# Convert degrees to one of the 32 points of the compass

The 32-point compass is... interesting, to say the least. By Denelson83 (Own work) [GFDL or CC-BY-SA-3.0], via Wikimedia Commons

Your challenge is to take a degree measure, and convert it into a direction on the 32-point compass.

Each direction is 11.25 (360 / 32) degrees farther than the previous. For example, N (north) is 0 degrees, NbE (north by east) is 11.25 degrees, NNE (north-northeast) is 22.5 degrees, etc.

As for how you're supposed to get the directions,

• 0 degrees is N, 90 degrees is E, 180 degrees is S, and 270 degrees is W.

• These are called cardinal directions.
• The halfway points between the cardinal directions are simply the cardinal directions they're between concatenated. N or S always go first, and W or E always are second.

• These are called ordinal directions.
• The halfway points between the cardinal and ordinal directions are the directions they're between concatenated, again, with a "-" in between. Cardinal directions go first, ordinal second.

• These are called secondary-intercardinal directions.
• The halfway points between secondary-intercardinal directions and other directions are the other directions "by" the cardinal direction they're closest to (other than the one directly next to them, of course).

• I have no idea what these are called :P

If all this explanation hurts your brain as much as mine, you can refer to this chart:

1   North               N
2   North by east       NbE
3   North-northeast     NNE
4   Northeast by north  NEbN
5   Northeast           NE
6   Northeast by east   NEbE
7   East-northeast      ENE
8   East by north       EbN
9   East                E
10  East by south       EbS
11  East-southeast      ESE
12  Southeast by east   SEbE
13  Southeast           SE
14  Southeast by south  SEbS
15  South-southeast     SSE
16  South by east       SbE
17  South               S
18  South by west       SbW
19  South-southwest     SSW
20  Southwest by south  SWbS
21  Southwest           SW
22  Southwest by west   SWbW
23  West-southwest      WSW
24  West by south       WbS
25  West                W
26  West by north       WbN
27  West-northwest      WNW
28  Northwest by west   NWbW
29  Northwest           NW
30  Northwest by north  NWbN
31  North-northwest     NNW
32  North by west       NbW


Here is a more detailed chart and possibly better explanation of the points of the compass.

Your challenge is to take input in degrees, and output the full name of the compass direction it corresponds to, along with its abbreviation.

Test cases:

Input  Output
0      North N
23.97  North-northeast NNE
33.7   Northeast by north NEbN
73.12  East-northeast ENE
73.13  East by north EbN
219    Southwest by south SWbS
275    West W
276    West by north WbN
287    West-northwest WNW


All capitalization must be preserved, as in the test cases. Maximum number of decimal places is 2. All input numbers will be greater than or equal to 0, and less than 360. If a decimal point is present, there will be digits on both sides (you don't have to handle .1 or 1.).

This is , so the shortest code in bytes wins.

• @WallyWest Hmm, this one allows arrays, has different capitalization, and has no "between," but I didn't notice that (possibly because of the... interesting title :P). I'll see what I can do to make it different enough... Feb 23, 2014 at 22:33
• @WallyWest There, now you have to output the abbreviation too. Along with all the other differences, that should be enough to make it a non-dup. (oh, also this one has dashes too) Feb 23, 2014 at 22:36
• @WallyWest There be no answers in "R" to your previous question (there weren't even any in "C"!) I hopes we'll see some this time, shipmates! Feb 25, 2014 at 9:33
• Would have been more fun if there were input from -360 to 360 degree (negative means anti-clockwise) and a bonus!. Feb 25, 2014 at 17:05
• For people not looking for a challenge, the easy solution is to find the possible output for which the distance is minimal from the input angle, using a lookup table angle <-> name. Feb 25, 2020 at 17:16

## Javascript 470453438434432421 404

s=String;s.prototype.a=s.prototype.replace;var a=prompt()/11.25,a=a+0.5|0,b,k,c=a,d=c%8,c=c/8|0,e=["north","east","south","west"],f,g,h;f=e[c];g=e[(c+1)%4];h=f==e|f==e?f+g:g+f;b="1;1 by 2;1-C;C by 1;C;C by 2;2-C;2 by 1".split(";")[d].a(1,f).a(2,g).a("C",h);k=b.a(/north/g,"N").a(/east/g,"E").a(/south/g,"S").a(/west/g,"W").a(/by/g,"b").a(/[\s-]/g,"");b=b.toUpperCase()+b.slice(1);alert(b+" "+k)


You can copy this code to your console and execute it. It prompts you for the degrees input and outputs the result with alert();

Ungolfed Javascript can be found at this fiddle: http://jsfiddle.net/AezL3/11

• +1 Nice but be careful : "All capitalization must be preserved, as in the test cases."
– user16229
Feb 25, 2014 at 10:15
• @BenH Which test case fails the capitalization check? Thanks for this, btw. This came in handy for my web interface I'm writing. Jan 5, 2016 at 5:56
• By the way, this does die for 355 degrees to 360 degrees. The fix is easy. Just make calcPoint(32) do what 0 does, so you can do this with %32 or any similar. Jan 5, 2016 at 6:01
• @StevenLu it took me a while to figure out what you means, but this line var name = calcPoint(input % 32); does the trick Oct 20, 2017 at 10:03

## Perl, 250 236 231 188 187

Edit: Some bytes off exploiting symmetry (as I've seen in @bazzargh solution)

+Edit: And some evil tricks...

+Edit: Back to where I started (working with list, not string), and exploiting more symmetry = 1 byte off and a whole lot uglier.

$_=((@_=(1,@_=qw(1b3 1-13 13b1 13 13b3 3-13 3b1),3,map{y/1/2/r}reverse@_)),map{y/312/421/r}@_)[int<>/11.25+.5];print ucfirst s/\w/(' by ',north,south,east,west)[$&]/ger,' ',y/1-4-/NSEW/dr


Pretty-printed:

$_=( (@_= ( 1, @_=qw(1b3 1-13 13b1 13 13b3 3-13 3b1), 3, map{y/1/2/r}reverse@_ ) ),map{y/312/421/r}@_ )[int<>/11.25+.5]; print ucfirst s/\w/(' by ',north,south,east,west)[$&]/ger,' ',y/1-4-/NSEW/dr


5.014 required because of r modifier.

• You have a typo in your code: sourth should be south (2nd statement that starts with s/b/ by ... Feb 25, 2014 at 13:39
• Those first 3 regexps can be replaced by y/NS/SN/; for 10 chars Feb 25, 2014 at 17:51
• @bazzargh, yes, and not only that ;-) Feb 25, 2014 at 17:54

Ended up converging on a solution like @VadimR's (and the symmetry's back!). Usage: h 219 outputs "Southwest by south SWbS"

d"N"="north"
d"S"="south"
d"E"="east"
d"W"="west"
d"b"=" by "
d"-"="-"
d(x:y)=d[x]++d y
e(x:y)=x:tail(d$x:y) k 'N'='S' k 'S'='N' k 'E'='W' k x=x r="N NbE N-NE NEbN NE NEbE E-NE EbN E EbS E-SE SEbE SE SEbS S-SE SbE " p=words$r++(map k r)
g x=p!!mod(round$x/11.25)32 h x=e(g x)++(filter(/='-')$' ':g x)


3 more chars gone, thanks @shiona.

• drop 1 is same as tail. Also if I'm not mistaken you can do e l@(x:_)=x:tail\$d l to shave still one more char. Feb 25, 2014 at 17:17
• can't believe I missed that. Thanks! Feb 25, 2014 at 17:41

# Python 3.8, 482438 424 bytes

lambda h:' '.join([b(h),a(a(a(b(h)),1),d={' by ':'b','-':''})])
L=str.lower
c={'North':'N','East':'E','South':'S','West':'W'}
a=lambda t,l=0,d=c:[*(t:=t.replace([i,L(i)][l],d[i])for i in[*d])][-1]
b=lambda h,k=[*c]:a('W|W by x|W-z|Z by w|Z|Z by x|X-z|X by w'.split('|')[int((q:=h*4/45+.5)%8)],d={'W':(W:=[*k][(v:=int(q//8)%4)]),'X':(X:=[*k][(v+1)%4]),'w':(w:=L(W)),'x':(x:=L(X)),'Z':(Z:=[W+x,X+w][W in'EastWest']),'z':L(Z)})


Try it online!

This is what I got after some golfing of tony goodwin's answer; posted at its own answer due to the TIO link being too long for a comment. If he chooses to update his answer to the above, I will delete this answer.

I am assuming that it's acceptable to submit a function as a solution rather than a full program. If not, here is a full program at 426 bytes.

I expect that much can still be done to shorten b.

Edit: Golfed off 44 bytes, courtesy of the majestic walrus. Still don't feel like b is done being golfed.

Edit2: Shaved off another 14 by unpacking dicts instead of using keys() and items().

# Python, 2103164711031034924889 848 bytes

Very late, I know. Thanks for the challenge, I am setting up a magnetometer for wind direction with my Pi, and wanted a 16 point of compass solution like this to feed into weather forecasting algorithms. All my code is in Python, so here is version of the javascript solution already posted in Python, but with an extra twist that you can specify either 32, 16, or 8 points of the compass at variable j, and I have changed the offset of degHead in the statement before it, depending on the number of points. I used a modified rename alogorithm (and used variables I could rename without corrupting the words!) to ensure I met the case requirements of the question.

I know this won't win, as Python is more wordy, and so am I.

Short version:

  def a(t,d,l):
for i,j in d.items():
if l:
i=i.lower()
t=t.replace(i,j)
return t
def b(h,q):
p=32
r=360
h=(h+(r/q/2))/(r/p)
j=int(int(int(h %8)%8/(p/q))*p/q)
h=int(h/8)%4
k=c.keys()
u=['W','W by x','W-z','Z by w','Z','Z by x','X-z','X by w']
d={}
d['W']=list(k)[h]
d['w']=d['W'].lower()
d['X']=list(k)[(h+1)%4]
d['x']=d['X'].lower()
if(d['W']=='North' or d['W']=='South'):
d['Z']=d['W']+d['x']
else:
d['Z']=d['X']+d['w']
d['z']=d['Z'].lower()
return a(u[j],d,0)
def g(n):
n=a(n,c,0)
n=a(n,c,1)
d={'by':'b',' ':'','-':''}
return a(n,d,0)
def v(m):
while True:
try:
return float(input(m))
except ValueError:
print("?")
c={'North':'N','East':'E','South':'S','West':'W'}
while True:
h=v("?")
n=b(h,32)
print(h,n,g(n))


Clear version

            import math
import sys

maxPoints=32
if points not in(8,16,32):
sys.exit("not a good question")
cardinal = ['North', 'East', 'South', 'West']
pointDesc = ['W', 'W by x', 'W-z', 'Z by w', 'Z', 'Z by x', 'X-z', 'X by w']#vars not compass points
X = cardinal[(degHead + 1) % 4]
w=W.lower()
x=X.lower()
if (W == cardinal or W == cardinal) :
Z =W + x
else:
Z =X + w
z=Z.lower()
return pointDesc[j].replace('W', W).replace('X', X).replace('w', w).replace('x', x).replace('Z', Z).replace('z', z);

def getShortName(name):
return name.replace('North', 'N').replace('East', 'E').replace('South', 'S').replace('West', 'W').replace('north', 'N').replace('east', 'E').replace('south', 'S').replace('west', 'W').replace('by', 'b').replace(' ', '').replace('-', '')

def input_number(msg, err_msg=None):
while True:
try:
return float(input(msg))
except ValueError:
sys.exit("not a number")

while True: