 14

Retina, 33 22 17 bytes

Husk, 16 14 bytes

Explanation:

Retina, 42 47 22 bytes

V, 37 bytes

Clojure, 88 bytes

J, 59 46 bytes

How it used to work

Charcoal, 29 bytes

Jelly, 16 bytes

Java 8, 46 bytes

Perl 5, 47 + 1 (-p) = 48 bytes

Python 2, 60 59 58 57 bytes

Ruby, 37 bytes

JavaScript (SpiderMonkey), 43 bytes

Jelly, 15 bytes

How it works

Linked

Related

Hot Network Questions