How drunk am I and when can I drive again?

Question

It's weekend and what are the cool guys doing on weekends? Drinking of course! But you know what's not so cool? Drinking and driving. So you decide to write a program that tells you how loaded you are and when you are gonna be able to drive again without getting pulled over by the cops and loosing your license.

The Challenge

Given a list of beverages you enjoyed this evening, calculate your blood alcohol level and the time you have to wait till you can hop into your car and get home.

Input

Input will be a list of drinks you had this night. This will look like this:

4 shots booze
1 glasses wine
2 bottles beer
3 glasses water

Containers will always be plural.

As you can see each entry consists of:

• The type of drink (booze, wine, beer, water)
• The container for the drink (shots, glasses, bottles)
• The amount x of the drinks you had of that type as integer with x > 0,

Each drink type adds a certain amount of alcohol to your blood:

booze ->  0.5 ‰ / 100 ml
beer  ->  0.1 ‰ / 100 ml
wine  ->  0.2 ‰ / 100 ml
water -> -0.1 ‰ / 100 ml

Water is the exception here, since it thins your blood out and lowers your alcohol level (would be so nice if that actually worked...).

Each container has a certain volume:

shots   -> 20 ml
glasses -> 200 ml
bottles -> 500 ml

Output

You have to output two numbers:

• The alcohol level in ‰
• The time in hours you have to wait until you reached 0.5 ‰ or less, so you can drive again. You loose 0.1 ‰ per hour.

Notes

• The alcohol level can never fall below zero.
• Same goes for the waiting time. If you have 0.5 ‰ or less, output zero.
• The order of drinks does not matter, so drinking water may lower the alcohol level below zero in the process of the calculation. If it remains there, you need to replace it with zero.

The alcohol level for the the example above would be calculated like this:

4 shots booze   -> 0.4 ‰
1 glasses wine  -> 0.4 ‰
2 bottles beer  -> 1.0 ‰
3 glasses water -> -0.6 ‰

=> 0.4 + 0.4 + 1 - 0.6 = 1.2 ‰

To reach 0.5 ‰ you need to loose 0.7 ‰. You loose 0.1 ‰ per hour, so you need to wait 7 hours to drive again.

Rules

• You can take the input in any format you want, but you have to use the exact strings as given above. You may take the numbers as integers.
• You can output the two numbers in any order, just make in clear which one you use in your answer.
• You may assume that the input will always have at least one entry.
• Function or full program allowed.
• Default rules for input/output.
• Standard loopholes apply.
• This is code-golf, so lowest byte-count wins. Tiebreaker is earlier submission.

Test cases

Input as list of strings. Outputs alcohol level first, values seperated by a comma.

["4 shots booze","1 glasses wine","2 bottles beer","3 glasses water"] -> 1.2, 7
["10 shots booze", "1 bottle water"] -> 0.5, 0
["3 glasses wine", "2 bottles booze"] -> 6.2, 57
["6 shots beer", "3 glasses water"] -> 0, 0
["10 glasses beer"] -> 2.0, 15

Happy Coding!

Answer 1

TSQL, 301, 299, 219, 206 Bytes

Input goes into temp table #I (you said any format :)

SELECT * INTO #I FROM (
  VALUES
    (4,'shots','booze')
   ,(1,'glasses','wine')
   ,(2,'bottles','beer')
   ,(3, 'glasses','water')
) A (Q, V, N)

Code:

SELECT IIF(L<0,0,L),IIF(10*L-.5<0,0,10*L-.5)FROM(SELECT SUM(Q*S*P)L FROM(VALUES('bo%',.5),('be%',.1),('wi%',.2),('wa%',-.1))A(W,S),(VALUES('s%',.2),('g%',2),('b%',5))B(X,P),#I WHERE N LIKE W AND V LIKE X)A;

Thanks for the ideas to improve it, Micky T :)

Answer 2

JavaScript (ES6), 131

a=>a.map(s=>([a,b,c]=s.split` `,t+=a*[50,20,2]['bgs'.search(b[0])]*~-'0236'['aeio'.search(c[1])]),t=0)&&[t>0?t/100:0,t>50?t/10-5:0]

Less golfed

a=>(
  t=0,
  a.map(s=>(
    [a,b,c] = s.split` `,
    t += a * [50,20,2]['bgs'.search(b[0])] // char 0 is unique
           * [-1,1,2,5]['aeio'.search(c[1])] // char 1 is unique 
           // golfed: add 1 to get a string of dingle digits, the sub 1 using ~-
    )
  ),
  [ t>0 ? t/100 : 0, t>50 ? t/10-5 : 0]
)

Answer 3

Javascript (ES6), 111

Takes input as arrays of arrays of strings/integers e.g.

[[4, "shots", "booze"],
[1, "glasses", "wine"],
[2, "bottles", "beer"],
[3, "glasses", "water"]]

Outputs to simple array e.g. [1.2, 7]

a=>(b=0,c={s:2,g:20,b:50,o:2,e:10,i:5,a:-10},a.map(([d,e,f])=>b+=d*c[e[0]]/c[f[1]]),g=b>0?b:0,[g/10,g>5?g-5:0])

#Explained

a => (
b = 0,                                           // Counter For Alcohol Level
c = {s:2, g:20, b:50, o:2, e:10, i:5, a:-10},    // Look up for values
a.map(                                           // Loops over array
  ([d, e, f]) =>                                 // Sets d,e,f to respective array indexes 
     b += d * c[e[0]] / c[f[1]]                  // Increases Level by values from lookup
  ),
g = b > 0 ? b : 0,                               // If Level is lower than 0 make it = 0
[g / 10, g > 5 ? g - 5 : 0])                     // Output: Level / 10 and Level - 5 bound to 0

Answer 4

Perl, 133 119 + 3 = 136 122 bytes

%u=(o,.5,e,.1,i,.2,a,-.1,g,2,b,5,s=>.2);/(\d+) (.).* .(.)/;$x+=$1*$u{$2}*$u{$3}}{$_=($x>0?$x:0).$".($x-.5>0?$x-.5:0)*10

To be run with perl -p. Takes line-oriented input on STDIN, produces output on STDOUT.

Less-golfed version:

# char->number conversion table
# declared using barewords except for 's', which can't be a bareword
# because it's a keyword
%u=(o,.5, e,.1, i,.2, a,-.1, g,2, b,5, s=>.2);

# extract the number, first letter of first word, second letter of
# second word
/(\d+) (.).* .(.)/;

# do unit conversion and multiply up all factors
$x += $1 * $u{$2} * $u{$3}

# hack for -p to produce an END { ... print } block
}{

# format output
$_ = ($x > 0 ? $x : 0) . $" . ($x-.5 > 0 ? $x-.5 : 0)*10

Thanks to dev-null for suggestions saving 11 bytes.

Answer 5

C (GCC), 176 166 bytes

-10 bytes thanks to @ceilingcat

float x;main(a){for(char b[9],c[9];~scanf("%d%s%s\n",&a,b,c);)x+=a*(*b-98?*b-103?.02:.2:.5)*(*c-98?c[1]-97?2:-1:c[1]-111?1:5);printf("%f %f",x>0?x:0,fmax(0,x*10-5));}

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