Temperature Converter

Question

This is a byte sized challenge where you have to convert an input temperature in one of the three units (Celsius, Kelvin and Fahrenheit) to the other two.

Input

You would be provided with a temperature as a number followed by a unit (separated by space). The temperature can be an integer or a floating point number (23 vs 23.0 or 23.678).

You can submit a function or a full program that reads the space separated string from STDIN/ARGV/function argument or the closest equivalent and prints the output to STDOUT or closest equivalent.

Output

Your output should be the temperature converted to the other two formats, separated by a newline and followed by the corresponding unit character on each line (optionally separated by a space). The order of the two units does not matter.

Output precision

• The converted number should be accurate to at least 4 decimal places without rounding.
• Trailing zeroes or decimal places are optional as long as first 4 decimal places (without rounding) are precise. You can also skip the 4 zeroes and/or the decimal point in case the actual answer has 4 zeroes after the decimal point.
• There should not be any leading zeroes
• Any number format is acceptable as long as it fulfills the above three requirements.

Unit representation

Unit of temperature can only be one of the following:

• C for Celsius
• K for Kelvin
• F for Fahrenheit

Examples

Input:

23 C

Output:

73.4 F
296.15 K

Input:

86.987 F

Output:

303.6983 K
30.5483 C

Input:

56.99999999 K

Output:

-216.1500 C
-357.0700 F

This is code-golf so shortest entry in bytes wins! Happy Golfing!

Leaderboard

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Answer 1

Python 3, 118 116 bytes

I=input()
t=eval(I[:-1])
u=ord(I[-1])%7
exec("u=-~u%3;t=[t*1.8-459.67,(t-32)/1.8,t+273.15][u];print(t,'FCK'[u]);"*2)

Performs the conversions in a rotary order K -> F -> C -> K twice.

Answer 2

CJam, 77 65 60 59 55 54 52 bytes

l~2|"459.67+1.8/'K 273.15-'C 1.8*32+'F"S/m<{~N2$}%6<

Try it online in the CJam interpreter.

How it works

l~    e# Read and evaluate the input: F, K, C -> 15, 20, 12
2|    e# Bitwise OR with 2: F, K, C -> 15, 22, 14 = 0, 1, 2 (mod 3)

"459.67+1.8/'K 273.15-'C 1.8*32+'F"S/

      e# Push ["459.67+1.8/K" "273.15-C" "1.8*32+F"].
      e# These commands convert from F to K, K to C and C to F.

m<    e# Rotate the array of commands by 15, 22 or 14 units to the left.
{     e# For each command:
  ~   e#     Execute it.
  N   e#     Push a linefeed.
  2$  e#     Copy the temperature for the next iteration.
}%    e# Collect all results in an array.
6<    e# Keep only the first 8 elements.

Answer 3

JavaScript (ES6), 127 130 132 bytes

Waiting for some amazing esolanguage answer, I did not find much to golf here.

Using template string, the 3 newlines are significant and counted.

Run snippet in Firefox to test.

F=x=>([t,u]=x.split(' '),z=273.15,y=9/5,u<'F'?`${t*y+32} F
${+t+z} K`:u<'K'?`${t=(t-32)/y} C
${t+z} K`:`${t-=z} C
${t*y+32} F`)

// Predefined tests

;['23 C','86.987 F','56.99999999 K']
.forEach(v=>O.innerHTML += v+' ->\n'+F(v)+'\n\n')

<input id=I><button onclick='O.innerHTML=F(I.value)'>-></button><br>
<pre id=O></pre>

Answer 4

Pip, 58 57 bytes

Ft"CFK"RMbP(a-(o:[32i273.15]Ab%7))*$/95@A[tb]+(oAt%7).s.t

Takes input from command-line arguments.

Formatted and slightly ungolfed:

o:[32 0 273.15]
F t ("CFK" RM b)
  P (a - o@(Ab % 7))
    * "95"@At / "95"@Ab
    + o@(At%7)
    . s . t

Explanation:

The general conversion formula between unit 1 and unit 2 is temp2 = (temp1 - offset1) * mult2 / mult1 + offset2. The offsets could be to absolute zero or to any other convenient temperature; let's use 0 °C.

Unit Mult Offset
C    5    0     
K    5    273.15
F    9    32    

We'll build lists of these values and index into them based on which unit we're dealing with. Pip doesn't have associative arrays/hashes/dictionaries, so we need to convert the characters into integer indices. Maybe there will be helpful patterns with the ASCII values.

Unit Asc A%2 Mult A%7 A%7%3 Offset
C    67  1   5    4   1     0
K    75  1   5    5   2     273.15
F    70  0   9    0   0     32

This looks promising. One more useful fact: indices in Pip wrap around. So we don't actually need the %3 as long as we're indexing into something of length 3. Defining o:[32 0 273.15] and using (o (Ab)%7) will do the trick. (A gets the ASCII value of a character. The (l i) syntax is used for indexing into iterables, as well as for function calls.)

For the multipliers, we only need the two numbers 9 and 5. Since numbers are the same as strings in Pip, and thus are indexable, getting the multiplier is as simple as 95@Ab (using the other method of indexing, the @ operator).

Using Pip's array-programming features, we can save a character over the naive method:

95@At/95@Ab
$/95@A[tb]    Make a list of t and b, apply 95@A(...) to each item, and fold on /

Finally, add the offset for the new unit, concatenate a space and the new unit's symbol onto the end, and print.

We do this for each t in "CFK" RM b, thus converting to each unit except the original.

Sample invocation:

C:\Golf> pip.py tempConv.pip 86.987 F
30.548333333333332 C
303.6983333333333 K

(For more on Pip, see the repository.)

Answer 5

dc, 102 bytes

I'm pretty sure this can be golfed more, but here's a start:

4k[9*5/32+dn[ F
]n]sf[459.67+5*9/dn[ K
]n]sk[273.15-1/dn[ C
]n]sc[rlfxlkxq]sC[lkxlcxq]sF?dC=CF=Flcxlfx

dc barely makes the grade on this one. Specifically dc string handling isn't really up to the job. But all we need to do is differentiate between "C", "F" and "K". Fortunately dc parses "C" and "F" as hex numerals 12 and 15. And fo "K" it just leaves 0 on the stack.

Output:

$ dc -f tempconv.dc <<< "23 C"
73.4000 F
296.1500 K
$ dc -f tempconv.dc <<< "86.987 F"
303.6983 K
30.5483 C
$ dc -f tempconv.dc <<< "56.99999999 K"
-216.1500 C
-357.0700 F
$

Answer 6

C, 160 bytes

float v,f[]={1,1.8,1},d[]={0,32,273.15};u,t,j;main(){scanf("%f %c",&v,&u);for(u=u/5-13,v=(v-d[u])/f[u];j<2;)t=(++j+u)%3,printf("%f %c\n",f[t]*v+d[t],"CFK"[t]);}

This reads from stdin. No output precision is specified, so it prints whatever printf() feels like printing, which is mostly more than 4 digits.

Ungolfed version:

#include <stdio.h>

float v, f[] = {1.0f, 1.8f, 1.0f}, d[] = {0.0f, 32.0f, 273.15f};
char u;
int t, j;

int main() {
    scanf("%f %c", &v, &u);
    u = u / 5 - 13;
    v = (v - d[u]) / f[u];
    for( ; j < 2; ) {
        t = (++j + u) % 3;
        printf("%f %c\n", f[t] * v + d[t], "CFK"[t]);
    }

    return 0;
}

Some explanations/remarks:

• This is based on lookup tables f and d that contains the multiplication factors and offsets to convert Celsius to any other unit (including Celsius itself, to avoid special cases).
• If u is the input unit, u / 5 - 13 maps C to 0, F to 1, and K to 2.
• Input is always converted to Celsius.
• Loop iterates over the two units that are not the input, and converts the Celsius value to that unit.
• The golfed version uses an int variable instead of a char variable to receive the scanf() input. This is a type mismatch, which will only produce correct results on little endian machines (which is almost all of them these days).

Answer 7

Python 2, 168 Bytes

I saw the Python 3 solution just as I was about to post this, but whatever, I'm just practicing golfing.

s=raw_input().split();v=eval(s[0]);c,t=(v-32)/1.8,v-273.15
print[["%fC\n%fF"%(t,t*1.8+32),"%fK\n%fC"%(c+273.15,c)][s[1]=="F"],"%fK\n%fF"%(v+273.15,v*1.8+32)][s[1]=="C"]

Ungolfed:

def conv(s):
    s = s.split()
    v = float(s[0])
    if s[1]=="C":
        print "%f K\n%f F"%(v+273.15,v*1.8+32)
    elif s[1]=="F":
        v = (v-32)/1.8
        print "%f K\n%f C"%(v+274.15,v)
    else:
        c = v-273.15
        print "%f C\n%f F"%(c,c*1.8+32)

Answer 8

Perl, 85 80 76

#!perl -nl
print$_=/C/?$_+273.15." K":/K/?$_*9/5-459.67." F":5/9*($_-32)." C"for$_,$_

Test me.

Answer 9

PHP, 161 153

function f($i){list($v,$u)=split(" ",$i);$u==C||print($v=$u==K?$v-273.15:($v-32)*5/9)."C\n";$u!=K&&print $v+273.15 ."K\n";$u!=F&&print $v*9/5+32 ."F\n";}

http://3v4l.org/CpvG7

Would be nice to golf out two more characters to get in the top ten...

Answer 10

Python 2, 167

s=raw_input();i=eval(s.split()[0]);d={"C":"%f K\n%f F"%(i+273,32+i*1.8),"K":"%f C\n%f F"%(i-273,i*1.8-459.4),"F":"%f C\n%f K"%(i/1.8-17.78,i/1.8+255.2)};print d[s[-1]]`

This is my first try on CodeGolf.

Answer 11

Pyth, 126 bytes

AYZczdJvY=Y*JK1.8
?%"%f F\n%f K",+32Y+J273.15qZ"C"?%"%f F\n%f C",-Y459.67-J273.15qZ"K"?%"%f C\n%f K",c-J32K*+J459.67c5 9qZ"F"

This feels really long to me...oh well.

Answer 12

R, 150 144 141 bytes

a=scan(,"");U=c("C","K","F");i=which(U==a[2]);Z=273.15;x=scan(t=a[1]);cat(paste(c(C<-c(x,x-Z,5*(x-32)/9)[i],C+Z,32+C*9/5)[-i],U[-i]),sep="\n")

Indented, with new lines:

a=scan(,"")
U=c("C","K","F")
i=which(U==a[2])
Z=273.15
x=as.double(a[1])
cat(paste(c(C<-c(x,x-Z,5*(x-32)/9)[i],C+Z,32+C*9/5)[-i],
            U[-i]),
    sep="\n")

Usage:

> a=scan(,"");U=c("C","K","F");i=which(U==a[2]);Z=273.15;x=scan(t=a[1]);cat(paste(c(C<-c(x,x-Z,5*(x-32)/9)[i],C+Z,32+C*9/5)[-i],U[-i]),sep="\n")
1: 23 C
3: 
Read 2 items
Read 1 item
296.15 K
73.4 F
> a=scan(,"");U=c("C","K","F");i=which(U==a[2]);Z=273.15;x=scan(t=a[1]);cat(paste(c(C<-c(x,x-Z,5*(x-32)/9)[i],C+Z,32+C*9/5)[-i],U[-i]),sep="\n")
1: 56.9999999 K
3: 
Read 2 items
Read 1 item
-216.1500001 C
-357.07000018 F
> a=scan(,"");U=c("C","K","F");i=which(U==a[2]);Z=273.15;x=scan(t=a[1]);cat(paste(c(C<-c(x,x-Z,5*(x-32)/9)[i],C+Z,32+C*9/5)[-i],U[-i]),sep="\n")
1: 86.987 F
3: 
Read 2 items
Read 1 item
30.5483333333333 C
303.698333333333 K

Thanks to @AlexA. & @MickyT. !

Answer 13

Javascript, 206 193 187 175 162 159 156

e=273.15,a=prompt(f=1.8).split(" "),b=+a[0],X=b-32,c=a[1];alert(c=="F"&&(X/f+"C\n"+(X/f+e+"K"))||(c=="C"?b*f+32+"‌​F\n"+(b+e+"K"):b*f-459.67+"F\n"+(b-e+"C")))

Thanks goes to Optimizer and Ismael Miguel for helping me golf it a little further.

Answer 14

Mathematica, 66

Hopefully correct this time.

Print[1##]&@@@Reduce[F==1.8K-459.67&&C==K-273.15&&#==#2,,#2];&@@#&

Example

%[37 C]   (* % represents the function above *)

310.15 K

98.6 F

Answer 15

Excel, 239 bytes

=IF(RIGHT(A1,1)="C",LEFT(A1,LEN(A1)-2)+273.15&"K
"&9*LEFT(A1,LEN(A1)-2)/5+32&"F",IF(RIGHT(A1,1)="K",LEFT(A1,LEN(A1)-2)-273.15&"C
"&(LEFT(A1,LEN(A1)-2)*9/5-459.67&"F",(LEFT(A1,LEN(A1)-2)+459.67)*5/9&"K
"&(LEFT(A1,LEN(A1)-2)-32)*5/9&"C"))

112 bytes of which are used to seperate Value and Unit. Anybody know a better solution?