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Bytes Lang Time Link
015Japt x180824T152157ZShaggy
032Brachylog180822T223644ZKroppeb
01505AB1E legacy180822T151247ZKevin Cr
044R180719T184445ZJayCe
025Japt180822T132434ZEtheryte
092R161009T120354Zrturnbul
030Actually160816T010957Zuser4594
062Javascript ES6160612T131034ZQwertiy
049Perl160612T121540ZDom Hast
062PowerShell v2+160613T154723ZAdmBorkB
063Python 2160612T043113Zxnor
146TIBASIC160612T053532ZConor O&
057Julia160612T151130Zlynn
103PHP160612T160611Zinsertus
014Jelly160612T060407ZDennis
012MATL160612T132942ZLuis Men
079Javascript ES6160612T130721ZQwertiy
015Pyth160612T042347ZLeaky Nu
054JavaScript ES6160612T101239Zedc65
384Java160612T100824ZKhaled.K
020CJam160612T091634Zaditsu q
020Pyth160612T044015ZMaltysen

Japt -x, 21 19 15 bytes

ä!Ìf í*UòÎËÎpDÊ

Try it

Brachylog, 34 32 bytes

~b≜ḅ⟨h⟨h≡^⟩l⟩ᵐs₂ᶠ{zh>₁&ttṅ|tt}ᵐ+

Try it online!

05AB1E (legacy), 17 16 15 bytes

ü.S…*-+sè‚ζJJ.E

-2 bytes thanks to @Emigna.

Try it online or verify all test cases.

Explanation:

ü                  # Pair-wise loop over the (implicit) input-list
                   #  i.e. [12,0,7,7] → [[12,0],[0,7],[7,7]]
 .S                # Calculate the signum of the pair (-1 for a<b; 0 for a==b; 1 for a>b)
                   #  i.e. [[12,0],[0,7],[7,7]] → [1,-1,0]
   …*-+sè          # Index over "*-+" (with wrap-around)
                   #  i.e. [1,-1,0] → ['-','+','*']
         ‚ζ        # Zip the two lists together (appending the operands to the numbers)
                   #  i.e. [12,0,7,7] and ['-','+','*','+']
                   #   → [[12,'-'],[0,'+'],[7,'*'],[7,' ']]
           JJ      # Join them together
                   #  [[12,'-'],[0,'+'],[7,'*'],[7,' ']] → '12-0+7*7 '
             .E    # Evaluate as Python code
                   #  i.e. '12-0+7*7' → 61

R, 120 44 bytes

r=rle(scan());c(1,sign(diff(r$v)))%*%r$v^r$l

Try it online!

The algorithm is similar to this answer's, but I only realized it after coding my answer. Much better than my original answer that was using eval(parse).

Fully leverages R's vectorized operations - does the * operation first using rle(x)$values ^ rle(x)$lenghts and dot-products this vector with sign( diff( rle(x)$values ) ) (predended with 1).

Japt, 25 bytes

Would like to cut it shorter, but I couldn't make an eval-less version work.

S+Uä!- ®"*+-"gZÎì)íU
OxU

Try it online!

R, 92 bytes

There's likely still some good golfing that can be done here.

eval(parse(t=paste(i<-scan(),c(ifelse(d<-diff(i),ifelse(d>0,"+","-"),"*"),""),collapse="")))

Ungolfed:

i = scan()                # Read list from stdin
d = diff(i)               # Calculate difference between each element of list
s = ifelse(d,             # If d!=0
             ifelse(d>0,  # And if d>1
                    "+",  # Return plus
                    "-"), # Else return minus
             "*")         # Else (i.e. d==0) return multiply.
s = c(s,"")               # Pad the list s with an empty string, so it's the same
                          # length as i
p = paste(i,s,collapse="")# Paste the elements of i and s into one long string.
eval(parse(t=p))          # Evaluate the string as a language object.

Actually, 30 bytes

;2@VpXdX`i-su"+*-"E`M' @o♀+εj≡

Unfortunately, because the eval () command only evaluates literals on TIO, this program does not work on TIO.

Explanation:

;2@VpXdX`i-su"+*-"E`M' @o♀+εj≡
;                               duplicate input
 2@V                            overlapping sublists of length <= 2
    pXdX                        discard first and last element (length-1 sublists)
        `i-su"+*-"E`M           map: for each pair of elements
         i-su                     flatten, subtract, sign, increment (results in a 0 if b < a, 1 if b == a, and 2 if b > a)
             "+*-"E               select the correct operation
                     ' @o       put a space at the beginning of the list to pad it properly
                         ♀+     pairwise addition (since addition is not defined for strings and integers, this just zips the lists and flattens the result into a single flat list)
                           εj   join with empty string
                             ≡  eval

Javascript ES6, 64 62 chars

a=>eval(a.map((x,i)=>x+('*+-'[x<a[++i]|(x>a[i])*2])).join``+1)

Perl, 49 bytes

48 bytes code + 1 for -p

s/\d+ (?=(\d+))/$&.qw(* - +)[$&<=>$1]/ge;$_=eval

Usage

perl -pe 's/\d+ (?=(\d+))/$&.qw(* - +)[$&<=>$1]/ge;$_=eval' <<< '12 0 7 7 29 10 2 2 1'
75

Notes

I learnt here that you can capture a lookahead in PCRE, although it's a little unintuitive ((?=(\d+)) instead of ((?=\d+))). It does make sense after reading though as you would be capturing a zero-length match (the lookahead) with the latter, and instead capture the match with the former).

Thanks to @ninjalj for saving 8 bytes!

PowerShell v2+, 62 bytes

-join($args|%{"$o"+'*+-'[($o-lt$_)+2*($o-gt$_)];$o=$_})+$o|iex

Takes input as space-separated command-line arguments, which gets converted into automatic array $args. We iterate through each element, using helper variable $o each iteration to remember what our previous entry was. We use a indexed-string to pull out the appropriate operator, done by performing math on the implicitly-converted Boolean values (e.g., if the previous entry is smaller, the [] evaluates to 1+2*0 so '*+-'[1] means the + is selected).

The concatenated strings are left on the pipeline. We collect all of those snippets together (e.g., 3-, 1+, 4-, etc.) with a -join operation, concatenate on the final number (implicitly converted to string), and pipe it to iex (alias for Invoke-Expression and similar to eval).

Python 2, 63 bytes

p=s='print-'
for x in input():s+='*+-'[cmp(x,p)]+`x`;p=x
exec s

Constructs and evals the expression string. The arithmetic symbol is chosen by comparing the previous number p to the current one x. The symbol is appended followed by the current number.

The first number is handled with a clever trick from Sp3000. The initial value of p is set to a string, which is bigger than any number and therefore causes a - before the first number. But, s is initialized to print- at the same time that makes the result start with print-- (thanks to xsot for saving 2 bytes by initializing with print.)

TI-BASIC, 146 bytes

I'll format it nicely when not on mobile. Sleep escapes me, so you get this. Enjoy.

Prompt L₁
"(→Str1
For(A,1,dim(L₁
{0,1→L₂
{0,L₁(A→L₃
LinReg(ax+b) L₁,L₃,Y₁
Equ►String(Y₁,Str2
sub(Str2,1,-1+inString(Str2,"X→Str2
If A>1
Then
L₁(A-1
2+(Ans>L₁(A))-(Ans<L₁(A
Str1+sub("+*-",Ans,1→Str1
End
Str1+Str2→Str2
End
expr(Str1

Julia, 76 57 bytes

!x=[[" ""-*+"[2+sign(diff(x))]...] x]'|>join|>parse|>eval

My first time golfing Julia, so maybe there are obvious improvements. Try it online!

Dennis saved a ton of bytes.

PHP, 103 bytes

Neat challenge. This got longer than expected. I think using array_map or similar won't improve the byte count, as anonymous functions are still expensive in PHP.

foreach(fgetcsv(STDIN)as$v)(0<=$p?$o.=$p.('*-+'[($p>$v)+2*($p<$v)]):_)&&$p=$v;echo eval("return$o$v;");

Runs from command line, will prompt for a comma separated list, like:

php array_to_math.php
12, 0, 7, 7, 29, 10, 2, 2, 1

Jelly, 18 16 15 14 bytes

I0;ð1g×⁹⁸œṗP€S

Uses no built-in eval. Try it online! or verify all test cases.

How it works

I0;ð1g×⁹⁸œṗP€S  Main link. Input: A (list)

I               Increments; compute the deltas of all adjacent items of A.
 0;             Prepend a 0.
   ð            Begin a new, dyadic chain.
                Left argument: D (0 plus deltas). Right argument: A
    1g          Compute the GCD of 1 and each item in D.
                This yields 1 for non-negative items, -1 for negative ones.
      ×⁹        Multiply each 1 or -1 with the corresponding item of A.
                This negates every item in A that follows a - sign.
        ⁸œṗ     Partition the result by D. This splits at occurrences of non-zero
                values of D, grouping items with identical absolute value.
           P€   Take the product of each group.
             S  Sum; add the results.

MATL, 12 bytes

Y'^l6MdZSh*s

This uses @aditsu's very nice idea of run-length encoding.

Try it online!

Explanation

       % Take input vector implicitly
Y'     % RLE. Produces two vectors: values and lengths
^      % Rise each value to the number of consecutive times it appears. This
       % realizes the product of consecutive equal values
l      % Push 1
6M     % Push vector of values again
d      % Consecutive differences
ZS     % Sign function. Gives 1 or -1 (no 0 in this case)
h      % Concatenate horizontally with previous 1
*      % Multiply. This gives plus or minus depending on increasing character
s      % Sum of vector. This realizes the additions or subtractions
       % Display implicitly

Javascript ES6, 79 chars

a=>eval(`${a}`.replace(/(\d+),(?=(\d+))/g,(m,a,b)=>a+('*+-'[+a<+b|(+a>+b)*2])))

Pyth, 31 26 19 17 16 15 bytes

Expressions with * won't evaluate online, but they would theoretically work.

2 bytes thanks to Maltysen.

vsm+@"*-+"._-~k

Test suite (with evaluation).

The other cases (without evaluation).

History

JavaScript (ES6), 54

p=>eval(0+p.map(v=>x+='*-+'[(p>v)+2*(p<v)]+(p=v),x=1))

eval receives a comma separated list of expressions and returns the value of the last one.

Test

f=p=>eval(0+p.map(v=>x+='*-+'[(p>v)+2*(p<v)]+(p=v),x=1))

t=p=>(0+p.map(v=>x+='*-+'[(p>v)+2*(p<v)]+(p=v),x=1))

function Test() {
  var a=I.value.match(/\d+/g).map(x=>+x) // convert input to a numeric array
  
  var x=f(a),y=t(a)
  O.textContent='Value '+x+'\n(no eval '+y+')'
}  

Test()
#I { width:80%}
<input value='12, 0, 7, 7, 29, 10, 2, 2, 1' id=I>
<button onclick='Test()'>Test</button>
<pre id=O></pre>

Java, 384 bytes

int s(int[]l){int n=l[0],m;for(int i=0;i<l.length-1;i++)if(l[i]<l[i+1])if(i<l.length-2&&l[i+1]!=l[i+2])n+=l[i+1];else{m=l[i+1];while(i<l.length-2&&l[i+1]==l[i+2])m*=l[(i++)+1];n+=m;}else if(l[i]>l[i+1])if(i<l.length-2&&l[i+1]!=l[i+2])n-=l[i+1];else{m=l[i+1];while(i<l.length-2&&l[i+1]==l[i+2])m*=l[(i++)+1];n-=m;}else{m=l[i];while(i<l.length-1&&l[i]==l[i+1])m*=l[i++];n+=m;}return n;}

Ungolfed try online

int s(int[] l)
{
    int n=l[0], m;

    for(int i=0; i<l.length-1; i++)
    {
        if(l[i] < l[i+1])
        {
            if (i<l.length-2 && l[i+1]!=l[i+2])
            {
                n += l[i+1];
            }
            else
            {
                m = l[i+1];
                while(i<l.length-2 && l[i+1]==l[i+2]) m *= l[(i++)+1];
                n += m;
            }
        }
        else if(l[i] > l[i+1])
        {
            if (i<l.length-2 && l[i+1]!=l[i+2])
            {
                n -= l[i+1];
            }
            else
            {
                m = l[i+1];
                while(i<l.length-2 && l[i+1]==l[i+2]) m *= l[(i++)+1];
                n -= m;
            }
        }
        else
        {
            m = l[i];
            while(i<l.length-1 && l[i]==l[i+1]) m *= l[i++];
            n += m;
        }
    }

    return n;
}

CJam, 20

q~e`{~_W-g\:W@#*}%:+

Try it online

Explanation:

q~       read and evaluate the input (array of numbers)
e`       RLE encode, obtaining [count number] pairs
{…}%     map each pair
  ~_     dump the count and number on the stack, and duplicate the number
  W-     subtract the previous number (W is initially -1 by default)
  g      get the sign of the result
  \      swap with the other copy of the number
  :W     store it in W (for next iteration)
  @#     bring the count to the top, and raise the number to that power
  *      multiply with the sign
:+       add all the results together

Pyth - 23 22 20 bytes

As with Kenny's, multiplication doesn't work online.

vs.i+\+@L"*+-"._M-Vt

Test Suite without doing eval.