Create a solar system

Question

Intro

This is based on an actual problem I recently faced while making a computer game and I thought it would make for a nice round of code-golf.

There are seven main spectral classes of star which put out varying amounts of heat. The geology of planets around a star are greatly influenced by the amount of heat received from the star, which is a factor of spectral class and distance from the star. Hence Mercury is practically molten, Neptune frozen.

The galaxy in my game is procedurally generated and randomly selecting planet types for given stars turned out to be a real 'if statement hell'!

The challenge

Your method should select one planet from a list of planet types appropriate for the class of star, based on a minimum heat threshold, a maximum heat threshold and a random number. For simplicity this challenge will only use a class G star, just like our sun.

Inputs

An integer heat in the range 4 to 11 representing the amount of heat received by the planet from the star.

Variables

This table shows the possible planets based on heat. Your method should first narrow the available choices based on the heat min and heat max, heat should fall on or between the two. E.g. with a heat of 10 passed in the only choices would be Desert, Iron and Lava.

Planet type    Heat min   Heat max   Random Chance
Gas Giant         4          9            15
Ice               4          6            10
Ice Giant         4          6            10
Gaia class        5          7            10
Dense Atmosphere  7          9            10
Desert            7          10           25
Iron              7          10           14
Lava             10          11           6

Next, the probability of a planet (in the remaining choices) being chosen is its random chances divided by the sum of the random chances of all the choices.

In the above example, the probability of Iron being chosen is 14/(25+14+6).

Output

Return the planet type as a string.

Do the best you can to avoid logic arrowheads. Shortest code wins, points all round for creativity. Happy golfing!

Answer 1

Jelly,  78  75 bytes

“'ĖøÆḳƙ’ḃ7ṣ6Ä+3r/ċ€×“½½½½©ÐÇı‘
“ŀỊẋ8ƒ³ẈRɼƈñqẋẏȧɱḌ<ṄỴḳ⁾ÆʋeẒĊ'@ƬØƓƝ}ḟ¬»Ỵx'ÇX

A monadic link accepting an integer (in [4,11]) which returns a list of characters.

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How?

Creates the heat ranges of the planets as a list of lists and counts the occurrences of the input heat in those lists to get a list of zeros and ones representing which planet types are possible, then multiplies by the likelihood numbers of the eight planet types to get the distribution. The distribution is used to repeat the planet type names, and finally, a uniform random choice is made.

“'ĖøÆḳƙ’ḃ7ṣ6+\+3r/ċ€×“½½½½©ÐÇı‘ - Link 1, getDistribution: integer
“'ĖøÆḳƙ’                        - base 250 integer = 39824688429662
        ḃ7                      - to bijective-base 7 = [1,1,2,4,7,1,4,4,6,2,2,2,2,1,5,3,3]
          ṣ6                    - split at sixes = [[1,1,2,4,7,1,4,4],[2,2,2,2,1,5,3,3]]
            +\                  - reduce by addition = [[1,1,2,4,7,1,4,4],[3,3,4,6,8,6,7,7]]
              +3                - add three = [[4,4,5,7,10,4,7,7],[6,6,7,9,11,9,10,10]]
                r/              - reduce by inclusive range = [[4,5,6],[4,5,6],[5,6,7],[7,8,9],[10,11],[4,5,6,7,8,9],[7,8,9,10],[7,8,9,10]]
                  ċ€            - count (input) in €ach e.g. for 5: [1, 1, 1, 0,0, 1, 0, 0]
                     “½½½½©ÐÇı‘ - list of code-page indices        [10,10,10,10,6,15,14,25]
                    ×           - multiply                         [10,10,10, 0,0,15, 0, 0]

“ ... »Ỵx'ÇX - Main link: integer
“ ... »      - compressed string = "Ice\nIce Giant\nGaia class\nDense Atmosphere\nLava\nGas Giant\nIron\nDesert"
       Ỵ     - split at new lines = ["Ice","Ice Giant","Gaia class","Dense Atmosphere","Lava","Gas Giant","Iron","Desert"]
          Ç  - call last link (1) as a monad e.g. for 5: [10,10,10,0,0,15,0,0]
         '   - spawn:
        x    -   times e.g. for 5: ["Ice","Ice","Ice","Ice","Ice","Ice","Ice","Ice","Ice","Ice","Ice Giant","Ice Giant","Ice Giant","Ice Giant","Ice Giant","Ice Giant","Ice Giant","Ice Giant","Ice Giant","Ice Giant"]
           X - a random choice from that list

Answer 2

R, 225 223 183 bytes

Thanks to Giuseppe for clever refactoring to take it down to 188 bytes; the remaining five were shaved off by using less redundant number representations.

i=scan()-4
sample(c("Gas Giant","Ice","Ice Giant","Gaia class","Dense Atmosphere","Desert","Iron","Lava")[l<-c(0,0,0,1,3,3,3,6)<=i&c(5,2,2,3,5,6,6,7)>=i],1,,c(3,2,2,2,2,5,2.8,1.2)[l])

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Answer 3

JavaScript 212

Edit 6 bytes save thx Jonathan Allan

h=>[963,640,640,649,667,1628,924,437].map((z,i)=>(z/8&7)+4>h|z%8+6<h?0:t=r.push(...Array(z>>6).fill(i)),r=[])&&"Gas Giant,Ice,Ice Giant,Gaia class,Dense Atmosphere,Desert,Iron,Lava".split`,`[r[t*Math.random()|0]]

less golfed

h=>( 
   r = [],
   // heat min,max and chance encoded in base 8 with offsets
   // min range 4 to 10, with offset 4, 0 to 6
   // max range 6 to 11, with offset 6, 0 to 5
   [(4-4)*8 + 9-6 + 15*64,
    (4-4)*8 + 6-6 + 10*64,
    (4-4)*8 + 6-6 + 10*64,
    (5-4)*8 + 7-6 + 10*64,
    (7-4)*8 + 9-6 + 10*64,
    (7-4)*8 + 10-6+ 25*64,
    (7-4)*8 + 10-6+ 14*64,
    (10-4)*8+ 11-6+  6*64]
   .forEach( (z,i) => (
      min = (z / 8 & 7) + 4, 
      max = z % 8 + 6,
      chance = z >> 6,
      min > h || max < h 
      ? 0 // out of range
      // add current position i repeated 'chance' times
      // array size in t
      : t = r.push(...Array(chance).fill(i))
   ),
   pos = r[t * Math.random() | 0],
   ["Gas Giant", "Ice", "Ice Giant", "Gaia class", "Dense Atmosphere", "Desert", "Iron", "Lava"][pos]
)

Test

var F=
h=>[963,640,640,649,667,1628,924,437].map((z,i)=>(z/8&7)+4>h|z%8+6<h?0:t=r.push(...Array(z>>6).fill(i)),r=[])&&"Gas Giant,Ice,Ice Giant,Gaia class,Dense Atmosphere,Desert,Iron,Lava".split`,`[r[t*Math.random()|0]]

function test()
{
   var heat=+H.value
   var i,result,hashtable={},rep=1e5
   for (i=0;i<rep;i++)
     result = F(heat),
     hashtable[result] = -~hashtable[result]
 
   console.log('Input',heat)
   for (i in hashtable)
   {
     console.log(i,(hashtable[i]/rep*100).toFixed(2),'%')
   }
}

<input id=H type=number min=1 max =15 value=10>
<button onclick='test()'>Test</button>

Answer 4

Coconut, 214 195 bytes

t->choice..sum([[n]*g(p)*(g(a)<t<g(b))for*n,a,b,p in'Gas Giant3AF_Ice37A_Ice Giant37A_Gaia class48A_Dense Atmosphere6AA_Desert6BP_Iron6BE_Lava9C6'.split('_')],[])
from random import*
g=int$(?,36)

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A Python port would be 203 200 bytes long:

lambda t:choice(sum([[n]*int(p,36)*(int(a)<t<int(b,36))for*n,a,b,p in'Gas Giant3AF_Ice37A_Ice Giant37A_Gaia class48A_Dense Atmosphere6AA_Desert6BP_Iron6BE_Lava9C6'.split('_')],[]))
from random import*

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Answer 5

Charcoal, 115 111 bytes

≔Ｉ⁻Ｎ³θＦ⁸«≔§⪪”↷＆∧⬤.YLφκ¦(⁼；σ≕]✂↙ζＣ” ιη¿›θη¿‹θ§η¹ＦＩ✂η²⊞υ黧⪪”↓(″1↨▷]Ｕ,&ζ^iＩ″ＲＳＹ≡´⍘'#﹪υＶw5Ｖu>D<U5r6⁰Ｑ▷Ｚ◨⌕⁸ΣεCＺ”¶‽υ

Try it online! Link is to verbose version of code. Edit: Saved 4 bytes thanks to @ASCII-only. Explanation:

≔Ｉ⁻Ｎ³θ

Subtract 3 from the input so that it can be compared against single digits.

Ｆ⁸«≔§⪪”↷＆∧⬤.YLφκ¦(⁼；σ≕]✂↙ζＣ” ιη

Split the string 0715 0410 0410 1510 3710 3825 3814 696 on spaces (spaces seem to compress better than commas but I didn't try any other characters) and loop over each portion.

¿›θη¿‹θ§η¹ＦＩ✂η²⊞υι»

Compare the input against the first and second digits and if it's between then push the loop index the given number of times to the predefined empty list, thus populating it.

§⪪”↓(″1↨▷]Ｕ,&ζ^iＩ″ＲＳＹ≡´⍘'#﹪υＶw5Ｖu>D<U5r6⁰Ｑ▷Ｚ◨⌕⁸ΣεCＺ”¶‽υ

Split the list of planets on newlines (again, better than commas for some reason) and select the element corresponding to a index chosen randomly from the list.

Answer 6

R, 196 193 190 175 171 bytes

sample(readLines(,8),1,,c(3,2,2,2,2,5,2.8,1.2)*((x=scan()-3)>c(0,0,0,1,3,3,3,6)&x<c(7,4,4,5,7,8,8,9)))
Gas Giant
Ice
Ice Giant
Gaia class
Dense Atmosphere
Desert
Iron
Lava

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Initially inspired by this solution by @rturnbull, however as both submissions have significantly evolved, this is now essentially a mix of ideas of the original author, @Giuseppe who has been very helpful in comments, and mine. Here is a summary of key points that helped to bring the byte count down:

• Encoding planet data as CSV Collecting names with readLines to avoid the large number of quotation characters around strings.

• Tweaking the heat params so that we could use < and > signs instead of <= and >=.

• Changing the heat data format from Heat min, Heat max to Heat min, Heat Delta to get rid of double digit numbers.
  Replaced by shifting all numbers by -3

• Dividing all planet probabilities by 5 which also results in a few less digits.

• Multiplying the vector of planet probabilities with the vector of Booleans (indicating whether our input satisfies the heat requirements) to nullify the probabilities of unsuitable planets.

Probably, a few more bytes could be gained by applying some sort of data compression.
I think, not anymore.

Answer 7

Octave with Statistics Package, 178 176 174 158 bytes

@(h)randsample(strsplit('Gas Giant,Ice,Ice Giant,Gaia class,Dense Atmosphere,Desert,Iron,Lava',','),1,1,('UPPPP_TL'-70).*(h>'IIIJLLLO'-70&h<'PMMNPQQR'-70)){1}

The code defines an anonymous function that inputs a number and outputs a string.

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Explanation

The code

@(h)

defines an anonymous function with input h.

The string

'Gas Giant,Ice,Ice Giant,Gaia class,Dense Atmosphere,Desert,Iron,Lava'

is split at commas using

strsplit(...,',')

The result is a cell array of strings, where each string is a planet class.

The code

'IIIJLLLO'-70

defines the shown string and subtracts 70 from the code points of its chars. This gives the array of minimum heat values minus 1, that is, [3 3 3 4 6 6 6 9].

Similarly,

'PMMNPQQR'-70

produces the array of maximum heat values plus 1, that is, [10 7 7 8 10 11 11 12].

The comparisons

h>...&h<...

give an array containing true or false indicating which planet classes are possible.

On the other hand,

'UPPPP_TL'-70

defines the array of random chance values, [15 10 10 10 10 25 14 6].

The operation

(...).*(...)

is the element-wise multiplication of the latter two arrays (true and false behave like 0 and 1 respectively). This gives an array where each planet class has either its random chance, or 0 if that class is not possible based on the input. This array will be used as weights in the random sampling

The function call

randsample(...,1,1,...)

selects one of the cells from the cell array of strings (first input argument), using the computed array of weights (fourth input argument). Specifically, the function randsample automatically normalizes the weights to probabilities, and then does the random selection with those probabilities. The result is a cell array containing a string. The code

{1}

is used to extract that string, which constitutes the function output.

Answer 8

Python 3, 263 bytes

from random import*
P=lambda h:"Gas Giant|Ice|Ice Giant|Gaia class|Dense Atmosphere|Desert|Iron|Lava".split("|")[choices(*list(zip(*filter(lambda x:h in range(*x[2:]),zip(*[[int(x,32)for x in"0f4a1a472a473a584a7a5p7b6e7b76ac"][a::4]for a in(0,1,2,3)]))))[:2])[0]]

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Answer 9

Python 3, 199 194 bytes

from random import*
lambda n:choices("Ice|Ice Giant|Gas Giant|Gaia class|Dense Atmosphere|Desert|Iron|Lava".split('|'),[(0x33b2a53d4a>>5*i&31)*(0xc07878380e3f0707>>8*i+n-4&1)for i in range(8)])

Splitting h into separate bit masks and random chance values (see explanation) saves a few bytes by eliminating an assignment to h and simplifying the range() in the list comprehension.

Previous solution

from random import*
h=0xc033c39e3270a0e51fbc1d40ea
lambda n:choices("Ice|Ice Giant|Gas Giant|Gaia class|Dense Atmosphere|Desert|Iron|Lava".split('|'),[(h>>i&31)*(h>>i+n+1&1)for i in range(0,104,13)])

Defines an anonymous function that takes an int and returns the planet type.

For each planet type, a 13-bit value was calculated. The top 8 bits define a bit mask of valid heat values for that planet type. The bottom 5 bits are the random chance for that planet type. For example, "Gaia class" is a valid type for heat values 4 to 7, so it has a mask of 0b00001111. It has a random chance of 10, or 0b01010. Combining them results it the 13-bit value 0b0000111101010 for the "Gaia class" type. The 13-bit values for each planet type are concatenated to get the value for h (the lowest 13 bits are for the "Ice" planet type). (The newer answer doesn't combine these values).

The list comprehension iterates over the 13-bit values to create a list of weights, where the weight is the random chance if the planet type is a valid choice for the given heat value, and zero otherwise. For each planet type, (h>>i&31) extracts the random chance for that planet type. (h>>i+n+1&1) evaluates to 1 if the planet type is a valid choice for the heat value n and evaluates to 0 otherwise.

The library function random.choices(choices, weights) selects an item from the list of choices based on the list of weights.

Answer 10

Python, 282 Bytes, 261 Bytes:

from random import*
i,p,l=input(),[('Gas Giant',3,11,15),("Ice",3,7,10),("Ice Giant",3,7,10),("Gaia Class",4,8,10),("Dense Atmosphere",6,10,10),("Desert",6,11,25),("Iron",6,11,14),("Lava",9,12,6)],[]
for x in p:exec"l+=x[0],;"*(x[1]<i<x[2])*x[3]
print choice(l)

Pretty simple - fairly sure it could be golfed more - Still looking for a better way to represent the planet range and probability data. If i is in range of the planet type, appends it to the list according to the probability, then randomly prints one.

EDIT: With credit to Jonathan Frech - redid the for loop to knock a few bytes off. Better way of appending items to the list

Answer 11

Perl 5 (-p), 230 bytes

@a=(['Gas Giant',4,9,15],[Ice,4,6,10],['Ice Giant',4,6,10],['Gaia class',5,7,10],['Dense Atmosphere',7,9,10],[Desert,7,10,25],[Iron,7,10,14],[Lava,10,11,6]);//;map{push@b,($$_[0])x($$_[3]*($$_[1]<=$'&&$'<=$$_[2]))}@a;$_=$b[rand@b]

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Answer 12

Nim, 314 298 294 bytes

import random,sequtils
proc c(h:int)=
 var a= @[""]
 a.del 0
 for n in[("Gas Giant",4,9,15),("Ice",4,6,10),("Ice Giant",4,6,10),("Gaia Class",5,7,10),("Dense Atmosphere",7,9,10),("Desert",7,10,25),("Iron",7,10,14),("Lava",10,11,6)]:(if h>=n[1]and h<=n[2]:a.add repeat(n[0],n[3]))
 echo random a

For loop now in one line, no return, less bytes to implicit type

4 spaces removed (thanks Kevin)

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Answer 13

05AB1E, 78 76 bytes

”Œï²°™Ä²° Gaia classêη•™Äµ‰Ÿ± Lava”#8äðýā<•ŒEŽuS,•2ôו9èÁnÇ∞Λ•SÌ2ôεŸIå}ÏSΩè

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Explanation

”Œï²°™Ä²° Gaia classêη•™Äµ‰Ÿ± Lava”
pushes the string Gas Giant Ice Giant Gaia class Dense Atmosphere Ice Desert Iron Lava

#                                          # split on spaces
 8ä                                        # divide into 8 parts
   ðý                                      # join each by spaces
     ā<                                    # push the range [0 ... 7]
       •ŒEŽuS,•                            # push 151010101025146
               2ô                          # split into pieces of 2
                                           # results in [15, 10, 10, 10, 10, 25, 14, 6]
                 ×                         # repeat each number in the range by these amounts
                                           # results in ['000000000000000', '1111111111', '2222222222', '3333333333', '4444444444', '5555555555555555555555555', '66666666666666', '777777']
                  •9èÁnÇ∞Λ•                # push 2724355724585889
                           S               # split to list of digits
                            Ì              # decrement each twice
                                           # results in [4,9,4,6,5,7,7,9,4,6,7,10,7,10,10,11]
                             2ô            # split into pieces of 2
                                           # results in [[4, 9], [4, 6], [5, 7], [7, 9], [4, 6], [7, 10], [7, 10], [10, 11]]
                               εŸIå}       # apply to each pair
                                Ÿ          # range [a ... b]
                                 Iå        # check if input is contained in the range
                                           # ex, for input 10: [0, 0, 0, 0, 0, 1, 1, 1]
                                    Ï      # keep only the indices which are true
                                           # ex, for input 10: ['5555555555555555555555555', '66666666666666', '777777']
                                     S     # split to list of digits
                                      Ω    # pick one at random
                                       è   # index into the list of strings with this

Answer 14

Ruby, 214 193 189 bytes

->h{'Gas Giant,Desert,Iron,Lava,Ice,Ice Giant,Gaia class,Dense Atmosphere'.split(?,).zip(31006330.digits,75449887.digits,[15,25,14,6]).flat_map{|n,m,x,r|m<h-3&&x>h-3?[n]*(r||10):[]}.sample}

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Answer 15

Haskell, 377 364 358 318 312 270 265 262 256 251 bytes

import System.Random
f h|x<-[n|(n,(a,b,c))<-zip(lines"Gas Giant\nIce\nIce Giant\nGaia class\nDense Atmosphere\n
Desert\nIron\nLava")$zip3[4,4,4,5,7,7,7,10][9,6,6,7,9,10,10,11][15,10,10,10,10,25,14,6],h<=
b,h>=a,_<-[1..c]]=(x!!)<$>randomRIO(0,length x-1)

(I've added linebreaks for nicer printout). The task says "return", not "print", so f is a function which returns the randomly selected planet name into the IO monad, f :: Int -> IO String.

The main is main = do {f 10 >>= print} (Haskell golfing tips says it doesn't count). Prints

"Iron"     -- or "Desert", or "Lava"

(edits: removed &'s base case; moved main out; changed to quadruples and unzip, and switched to pattern guards and >>= following suggestions from Laikoni, thanks!; implemented the approach from the Jelly solution instead, repeating the names; explicit type is no longer needed; another advice by Laikoni saves 3 more bytes; made it an IO function; implemented advice from the chat room).

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Answer 16

Java 8, 398 384 bytes

n->{String r="",a[];for(String x:"456789~Gas Giant~15;456~Ice~10;456~Ice Giant~10;567~Gaia class~10;789~Dense Atmosphere~10;78910~Desert~25;78910~Iron~14;1011~Lava~6".split(";"))if(x.split("~")[0].contains(n))r+=x+";";long t=0,u=0;for(String x:(a=r.split(";")))t+=new Long(x.split("~")[2]);t*=Math.random();for(String x:a)if((u+=new Long((a=x.split("~"))[2]))>t)return a[1];return"";}

It can definitely be golfed some more, but the probability in combination with the Strings isn't very easy in Java.

Explanation:

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n->{                // Method with String as both parameter and return-type
  String r="",      //  Temp-String, starting empty
         a[];       //  Temp String-array
  for(String x:"456789~Gas Giant~15;456~Ice~10;456~Ice Giant~10;567~Gaia class~10;789~Dense Atmosphere~10;78910~Desert~25;78910~Iron~14;1011~Lava~6".split(";"))
                    //  Loop over the String-parts in the format "heats~type~probability"
    if(x.split("~")[0].contains(n))
                    //   If the heats contains the input
      r+=x+";";     //    Append this entire String-part to the temp-String `r`
  long t=0,u=0;     //  Temp numbers, both starting empty
  for(String x:(a=r.split(";")))
                    //  Loop over the temp-String parts:
    t+=new Long(x.split("~")[2]);
                    //   Sum their probabilities
  t*=Math.random(); //  Get a random number in the range [0,sum_of_probabilities)
  for(String x:a)   //  Loop over the temp-String parts again
    if((u+=new Long((a=x.split("~"))[2]))>t)
                    //   The moment the current probability-sum is > the random number
      return a[1];  //    Return the Type of planet
  return"";}        //  Mandatory return we won't encounter (which returns nothing)

Answer 17

Min, 280 277 bytes

:a ' =b (("Gas Giant" 4 9 15) ("Ice" 4 6 10) ("Ice Giant" 4 6 10) ("Gaia Class" 5 7 10) ("Dense Atmosphere" 7 9 10) ("Desert" 7 10 25) ("Iron" 7 10 14) ("Lava" 10 11 6)) (=n (a n 1 get >= a n 2 get <= and) ((n 0 get b append #b) n 3 get times) when) foreach b b size random get

Starts with heat on the stack, leaves a string on the stack. Same general process as Python 2 answer.

Explanation

Note that min is concatenative

:a ' =b                               ;Set the value on the stack (heat) to a, set empty quot to b
(("Gas Giant" 4 9 15) ("Ice" 4 6 10) ("Ice Giant" 4 6 10) ("Gaia Class" 5 7 10) ("Dense Atmosphere" 7 9 10) ("Desert" 7 10 25) ("Iron" 7 10 14) ("Lava" 10 11 6)) ;Data to be iterated over
(=n                                   ;  set n to current item
 (a n 1 get >= a n 2 get <= and)      ;    check if n is between the min (2nd elment of n) and max (3rd element of n) heat
 (
  (n 0 get b append #b) n 3 get times ;      insert the name(1st element of n) into the quot of names (b) a number of times corresponding to the 4th element of n
 ) when                               ;    when the previous check is true
) foreach                             ;  for every quot in previous data
b b size random get                   ;choose a random element from the list of names

Answer 18

PowerShell, 56 + 135 (CSV file) + 1 (file name) = 192 bytes

param($z)ipcsv a|?{$z-in$_.m..$_.x}|%{,$_.p*$_.r}|Random

Try it online! _{(this is a slightly modified version that creates the temporary CSV file described below)}

Imports a CSV file using ipcsv (short for Import-CSV) named a in the local directory that contains the following:

P,m,x,r
Gas Giant,4,9,15
Ice,4,6,10
Ice Giant,4,6,10
Gaia class,5,7,10
Dense Atmosphere,7,9,10
Desert,7,10,25
Iron,7,10,14
Lava,10,11,6

That automatically creates an iterable hashtable of things like the following:

@{P=Gas Giant; m=4; x=9; r=15}
@{P=Ice; m=4; x=6; r=10}
...

We then use Where-Object (?) to pull out those entries where our input integer $z is -in the range $_.m to $_.x (i.e., it's in the heat range). We then pump those into a Foreach-Object loop (%) that creates an array of strings of names based on the random chance of those names. For example, this will create an array of 15 "Gas Giant" strings if that heat matches. We then put those into Get-Random which will pull out the appropriate string with the appropriate weighting.