Detect heat waves

Question

Background

The Royal Netherlands Meteorological Institute defines a heat wave* as a series of at least 5 consecutive days of ≥25°C weather (“summery weather”), such that at least 3 of those days are ≥30°C (“tropical weather”).

The tropical weather doesn't have to be measured consecutively: for example: 30, 25, 30, 26, 27, 28, 32, 30 is a 8-day long heat wave with 4 days of tropical weather.

^{*(Well, by Dutch standards.)}

Challenge

Given a non-empty list of positive integers representing Celsius temperature measurements from successive days, decide whether that list contains a heat wave (as per the above definition).

The shortest answer in bytes wins.

Test cases

Falsey:

[30]
[25, 30, 25, 60, 25]
[29, 29, 29, 47, 30]
[31, 29, 29, 28, 24, 23, 29, 29, 26, 27, 33, 20, 26, 26, 20, 30]
[23, 31, 29, 26, 30, 24, 29, 29, 25, 27, 24, 28, 22, 20, 34, 22, 32, 24, 33]
[23, 24, 25, 20, 24, 34, 28, 32, 22, 20, 24]
[24, 28, 21, 34, 34, 25, 24, 33, 23, 20, 32, 26, 29, 29, 25, 20, 30, 24, 23, 21, 27]
[26, 34, 21, 32, 32, 30, 32, 21, 34, 21, 34, 31, 23, 27, 26, 32]
[29, 24, 22, 27, 22, 25, 29, 26, 24, 24, 20, 25, 20, 20, 24, 20]
[23, 33, 22, 32, 30]
[28, 21, 22, 33, 22, 26, 30, 28, 26, 23, 31, 22, 31, 25, 27, 27, 25, 28]
[27, 23, 42, 23, 22, 28]
[25, 20, 30, 29, 32, 25, 22, 21, 31, 22, 23, 25, 22, 31, 23, 25, 33, 23]

Truthy:

[30, 29, 30, 29, 41]
[1, 1, 25, 30, 25, 30, 25, 25, 25, 25, 25, 25, 25, 25, 40, 1, 1]
[31, 34, 34, 20, 34, 28, 28, 23, 27, 31, 33, 34, 29, 24, 33, 32, 21, 34, 30, 21, 29, 22, 31, 23, 26, 32, 29, 32, 24, 27]
[26, 29, 22, 22, 31, 31, 27, 28, 32, 23, 33, 25, 31, 33, 34, 30, 23, 26, 21, 28, 32, 22, 30, 34, 26, 33, 20, 27, 33]
[20, 31, 20, 29, 29, 33, 34, 33, 20]
[25, 26, 34, 34, 41, 28, 32, 30, 34, 23, 26, 33, 30, 22, 30, 33, 24, 20, 27, 23, 30, 23, 34, 20, 23, 20, 33, 20, 28]
[34, 23, 31, 34, 34, 30, 29, 31, 29, 21, 25, 31, 30, 29, 29, 28, 21, 29, 33, 25, 24, 30]
[22, 31, 23, 23, 26, 21, 22, 20, 20, 28, 24, 28, 25, 31, 31, 26, 33, 31, 27, 29, 30, 30]
[26, 29, 25, 30, 32, 28, 26, 26, 33, 20, 21, 32, 28, 28, 20, 34, 34]
[34, 33, 29, 26, 34, 32, 27, 26, 22]
[30, 31, 23, 21, 30, 27, 32, 30, 34, 29, 21, 31, 31, 31, 32, 27, 30, 26, 21, 34, 29, 33, 24, 24, 32, 27, 32]
[25, 33, 33, 25, 24, 27, 34, 31, 29, 31, 27, 23]

Answer 1

C (gcc), 88 75 73 bytes

• Saved two bytes thanks to ceilingcat.

h,e,a;t(int*_){for(h=e=a=0;*_;h+=e>4&a>2)e=*_<25?a=0:++e,a+=*_++>29;e=h;}

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Answer 2

Jelly, 15 bytes

:5_5Ṡ‘ẆP«LƊ€>4Ṁ

A monadic link accepting a list of numbers which returns 1 if a heatwave was detected else 0.

Try it online! or see the test-suite.

How?

The criteria is the existence of a run of more than four values greater than or equal to 25, of which more than two must be greater than or equal to 30.

If we divide through by five the criteria becomes the existence of a run of more than four values greater than or equal to five, of which more than two must be greater than or equal to six.

If we subtract five from these values the criteria becomes the existence of a run of more than four values greater than or equal to zero, of which more than two must be greater than or equal to one.

If we take the sign of these values (getting -1, 0, or 1) the criteria becomes the existence of a run of more than four values not equal to -1, of which more than two must be equal to one.

If we add one to these values (getting 0, 1, or 2) the criteria becomes the existence of a run of more than four values not equal to zero, of which more than two must be equal to two.

The product of a list containing any zeros is zero and the product of a list containing more than two twos (and the rest being ones) is more than four. This means that the criteria on this adjusted list becomes that the minimum of the product and the length is greater than 4.

:5_5Ṡ‘ẆP«LƊ€>4Ṁ - Link: list of numbers
:5              - integer divide by five (vectorises)
  _5            - subtract five (vectorises)
    Ṡ           - sign {negatives:-1, zero:0, positives:1} (vectorises)
     ‘          - increment (vectorises)
      Ẇ         - all sublists
          Ɗ€    - last three links as a monad for €ach:
       P        -   product
         L      -   length
        «       -   minimum
            >4  - greater than four? (vectorises) -- 1 if so, else 0
              Ṁ - maximum -- 1 if any are 1, else 0

Answer 3

C (clang), 64 bytes

h;o(*t){for(h=1;*t;++t)h=h&&*t<25?1:h*(*t<30?2:6)%864;return!h;}

The function o() returns 1 for a heatwave or 0 else.

Thanks to the magic number 864 and to Udo Borkowski and Mathis for their ideas.

How does if work? Each sequence of numbers is iterated with a reduce operation starting at the reduce value 1. If a number >= 25 is seen the reduce is multiplied by 2. If a number >= 30 is seen the reduce is multiplied by 2 and by 3 = 6. If a number < 25 is seen the reduce starts again at 1. If the reduce is divisible by 864=2*2*2*2*2*3*3*3 then a heatwave is found, and the result of the modulo operation is 0 which results in a reduce value of 0 and in a return value of true.

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Answer 4

Haskell, 73 72 71 69 67 66 bytes

any(\a->sum[1|x<-a,x>29,take 4a<a]>2).scanl(\a t->[0|t>24]>>t:a)[]

Thanks to @flawr and @Laikoni for two bytes each and @xnor for a byte!

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Equal length:

any(\a->take 4a<a&&sum a>2).scanl(\a t->[0|t>24]>>sum[1|t>29]:a)[]

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Answer 5

Python 3, 79 bytes

lambda T:any(len(s)>4<sum(s+s)for s in bytes(t>29or(t<25)*9for t in T).split())

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Answer 6

APL (Dyalog Classic), 21 20 bytes

1∊8≤4↓⍉×\25 30⍸↑,⍨\⎕

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uses ⎕io←1

25 30⍸x is 0 if x<25, 1 if 25≤x<30, or 2 otherwise

we compute cumulative products of these starting from (or equivalently: ending at) all possible locations, discard the first 4 products, and detect the presence of products ≥8 (which is 2³)

Answer 7

Japt, 19 18 bytes

ô<25 d_ʨ5©3§Zè¨30
ô                  // Partition the input at every item
 <25               // where the value is less than 25.
     d_            // Then, return whether any resulting subarray
       ʨ5         // is at least five items long
          ©        // and
           3§      // has at least three items
             Zè¨30 // with a value of at least 30.

I hope I got all the discussions in the comments correctly.
Shaved off one byte thanks to Shaggy.

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Answer 8

PowerShell, 121 bytes

param($a)$b="";($a|%{if($_-ge25){$b+="$_ "}else{$b;$b=""}})+$b|?{(-split$_).count-ge5-and(-split$_|?{$_-ge30}).count-ge3}

Try it online! or Verify all test cases

PowerShell doesn't have the equivalent of a .some or .every or the like, so this is rolled by hand.

We take input $a as an array of integers. Set helper variable $b to the empty string. Then, loop through every integer in $a. Inside the loop, if the integer is -greaterthanorequal to 25, add it to our potential string $b, otherwise put $b on the pipeline and set it to the empty string.

Once outside the loop, array-concatenate the pipeline results with $b, and put those through a Where-Object clause |?{...}. This pulls out those strings that have an element length of -ge5 (based on splitting on whitespace) and a count of temps greater than 30 being -ge3. Those strings are left on the pipeline, so a truthy value is non-empty (see the "verify all test cases" link for truthy/falsey distinction).

Answer 9

JavaScript (ES6), 63 51 bytes

Returns a boolean.

a=>a.some(n=>(n>24?y+=++x&&n>29:x=y=0)>2&x>4,x=y=0)

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Answer 10

Jelly, 17 16 bytes

:5_5Ṡṣ-ḤS«LƊ€Ṁ>4

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How it works

:5_5Ṡṣ-ḤS«LƊ€Ṁ>4  Main link. Argument: T (array of temperatures)

:5                Divide each item of T by 5 (integer division).
  _5              Subtract 5 from each quotient.
    Ṡ             Take the signs.
                  This maps (-oo,25) to -1, [25,30) to 0, and [30,+oo) to 1.
     ṣ-           Split at occurrences of -1.
       Ḥ          Double, replacing 1's with 2's.
           Ɗ€     Map the three links to the left over each chunk.
        S             Take the sum.
          L           Take the length.
         «            Take the minimum of the results.
             Ṁ    Take the maximum.
              >4  Test if it's larger than 4.
                  Note that the sum is larger than 4 if and only if there are more
                 than two 2's, which correspond to temperatures in [30,+oo).

Answer 11

Python 2, 86 bytes

lambda l:any(2<s.count('2')*(len(s)>14)for s in`[(t>24)+(t>29)for t in l]`.split('0'))

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Answer 12

05AB1E, 20 bytes

Œʒ24›DPsO4›*}29›O2›Z

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Explanation

Π                    # push sublists of input
 ʒ          }         # filter, keep the lists where:
           *          # the product of:
     DP               # the product and
       sO4›           # the sum compared using greater-than to 4
  24›                 # for the elements greater than 24
                      # is true
                      # the result is:
                   Z  # the maximum from the remaining lists where
                O     # the sum of 
             29›      # the elements greater than 29
                 2›   # is greater than 2

Answer 13

Batch, 119 bytes

@set h=0
@for %%t in (0 %*)do @if %%t lss 25 (set/as=5,t=3)else set/a"t+=!!t*(29-%%t)>>9,s-=!!s,h+=!(s+t+h)
@echo %h%

Takes input as command-line arguments and outputs 1 for a heatwave otherwise 0.

Answer 14

Python, 67 bytes

f=lambda l:l>l[:4]and(min(l)>24<sorted(l)[~2]-5)|f(l[1:])|f(l[:-1])

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Times out on the longer test cases due to exponential growth. Finds contiguous sublists by repeatedly chopping the first or last element. That 3 days are ≥30°C is checked by looking at the third-largest value sorted(l)[~2]. The base cases could perhaps be shorter by taking advantage of truthy/falsey or terminating with error.

Answer 15

JavaScript (Node.js), 48 bytes

a=>a.some(x=>x>24?++A>4&(B+=x>29)>2:A=B=0,A=B=0)

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quite bad one

Answer 16

Haskell, 64 bytes

f l=or$drop 4l>>[sum[1|x<-l,x>29,all(>24)l]>2,f$tail l,f$init l]

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Answer 17

APL (Dyalog Unicode), 29 bytes

∨/(5≤≢¨a)∧3≤+/30≤↑a←e⊆⍨25≤e←⎕

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∨/ are there any elements such that

(5≤≢¨a) 5 < the tally ≢ of days in each series (a has all the possible series of days)

∧ and

3≤+/30≤ 3 ≤ the total +/ number of elements that are ≥ 30 in

↑a← the matrix formed by

e⊆⍨25≤e←⎕ the series of consecutive elements that are ≥ 25

Answer 18

Python 2, 66 63 bytes

lambda a:reduce(lambda b,c:(b*(6,2)[c<30]%864,1)[b*25>b*c],a,1)

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-3 bytes thanks to Lynn

How does if work? Each sequence of numbers is iterated with a reduce operation starting at the reduce value 1. If a number >= 25 is seen the reduce is multiplied by 2. If a number >= 30 is seen the reduce is multiplied by 2 and by 3 = 6. If a number < 25 is seen the reduce starts again at 1. If the reduce is divisible by 864=2*2*2*2*2*3*3*3 then a heatwave is found, and the result of the modulo operation is 0 which results in a reduce value of 0. Only when a heat wave was found the reduce can become 0. Once the reduce value is 0 it will be 0 for all future reduces, i.e. also for the end result.

A more readable, but longer version looks like this:

lambda a:reduce((lambda b,c: 1 if b>0 and c<25 else b*(2 if c<30 else 6)%864), a, 1)

Removing extra spaces/parenthesis and replacing x if cond else y by (y,x)[cond] gives

lambda a:reduce(lambda b,c:(b*(6,2)[c<30]%864,1)[b>0and c<25],a,1)

Lynn suggested to shorten the condition b>0and c<25:

b>0and c<25 --> b*25>0 and b*c<b*25 --> b*25>0 and b*25>b*c --> b*25>b*c

resulting in

lambda a:reduce(lambda b,c:(b*(6,2)[c<30]%864,1)[b*25>b*c],a,1)

Answer 19

Kotlin, 57 bytes

{var r=1;it.any{r*=2;if(it>29)r*=3;if(it<25)r=1;r%864<1}}

(-1 Byte by replacing the explicit Parameter v-> with the implicit parameter it)

{var r=1;it.any{v->r*=2;if(v>29)r*=3;if(v<25)r=1;r%864<1}}

(-16 bytes using the any{} operation as seen in the Ruby Solution by G B)

{it.stream().reduce(1){r,v->if(r*25>r*v)1 else(r*if(v<30)2 else 6)%864}<1}

(-1 byte thanks Lynn: replaced r>0&&v<25 with r*25>r*v)

{it.stream().reduce(1){r,v->if(r>0&&v<25)1 else(r*if(v<30)2 else 6)%864}<1}

This lambda expression takes a List and returns true for a heatwave or false else.

Thanks to the magic number 864 and to Udo Borkowski and Mathis for their ideas.

How does if work? Each sequence of numbers is iterated with an any{} operation starting at the reduce value 1. The reduce is multiplied by 2 and multiplied with 3 (2*3=6) if the number is greater or equal 30. If a number < 25 is seen the reduce starts again at 1. If the reduce is divisible by 864=2*2*2*2*2*3*3*3 then a heatwave is found, and the result of the modulo operation is 0 which results in a true return value in the inner lambda called from the any{} operation which then stops iterating and returns value of true.

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Answer 20

Wonder, 34 bytes

(/>@(& <2!> '<29#0)*> '<24#0).cns5

Usage example:

((/>@(& <2!> '<29#0)*> '<24#0).cns5) [25 33 33 25 24 27 34 31 29 31 27 23]

Explanation

Verbose version:

(some x\\(and <2 (fltr <29) x) (every <24) x) . (cns 5)

Take overlapping sequences of 5 consecutive items, then check if any of the sequences have all items > 25 and more than 2 items > 30.

Answer 21

Jelly, 21 bytes

L5r⁸ṡẎ<25Ẹ$Ðḟ29<S€2<Ẹ

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Answer 22

Stax, 23 bytes

Æ7)║▄░Ä╟═╙hⁿ╧\ßY8÷K▌µ½x

Run and debug it at staxlang.xyz! This takes a long time to run, so I disabled auto-run.

Unpacked (28 bytes) and explanation

:efc%4>nc{24>f=a{29>f%2>|&|&
:e                              Set of all contiguous subarrays
  f                             Filter, using the rest of the program as a predicate:
   c                              Copy subarray on the stack
    %4>                           Five or more elements?
                        |&        AND
       nc                         Copy subarray twice to top
         {   f                    Filter:
          24>                       Greater than 24?
              =                   Equals the original subarray?
                          |&      AND
               a                  Move subarray to top
                {   f             Filter:
                 29>                Greater than 30?
                     %2>          Length greater than two?
                                  Implicit print if all three conditions are met

This'll print all subarrays that can be counted as heat waves, which will be falsy if and only if none exist.

Answer 23

Ruby, 89 bytes

->a{(0..a.size).map{|i|(b=a[i..-1].take_while{|t|t>24}).size>4&&b.count{|t|t>29}>2}.any?}

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Answer 24

Husk, 19 bytes

Vo≥3#≥30fo≥5Lġ(±≥25

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Using filter (f) is one byte shorter than using checking with a logical and (&), also it would be really nice to get rid of the ± - costing 2 bytes :(

Explanation

V(≥3#≥30)f(≥5L)ġ(±≥25)  -- example input: [12,25,26,27,28,29,18,24,32]
               ġ(    )  -- group by
                ( ≥25)  -- | greater or equal to 25: [0,1,2,3,4,5,6,0,0,8]
                (±   )  -- | sign: [0,1,1,1,1,1,1,0,0,1]
                        -- : [[12],[25,26,27,28,29,30],[18,24],[32]]
         f(   )         -- filter by
          (  L)         -- | length: [1,6,2,1]
          (≥5 )         -- | greater or equal to 5: [0,2,0,0]
                        -- : [[25,26,27,28,29,30]]
V(      )               -- does any element satisfy
 (  #   )               -- | count occurences where
 (   ≥30)               -- | | elements greater or equal to 30
 (      )               -- | : [1]
 (≥3    )               -- | greater or equal to 3: [0]
                        -- : 0

Answer 25

R, 111 93 71 67 66 bytes

!Reduce(function(i,j)"if"(j<25,!!i,(i*(2+4*!j<30))%%864),scan(),1)

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Shameless port of Roland Schmitz's answers. -4 bytes thanks to Roland and -1 thanks to Giuseppe.

TIO links to functional version.

Previous version extracted consecutive days>25 using rle and saved a whopping 18 bytes thanks to Giuseppe!

Answer 26

Swift 4, 50 bytes

{$0.reduce(1){$0>0&&$1<25 ?1:$0*($1<30 ?2:6)%864}}

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The closure expression returns 0 for a heatwave or >0 else.

Created in collaboration with Roland Schmitz and Mathis.

How does if work? Each sequence of numbers is iterated with a reduce operation starting at the reduce value 1. If a number >= 25 is seen the reduce is multiplied by 2. If a number >= 30 is seen the reduce is multiplied by 2 and by 3 = 6. If a number < 25 is seen the reduce starts again at 1. If the reduce is divisible by 864=2*2*2*2*2*3*3*3 then a heatwave is found, and the result of the modulo operation is 0 which results in a reduce value of 0. Only when a heat wave was found the reduce can become 0. Once the reduce value is 0 it will be 0 for all future reduces, i.e. also for the end result.

Answer 27

[JavaScript (ES6)], 49 bytes

a=>a.reduce((b,c)=>b>0&c<25?1:b*(c<30?2:6)%864,1)

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The arrow function returns 0 for a heatwave or >0 else.

(Port of the Swift solution to JavaScript)

How does if work? Each sequence of numbers is iterated with a reduce operation starting at the reduce value 1. If a number >= 25 is seen the reduce is multiplied by 2. If a number >= 30 is seen the reduce is multiplied by 2 and by 3 = 6. If a number < 25 is seen the reduce starts again at 1. If the reduce is divisible by 864=2*2*2*2*2*3*3*3 then a heatwave is found, and the result of the modulo operation is 0 which results in a reduce value of 0. Only when a heat wave was found the reduce can become 0. Once the reduce value is 0 it will be 0 for all future reduces, i.e. also for the end result.

Answer 28

Java (JDK 10), 60 bytes

h->h.stream().reduce(1,(r,v)->r*25>r*v?1:r*(v<30?2:6)%864)<1

(-1 byte thanks Lynn: replaced r>0&&v<25 with r*25>r*v)

h->h.stream().reduce(1,(r,v)->r>0&&v<25?1:r*(v<30?2:6)%864)<1

The lambda expression returns true for a heatwave or false else.

Thanks to the magic number 864 and to Udo Borkowski and Mathis for their ideas.

How does if work? Each sequence of numbers is iterated with a reduce operation starting at the reduce value 1. If a number >= 25 is seen the reduce is multiplied by 2. If a number >= 30 is seen the reduce is multiplied by 2 and by 3 = 6. If a number < 25 is seen the reduce starts again at 1. If the reduce is divisible by 864=2*2*2*2*2*3*3*3 then a heatwave is found, and the result of the modulo operation is 0 which results in a reduce value of 0 and in a return value of true.

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Answer 29

Ruby, 50 48 bytes

->a{b=1;a.any?{|r|1>b=[b*2,1,b*6][5<=>r/5]%864}}

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Thanks to Roland Schmitz and Udo Borkowski for the idea and the magic number 864. And again to Roland for saving 2 bytes.

Answer 30

Haskell, 60 bytes

f q|(a,~(_:t))<-span(>24)q=sum[1|x<-a,x>29,take 4a<a]>2||f t

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This combines some ideas from previous Haskell answers in a new form along with exit by error code to come out ahead.

How?

(a,~(_:t))<-span(>24)q puts the prefix of the list which is entirely >24 into a, and then tries to match the remainder of the list to the pattern h:t (dropping one value <= 24 and putting the remaining suffix into t). This pattern match might fail even on truthy inputs, if the heat wave is at the end of the list, which is where the ~ comes in: it forces Haskell to match the rest of the pattern even if _:t can't be matched (an "irrefutable pattern"), in which case t is undefined (i.e. set to _|_) and will error out if we ever try to use it.

Since Haskell has the usual short circuiting on ||, we will only try to evaluate f t on undefined t if there was no heat wave in the list, in which case we exit high. If there was a heat wave, we'll find it and short circuit out, returning True and exiting low.

The irrefutable pattern saves one byte compared to tail, which will similarly fail if called on an empty list:

Haskell, 61 bytes

f q|(a,b)<-span(>24)q=sum[1|x<-a,x>29,take 4a<a]>2||f(tail b)

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