[Question]
  [
*This is my first code golf so please let me know if it's too broad or if I'm missing any information for a good puzzle!*
# Challenge
In Ontario and possibly other areas of the world, electricity is billed using [Time-Of-Use (TOU) pricing](https://hydroottawa.com/residential/time-of-use/rate-periods/), which varies the cost per kilowatt-hour according to when you use power.
Given a date and time, I want to know whether I'm in an on-peak (red), mid-peak (yellow), or off-peak (green) time period.
# Input
Assume that input is provided in an acceptable timezone-less [ISO 8601 date-time format](https://en.wikipedia.org/wiki/ISO_8601) with the minimum precision of hours: `YYYY-MM-DDThh[:mm[:ss]]` (the T is literal).
### Examples
* 2014-09-01T14
* 2014-09-01T17:30
* 2014-09-01T17:30:02
# Output
The output should be a string `On`, `Mid`, or `Off`.
# Rules
* Shortest code wins
* For the purposes of this challenge, ignore statutory holidays
* Assume the information found in this post. The actual rules of time-of-use pricing might change in the future by the Ontario Ministry of Energy.
# Information
## Summer weekdays (May 1st to October 31st)
![Time-of-use clock for summer weekdays](https://static.hydroottawa.com/images/residential/time-of-use/tou-circle-summer_en.png)
* Off-peak: 19h00 - 07h00
* Mid-peak: 07h00 - 11h00 and 17h00 - 19h00
* On-peak: 11h00 - 17h00
## Winter weekdays (November 1st to April 30th)
![Time-of-use clock for winter weekdays](https://static.hydroottawa.com/images/residential/time-of-use/tou-circle-winter_en.png)
* Off-peak: 19h00 - 07h00
* Mid-peak: 11h00 - 17h00
* On-peak: 07h00 - 11h00 and 17h00 - 19h00
## Weekends
![Time-of-use clock on weekends](https://static.hydroottawa.com/images/residential/time-of-use/tou-circle-weekend_en.png)
* Off-peak: All day
  
[Answer]
# Python 2 - 164
```
from datetime import*
d,t=input().split('T')
y,m,d=map(int,d.split('-'))
t=int(t[:2])
print'OMOfinfd'[(1+((10<t<17)==(4<m<11)))*(date(y,m,d).weekday()<5<6<t<19)::3]
```
---
If needed, below is a explanation of the logic in the final line:
The final line prints a slice of `'OMOfinfd'` depending on the evaluation of its conditionals. 
* First, evaluate the operation `1+((10<t<17)==(4<m<11))`.
If the XNOR between the conditions `10<t<17` and `4<m<11` is `False`, this will evaluate to `1+False => 1+0 => 1`. Otherwise, the operation will evaluate to `1+True => 1+1 => 2`.
* Finally, multiply that result of the above operation by whether the day is a weekday and whether the time is between 6am-7pm.
If this is `False`, either the day is a weekend or the time is between 7pm-6am, and the result will be `(1|2)*0 => 0`. Otherwise the result will be `(1|2)*1 => 1|2`.
A result of `0` will print `Off`, `1` will print `Mid`, and `2` will print `On`.
[Answer]
## C# - ~~240~~ 220 chars
```
string x(string s){var d=DateTime.Parse((s+":00:00").Substring(0,19));int h=d.Hour,i=(int)d.DayOfWeek,x=d.Month;string o="off",m="mid",f="on";return i==6|i==0?o:x>=5&x<11?h>18|h<7?o:h>10&h<17?f:m:h>18|h<7?o:h>10&h<17?m:f;}
```
Nothing special. Straight forward coding.
Thanks to w0lf :)
[Answer]
## Ruby - 135
Abuses the Time module. Input by command line argument.
```
d=Time.new(*$*[0].scan(/\d+/)[0..3])
o,m,f=%w{On Mid Off}
o,m=m,o if (d.mon-5)%10<6
p d.wday%6<1||(12>h=(d.hour+5)%24)?f:15<h&&h<22?m:o
```
**Edit:** Thanks w0lf for Time which helped shorten and solve a bug.
[Answer]
## Ruby - ~~147 144 143 141 137~~ 135
```
x=->s{y,m,d,h=s.scan(/\d+/).map &:to_i
g=Time.new(y,m,d).wday%6<1?0:[0..11,4..9].count{|r|r===h-7}
%W{Off Mid On}[m<5||m>10?(3-g)%3:g]}
```
This represents a function which takes a string as a parameter and returns a string.
Here's an online demo with some test cases: **<http://ideone.com/wyIydw>**
[Answer]
## Groovy - 621 534 524 491 chars
Some further golfing to do, but pretty simple when leveraging Joda-Time
```
@Grab(group='joda-time',module='joda-time',version='2.3')
f={p,a->org.joda.time.format.DateTimeFormat.forPattern(p).parseDateTime a}
g={p,a->def x=null;try{x=f p,a}catch(Exception e){}}
a=args[0]
d=["",":mm",":mm:ss"].collect{g "yyyy-MM-dd'T'HH$it",a}.find{it}
r="Off"
m="Mid"
j={d,a,b->d.hourOfDay>a&&d.hourOfDay<b}
k={j(it,6,11)||(j(it,16,19))}
if(d.dayOfWeek<6){x=d.monthOfYear;if(x>4&&x<12){if(j(d,10,17))r="On";if(k(d))r=m}else if(x<5||x>10){if(j(d,10,17))r=m;if(k(d))r="On"}}
println r
```
sample runs:
```
bash-3.2$ ./run.peak.sh 
groovy Peak.groovy 2014-08-26T19
Off
groovy Peak.groovy 2014-08-26T07:00
Mid
groovy Peak.groovy 2014-08-26T18:00:00
Mid
groovy Peak.groovy 2014-08-26T12:30:30
On
groovy Peak.groovy 2014-11-01T00
Off
groovy Peak.groovy 2014-02-05T11:11:11
Mid
groovy Peak.groovy 2014-01-05T08:08
Off
groovy Peak.groovy 2014-12-18T18:59:59
On
groovy Peak.groovy 2014-08-31T14
Off
```
Ungolfed:
```
@Grab(group='joda-time',module='joda-time',version='2.3')
f = { p,a -> org.joda.time.format.DateTimeFormat.forPattern(p).parseDateTime a}
g = { p,a -> def x=null; try{x=f p,a} catch(Exception e) {} }
a=args[0]
d = ["",":mm",":mm:ss"].collect{g "yyyy-MM-dd'T'HH$it",a}.find{it}
r = "Off"
m = "Mid"
j = {d,a,b -> d.hourOfDay > a && d.hourOfDay < b}
k = { j(it,6,11) || (j(it,16,19)) }
if (d.dayOfWeek<6) {
    x = d.monthOfYear;
    if ( x>4 && x<12 ) {
        if (j(d,10,17)) r="On";
        if (k(d)) r=m
    } else if (x<5||x>10) {
        if (j(d,10,17)) r=m;
        if (k(d)) r="On"
    }
}
println r
```
[Answer]
## R, ~~243~~ 204 characters
```
b=strptime(scan(,""),"%Y-%m-%dT%H");i=function(x)as.integer(format(b,x));h=i("%H");d=i("%m%d");w=d>1100|d<430;f=ifelse;cat(c("Off","Mid","On")[f(i("%u")%in%5:6|h<7|h>19,1,f(h>11&h<17,f(w,2,3),f(w,3,2)))])
```
Indented and commented:
```
b=strptime(scan(,""),"%Y-%m-%dT%H") #Takes stdin and converts into POSIXct
i=function(x)as.integer(format(b,x)) #Format the POSIXct and convert it to integer
h=i("%H")      #Format to hours
d=i("%m%d")    #Format to Month/Day
w=d>1100|d<430 #True if winter time, false if summer
f=ifelse
cat(c("Off","Mid","On")[f(i("%u")%in%5:6|h<7|h>19, #If weekend or night
                          1,                       #Case 1
                          f(h>11&h<17,            #Else if mid-day
                             f(w,3,2),             #Case 2 in winter, case 3 in summer
                             f(w,2,3)))])          #else vice versa
```
Examples:
```
> b=strptime(scan(,""),"%Y-%m-%dT%H");i=function(x)as.integer(format(b,x));h=i("%H");d=i("%m%d");w=d>1100|d<430;f=ifelse;cat(c("Off","Mid","On")[f(i("%u")%in%5:6|h<7|h>19,1,f(h>11&h<17,f(w,3,2),f(w,2,3)))])
1: 2014-08-26T15
2: 
Read 1 item
On
> b=strptime(scan(,""),"%Y-%m-%dT%H");i=function(x)as.integer(format(b,x));h=i("%H");d=i("%m%d");w=d>1100|d<430;f=ifelse;cat(c("Off","Mid","On")[f(i("%u")%in%5:6|h<7|h>19,1,f(h>11&h<17,f(w,3,2),f(w,2,3)))])
1: 2014-12-10T15
2: 
Read 1 item
Mid
> b=strptime(scan(,""),"%Y-%m-%dT%H");i=function(x)as.integer(format(b,x));h=i("%H");d=i("%m%d");w=d>1100|d<430;f=ifelse;cat(c("Off","Mid","On")[f(i("%u")%in%5:6|h<7|h>19,1,f(h>11&h<17,f(w,3,2),f(w,2,3)))])
1: 2014-08-26T23
2: 
Read 1 item
Off
```
[Answer]
## Bash, 286
this is a simple bash answer using the date program
```
d(){ date -d $1 +%$2; };D=$(echo $1|sed 's/\(T..\)$/\1:00/');H=$(d $D H);M=$(d $D m);if [ $(d $D u) -gt 5 ]||[ $H -lt 7 ]||[ $H -gt 18 ];then echo Off;exit;fi;if [ $M -gt 4 ]&&[ $M -lt 11 ];then I=On;O=Mid;else I=Mid;O=On;fi;if [ $H -gt 10 ]&&[ $H -lt 17 ];then echo $I;else echo $O;fi
```
[Answer]
Here goes another one!
# JavaScript, ~~175~~ 171
```
function f(x){d=new Date(x.slice(0,10));t=x.slice(11,13),m=(x.slice(5,7)+1)%12;return(t<8||t>18||!(d.getUTCDay()%6)?'off':((t<11||t>17)?(m<5?'on':'mid'):(m<5?'mid':'on'))}
```
Unminified:
```
function f(x) {
  d = new Date(x.slice(0, 10));
  t = x.slice(11, 13), m = (x.slice(5, 7) + 1) % 12;
  return (t < 8 || t > 18 || !(d.getUTCDay() % 6) ? 'off' : ((t < 11 || t > 17) ? (m < 5 ? 'on' : 'mid') : (m < 5 ? 'mid' : 'on'))
}
```
Only works on interpreters where an ISO8601 date string can be passed into the [`Date`](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Date) constructor.
# CoffeeScript, ~~192~~ 189
Surprisingly, it's longer in CoffeeScript because there's no ternary operator in that language (which as you can see from my JavaScript, I heavily relied on).
```
f=(x)->d=new Date(x.slice(0,10));t=x.slice(11,13);m=(x.slice(5,7)+1)%12;return'off'if(t<8||t>18||!(d.getUTCDay()%6));if(t<11||t>17)then(if m<5then'on'else'mid')else(if m<5then'mid'else'on')
```
[Answer]
# ES6 - 146
This is in function form, uses a couple nasty hacks.
```
let y,m,d,h,c=s=>[y,m,d,h]=s.split(/[T:-]/g).map(i=>+i),new Date(y,m,d).getDay()in{0:1,6:1}||h<20||8>h?'Off':['Mid','On'][(10>h&&h>16)^(m>5||m<8)]
```
Explained:
```
// These variables are declared outside of the function scope to save
// characters.
let y, // year
    m, // month
    d, // day
    h, // hour
    c = s => // c for check, called as c(isoString)
      [y, m, d, h] = // Assign to fields
        s.split(/[T:-]/g) // Split at delimiters
         .map(i => +i), // Convert all to numbers
                        // Comma used to keep it as a single statement.
      new Date(y, m, d).getDay() // Create Date to get day of week
        in {0:1, 6:1} // Check if it is good (0 = Sunday, 6 = Saturday). It
                      // is shorter to use an object literal than to
                      // do individual checks.
      ||
      h < 20 || 8 > h ? // h is between 7 and 19 (7pm)
       'Off' : // Return 'Off'
       ['Mid','On'][ // Two other outputs
        (10 > h && h > 16) ^ // Check if it is 11-16 (5pm)
        (m > 5 || m < 8)] // Invert that if the month is in summer/fall. Note
                          // that this returns a number, either 1 or 0. This
                          // is an ugly hack using the bitwise xor operator.
```
[Answer]
# Python 3 - 352 chars
```
import datetime as dt
t=input()
i=int
p=print
a='on'
b='mid'
c='off'
m=i(t[5:7])
h=i(t[11:13])
d=dt.date(i(t[0:4]),m,i(t[8:10])).weekday()
if 5==d or 6==d:p(c)
elif h>=19 and h<7:p(c)
elif m<=10 and m>=4:
 if h>=7 and h<11:p(b)
 if h>=11 and h<17:p(a)
 if h>=17 and h<19:p(b)
else:
 if h>=7 and h<11:p(a)
 if h>=11 and h<17:p(b)
 if h>=17 and h<19:p(a)
```
[Answer]
## Java - ~~426~~ 309 / 301? (see comments)
```
String z(String a){
    DateParser b=new DateParser(a);
    boolean c=b.parseEcmaDate();
    Integer[]d=b.getDateFields();
    GregorianCalendar e=new GregorianCalendar(d[0],d[1]-(c?0:1),d[2]);
    e.setTimeZone(new SimpleTimeZone(0,""));
    return(124>>e.get(7)&1)>0&d[3]>6&d[3]<19?
           d[3]>10&d[3]<17^(1008>>e.get(2)&1)>0?"Mid":"On":"Off";
}
```
Example output:
```
2014-03-02T00   Off
2014-03-02T06   Off
2014-03-02T07   Off
2014-03-02T10   Off
2014-03-02T11   Off
2014-03-02T16   Off
2014-03-02T17   Off
2014-03-02T18   Off
2014-03-02T19   Off
2014-03-02T23   Off
2014-04-02T00   Off
2014-04-02T06   Off
2014-04-02T07   On
2014-04-02T10   On
2014-04-02T11   Mid
2014-04-02T16   Mid
2014-04-02T17   On
2014-04-02T18   On
2014-04-02T19   Off
2014-04-02T23   Off
2014-05-02T00   Off
2014-05-02T06   Off
2014-05-02T07   Mid
2014-05-02T10   Mid
2014-05-02T11   On
2014-05-02T16   On
2014-05-02T17   Mid
2014-05-02T18   Mid
2014-05-02T19   Off
2014-05-02T23   Off
```
I used the same EXOR trick as the Python submission. ~~I also used a `+` as an OR function, for when it's `weekend OR night`.~~
My other big trick: bit masks.
For example, to see if a number is between 2 and 6 (Monday to Friday), first create a bit pattern where the interesting values are 1:
```
  6   2 
0b1111100 = 124
```
Then, use bit shift to get the interesting bit to the LSB and extract it:
```
(124 >> day_of_week) & 1
```
Similarly, I made bit patterns for months and hours:
```
  9    4
0b1111110000 = 1008
      76          76
0b111110000000000001111111 = 16253055
0b000000011111100000000000 = 129024
```
Unfortunately, it turns out simply `x>y&x<z` is shorter in most cases, so I didn't use it in some places.
And finally, a bit of hackery (highly implementation dependent) with `jdk.nashorn.internal.parser.DateParser`:
When `parseEcmaDate()` doesn't completely parse a date (like when it reads hour and hits end of string), it returns false.
* When it finishes properly, the date is stored in an `Integer[]` (auto-unboxing ftw), with the month fixed to be base-0 (like other Java classes).
* When it returns false, it aborted half-way and didn't do this fixing. It however still put whatever it parsed into the array, which is readily available. Hence the `-(c?0:1)`.
[Answer]
Nothing stops me from entering my own, and there's other shorter ones here anyways, so…
# PHP 5.4+, 194
```
<?function f($x){$t=substr($x,11,2);$m=(substr($x,5,2)+1)%12;if($t<8||$t>18||(new DateTime(substr($x,0,10)))->format('N')>5)return'off';return($t<11||$t>17)?($m<5?'on':'mid'):($m<5?'mid':'on');}
```
Unminified and commented:
```
<?
function f($x) {
  $t = substr($x,11,2); // hour
  $m = (substr($x,5,2) + 1) % 12; // month shifted up by 1
  if ($t < 8 || $t > 18 || (new DateTime(substr($x,0,10)))->format('N') > 5)
    return 'off'; // evenings and weekends
  return ($t < 11 || $t > 17)
    ? ($m < 5 ? 'on' : 'mid') // morning and mid-afternoon
    : ($m < 5 ? 'mid' : 'on'); // afternoon
}
```
Also note that the `date.timezone` directive in php.ini must be set, otherwise an Exception will be thrown.
]