Randomly distribute numbers to lists

up vote
2
down vote

favorite

I have 4 empty lists. I would like to randomly distribute numbers from 1 to 10 to these 4 empty lists. How can I add integers to lists randomly?

For example:

m = ;
n = ;
p = ;
r = ;
listz = Range[10];

1->m , 2->n , 3->p , 4->m , 5->r , 6->r, 7->n , 8->m, 9->p , 10->n (Randomly distributed)

Here is the code I tried:

m = ;
n = ;
p = ;
r = ;
listz = Range[10];
s = RandomChoice[m, n, p, r, 4];
Do[listm = AppendTo[s, listz[i]];, i, 1, 10]
listm

Wanted Solution:(Randomly distributed)

m = 1, 5, 8
n = 4, 6, 9
p = 7, 10
r = 2, 3

asked 2 hours ago

ithilquessirr

283

New contributor

Why not just generate four random lists in the first place?
â€“Â gothicVI
2 hours ago

add a commentÂ |Â

up vote
2
down vote

favorite

I have 4 empty lists. I would like to randomly distribute numbers from 1 to 10 to these 4 empty lists. How can I add integers to lists randomly?

For example:

m = ;
n = ;
p = ;
r = ;
listz = Range[10];

1->m , 2->n , 3->p , 4->m , 5->r , 6->r, 7->n , 8->m, 9->p , 10->n (Randomly distributed)

Here is the code I tried:

m = ;
n = ;
p = ;
r = ;
listz = Range[10];
s = RandomChoice[m, n, p, r, 4];
Do[listm = AppendTo[s, listz[i]];, i, 1, 10]
listm

Wanted Solution:(Randomly distributed)

m = 1, 5, 8
n = 4, 6, 9
p = 7, 10
r = 2, 3

asked 2 hours ago

ithilquessirr

283

New contributor

Why not just generate four random lists in the first place?
â€“Â gothicVI
2 hours ago

add a commentÂ |Â

up vote
2
down vote

favorite

I have 4 empty lists. I would like to randomly distribute numbers from 1 to 10 to these 4 empty lists. How can I add integers to lists randomly?

For example:

m = ;
n = ;
p = ;
r = ;
listz = Range[10];

1->m , 2->n , 3->p , 4->m , 5->r , 6->r, 7->n , 8->m, 9->p , 10->n (Randomly distributed)

Here is the code I tried:

m = ;
n = ;
p = ;
r = ;
listz = Range[10];
s = RandomChoice[m, n, p, r, 4];
Do[listm = AppendTo[s, listz[i]];, i, 1, 10]
listm

Wanted Solution:(Randomly distributed)

m = 1, 5, 8
n = 4, 6, 9
p = 7, 10
r = 2, 3

asked 2 hours ago

ithilquessirr

283

New contributor

I have 4 empty lists. I would like to randomly distribute numbers from 1 to 10 to these 4 empty lists. How can I add integers to lists randomly?

For example:

m = ;
n = ;
p = ;
r = ;
listz = Range[10];

1->m , 2->n , 3->p , 4->m , 5->r , 6->r, 7->n , 8->m, 9->p , 10->n (Randomly distributed)

Here is the code I tried:

m = ;
n = ;
p = ;
r = ;
listz = Range[10];
s = RandomChoice[m, n, p, r, 4];
Do[listm = AppendTo[s, listz[i]];, i, 1, 10]
listm

Wanted Solution:(Randomly distributed)

m = 1, 5, 8
n = 4, 6, 9
p = 7, 10
r = 2, 3

list-manipulation random education

asked 2 hours ago

ithilquessirr

283

New contributor

asked 2 hours ago

ithilquessirr

283

New contributor

asked 2 hours ago

ithilquessirr

283

New contributor

asked 2 hours ago

ithilquessirr

283

asked 2 hours ago

ithilquessirr

283

New contributor

ithilquessirr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Why not just generate four random lists in the first place?
â€“Â gothicVI
2 hours ago

add a commentÂ |Â

Why not just generate four random lists in the first place?
â€“Â gothicVI
2 hours ago

Why not just generate four random lists in the first place?
â€“Â gothicVI
2 hours ago

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
2
down vote

on Mathematica 11.3

(R=RandomSample)[Range@10,10]~TakeList~R@RandomChoice[Select[IntegerPartitions@10,Length@#==4&]]

Also I tried to fix your code

a = ;
b = ;
c = ;
d = ;
Listz = a, b, c, d;
Do[t=RandomChoice@Range@4;Listz[[t]]=AppendTo[Listz[[t]],i],i,1,10]
Listz

edited 59 mins ago

answered 1 hour ago

J42161217

2,447218

Have you tried that with a list of 100 elements?
â€“Â Henrik Schumacher
1 hour ago

how about 10^100?
â€“Â J42161217
1 hour ago

@J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
â€“Â Okkes Dulgerci
1 hour ago

@OkkesDulgerci fixed!
â€“Â J42161217
1 hour ago

add a commentÂ |Â

up vote
1
down vote

n = 100;
m = 4;
list = Range[n];
s = RandomChoice[Range[m], n];

SparseArray[Transpose[s, list] -> _, m, n]["AdjacencyLists"]

Or slower, but easier to understand:

Lookup[GroupBy[Transpose[s, list], First -> Last], Range[m]]

edited 1 hour ago

answered 2 hours ago

Henrik Schumacher

43.9k263129

Is this supposed to spew errors in 11.3?
â€“Â user6014
2 hours ago

Ah, sorry. Please try again.
â€“Â Henrik Schumacher
2 hours ago

Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
â€“Â ithilquessirr
2 hours ago

Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
â€“Â Henrik Schumacher
1 hour ago

add a commentÂ |Â

up vote
1
down vote

Edit: I realized that my previous solution doesn't produce the same length partitions. Here is the desired solution.

s = RandomSample[Range@10, 10];
t = RandomSample@RandomChoice@IntegerPartitions[10, 4]
m, n, p, r = TakeList[s, t]

(*or m, n, p, r = FoldPairList[TakeDrop, s, t]*)

1, 5, 10, 3, 7, 2, 9, 6, 8, 4

Original answer:

s = RandomSample[Range@10, 10];
t = RandomSample[Range@4, 4];
m, n, p, r = FoldPairList[TakeDrop, s, t]

6, 2, 9, 5, 1, 7, 10, 3, 8, 4

edited 1 hour ago

answered 2 hours ago

Okkes Dulgerci

3,4061716

add a commentÂ |Â

up vote
1
down vote

listz = CharacterRange["A", "K"];
k = 4;

SeedRandom[1]
rc = RandomChoice[Range @ k, Length @ listz];
m, n, p, r = Pick[listz, rc, #]& /@ Range[k];

"C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"

Also

Lookup[GroupBy[Transpose[rc, listz], First -> Last], #, ]& /@ Range[k]

"C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"

Update: To get k non-empty sublists:

SeedRandom[1]
rc = Module[r, While[Length[Counts[r = RandomChoice[Range@k, Length @ listz]]] < k]; r];
Pick[listz , rc, # ] & /@ Range[k ]

"A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"

Lookup[GroupBy[Transpose[rc , listz], First -> Last], #, ] & /@ Range[k]

"A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"

edited 8 mins ago

answered 57 mins ago

kglr

169k8192396

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
2
down vote

on Mathematica 11.3

(R=RandomSample)[Range@10,10]~TakeList~R@RandomChoice[Select[IntegerPartitions@10,Length@#==4&]]

Also I tried to fix your code

a = ;
b = ;
c = ;
d = ;
Listz = a, b, c, d;
Do[t=RandomChoice@Range@4;Listz[[t]]=AppendTo[Listz[[t]],i],i,1,10]
Listz

edited 59 mins ago

answered 1 hour ago

J42161217

2,447218

Have you tried that with a list of 100 elements?
â€“Â Henrik Schumacher
1 hour ago

how about 10^100?
â€“Â J42161217
1 hour ago

@J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
â€“Â Okkes Dulgerci
1 hour ago

@OkkesDulgerci fixed!
â€“Â J42161217
1 hour ago

add a commentÂ |Â

up vote
2
down vote

on Mathematica 11.3

(R=RandomSample)[Range@10,10]~TakeList~R@RandomChoice[Select[IntegerPartitions@10,Length@#==4&]]

Also I tried to fix your code

a = ;
b = ;
c = ;
d = ;
Listz = a, b, c, d;
Do[t=RandomChoice@Range@4;Listz[[t]]=AppendTo[Listz[[t]],i],i,1,10]
Listz

edited 59 mins ago

answered 1 hour ago

J42161217

2,447218

Have you tried that with a list of 100 elements?
â€“Â Henrik Schumacher
1 hour ago

how about 10^100?
â€“Â J42161217
1 hour ago

@J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
â€“Â Okkes Dulgerci
1 hour ago

@OkkesDulgerci fixed!
â€“Â J42161217
1 hour ago

add a commentÂ |Â

up vote
2
down vote

on Mathematica 11.3

(R=RandomSample)[Range@10,10]~TakeList~R@RandomChoice[Select[IntegerPartitions@10,Length@#==4&]]

Also I tried to fix your code

a = ;
b = ;
c = ;
d = ;
Listz = a, b, c, d;
Do[t=RandomChoice@Range@4;Listz[[t]]=AppendTo[Listz[[t]],i],i,1,10]
Listz

edited 59 mins ago

answered 1 hour ago

J42161217

2,447218

on Mathematica 11.3

(R=RandomSample)[Range@10,10]~TakeList~R@RandomChoice[Select[IntegerPartitions@10,Length@#==4&]]

Also I tried to fix your code

a = ;
b = ;
c = ;
d = ;
Listz = a, b, c, d;
Do[t=RandomChoice@Range@4;Listz[[t]]=AppendTo[Listz[[t]],i],i,1,10]
Listz

edited 59 mins ago

answered 1 hour ago

J42161217

2,447218

edited 59 mins ago

answered 1 hour ago

J42161217

2,447218

answered 1 hour ago

J42161217

2,447218

answered 1 hour ago

J42161217

2,447218

Have you tried that with a list of 100 elements?
â€“Â Henrik Schumacher
1 hour ago

how about 10^100?
â€“Â J42161217
1 hour ago

@J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
â€“Â Okkes Dulgerci
1 hour ago

@OkkesDulgerci fixed!
â€“Â J42161217
1 hour ago

add a commentÂ |Â

Have you tried that with a list of 100 elements?
â€“Â Henrik Schumacher
1 hour ago

how about 10^100?
â€“Â J42161217
1 hour ago

@J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
â€“Â Okkes Dulgerci
1 hour ago

@OkkesDulgerci fixed!
â€“Â J42161217
1 hour ago

Have you tried that with a list of 100 elements?
â€“Â Henrik Schumacher
1 hour ago

how about 10^100?
â€“Â J42161217
1 hour ago

@J42161217 Because Select[IntegerPartitions@10, Length@# == 4 &] sorted, you cannot get Lenght=1or 2 set for m.
â€“Â Okkes Dulgerci
1 hour ago

@OkkesDulgerci fixed!
â€“Â J42161217
1 hour ago

add a commentÂ |Â

up vote
1
down vote

n = 100;
m = 4;
list = Range[n];
s = RandomChoice[Range[m], n];

SparseArray[Transpose[s, list] -> _, m, n]["AdjacencyLists"]

Or slower, but easier to understand:

Lookup[GroupBy[Transpose[s, list], First -> Last], Range[m]]

edited 1 hour ago

answered 2 hours ago

Henrik Schumacher

43.9k263129

Is this supposed to spew errors in 11.3?
â€“Â user6014
2 hours ago

Ah, sorry. Please try again.
â€“Â Henrik Schumacher
2 hours ago

Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
â€“Â ithilquessirr
2 hours ago

Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
â€“Â Henrik Schumacher
1 hour ago

add a commentÂ |Â

up vote
1
down vote

n = 100;
m = 4;
list = Range[n];
s = RandomChoice[Range[m], n];

SparseArray[Transpose[s, list] -> _, m, n]["AdjacencyLists"]

Or slower, but easier to understand:

Lookup[GroupBy[Transpose[s, list], First -> Last], Range[m]]

edited 1 hour ago

answered 2 hours ago

Henrik Schumacher

43.9k263129

Is this supposed to spew errors in 11.3?
â€“Â user6014
2 hours ago

Ah, sorry. Please try again.
â€“Â Henrik Schumacher
2 hours ago

Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
â€“Â ithilquessirr
2 hours ago

Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
â€“Â Henrik Schumacher
1 hour ago

add a commentÂ |Â

up vote
1
down vote

n = 100;
m = 4;
list = Range[n];
s = RandomChoice[Range[m], n];

SparseArray[Transpose[s, list] -> _, m, n]["AdjacencyLists"]

Or slower, but easier to understand:

Lookup[GroupBy[Transpose[s, list], First -> Last], Range[m]]

edited 1 hour ago

answered 2 hours ago

Henrik Schumacher

43.9k263129

n = 100;
m = 4;
list = Range[n];
s = RandomChoice[Range[m], n];

SparseArray[Transpose[s, list] -> _, m, n]["AdjacencyLists"]

Or slower, but easier to understand:

Lookup[GroupBy[Transpose[s, list], First -> Last], Range[m]]

edited 1 hour ago

answered 2 hours ago

Henrik Schumacher

43.9k263129

edited 1 hour ago

answered 2 hours ago

Henrik Schumacher

43.9k263129

answered 2 hours ago

Henrik Schumacher

43.9k263129

answered 2 hours ago

Henrik Schumacher

43.9k263129

Is this supposed to spew errors in 11.3?
â€“Â user6014
2 hours ago

Ah, sorry. Please try again.
â€“Â Henrik Schumacher
2 hours ago

Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
â€“Â ithilquessirr
2 hours ago

Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
â€“Â Henrik Schumacher
1 hour ago

add a commentÂ |Â

Is this supposed to spew errors in 11.3?
â€“Â user6014
2 hours ago

Ah, sorry. Please try again.
â€“Â Henrik Schumacher
2 hours ago

Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
â€“Â ithilquessirr
2 hours ago

Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
â€“Â Henrik Schumacher
1 hour ago

Is this supposed to spew errors in 11.3?
â€“Â user6014
2 hours ago

Ah, sorry. Please try again.
â€“Â Henrik Schumacher
2 hours ago

Sorry, I couldn't understand what sparsearray function does here. But thanks this code is working.
â€“Â ithilquessirr
2 hours ago

Well, you might inspect MatrixForm@Normal[SparseArray[Transpose[s, listz] -> _, mm, nn]] to find that out.
â€“Â Henrik Schumacher
1 hour ago

add a commentÂ |Â

up vote
1
down vote

Edit: I realized that my previous solution doesn't produce the same length partitions. Here is the desired solution.

s = RandomSample[Range@10, 10];
t = RandomSample@RandomChoice@IntegerPartitions[10, 4]
m, n, p, r = TakeList[s, t]

(*or m, n, p, r = FoldPairList[TakeDrop, s, t]*)

1, 5, 10, 3, 7, 2, 9, 6, 8, 4

Original answer:

s = RandomSample[Range@10, 10];
t = RandomSample[Range@4, 4];
m, n, p, r = FoldPairList[TakeDrop, s, t]

6, 2, 9, 5, 1, 7, 10, 3, 8, 4

edited 1 hour ago

answered 2 hours ago

Okkes Dulgerci

3,4061716

add a commentÂ |Â

up vote
1
down vote

Edit: I realized that my previous solution doesn't produce the same length partitions. Here is the desired solution.

s = RandomSample[Range@10, 10];
t = RandomSample@RandomChoice@IntegerPartitions[10, 4]
m, n, p, r = TakeList[s, t]

(*or m, n, p, r = FoldPairList[TakeDrop, s, t]*)

1, 5, 10, 3, 7, 2, 9, 6, 8, 4

Original answer:

s = RandomSample[Range@10, 10];
t = RandomSample[Range@4, 4];
m, n, p, r = FoldPairList[TakeDrop, s, t]

6, 2, 9, 5, 1, 7, 10, 3, 8, 4

edited 1 hour ago

answered 2 hours ago

Okkes Dulgerci

3,4061716

add a commentÂ |Â

up vote
1
down vote

Edit: I realized that my previous solution doesn't produce the same length partitions. Here is the desired solution.

s = RandomSample[Range@10, 10];
t = RandomSample@RandomChoice@IntegerPartitions[10, 4]
m, n, p, r = TakeList[s, t]

(*or m, n, p, r = FoldPairList[TakeDrop, s, t]*)

1, 5, 10, 3, 7, 2, 9, 6, 8, 4

Original answer:

s = RandomSample[Range@10, 10];
t = RandomSample[Range@4, 4];
m, n, p, r = FoldPairList[TakeDrop, s, t]

6, 2, 9, 5, 1, 7, 10, 3, 8, 4

edited 1 hour ago

answered 2 hours ago

Okkes Dulgerci

3,4061716

Edit: I realized that my previous solution doesn't produce the same length partitions. Here is the desired solution.

s = RandomSample[Range@10, 10];
t = RandomSample@RandomChoice@IntegerPartitions[10, 4]
m, n, p, r = TakeList[s, t]

(*or m, n, p, r = FoldPairList[TakeDrop, s, t]*)

1, 5, 10, 3, 7, 2, 9, 6, 8, 4

Original answer:

s = RandomSample[Range@10, 10];
t = RandomSample[Range@4, 4];
m, n, p, r = FoldPairList[TakeDrop, s, t]

6, 2, 9, 5, 1, 7, 10, 3, 8, 4

edited 1 hour ago

answered 2 hours ago

Okkes Dulgerci

3,4061716

edited 1 hour ago

answered 2 hours ago

Okkes Dulgerci

3,4061716

answered 2 hours ago

Okkes Dulgerci

3,4061716

answered 2 hours ago

Okkes Dulgerci

3,4061716

add a commentÂ |Â

up vote
1
down vote

listz = CharacterRange["A", "K"];
k = 4;

SeedRandom[1]
rc = RandomChoice[Range @ k, Length @ listz];
m, n, p, r = Pick[listz, rc, #]& /@ Range[k];

"C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"

Also

Lookup[GroupBy[Transpose[rc, listz], First -> Last], #, ]& /@ Range[k]

"C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"

Update: To get k non-empty sublists:

SeedRandom[1]
rc = Module[r, While[Length[Counts[r = RandomChoice[Range@k, Length @ listz]]] < k]; r];
Pick[listz , rc, # ] & /@ Range[k ]

"A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"

Lookup[GroupBy[Transpose[rc , listz], First -> Last], #, ] & /@ Range[k]

"A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"

edited 8 mins ago

answered 57 mins ago

kglr

169k8192396

add a commentÂ |Â

up vote
1
down vote

listz = CharacterRange["A", "K"];
k = 4;

SeedRandom[1]
rc = RandomChoice[Range @ k, Length @ listz];
m, n, p, r = Pick[listz, rc, #]& /@ Range[k];

"C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"

Also

Lookup[GroupBy[Transpose[rc, listz], First -> Last], #, ]& /@ Range[k]

"C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"

Update: To get k non-empty sublists:

SeedRandom[1]
rc = Module[r, While[Length[Counts[r = RandomChoice[Range@k, Length @ listz]]] < k]; r];
Pick[listz , rc, # ] & /@ Range[k ]

"A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"

Lookup[GroupBy[Transpose[rc , listz], First -> Last], #, ] & /@ Range[k]

"A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"

edited 8 mins ago

answered 57 mins ago

kglr

169k8192396

add a commentÂ |Â

up vote
1
down vote

listz = CharacterRange["A", "K"];
k = 4;

SeedRandom[1]
rc = RandomChoice[Range @ k, Length @ listz];
m, n, p, r = Pick[listz, rc, #]& /@ Range[k];

"C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"

Also

Lookup[GroupBy[Transpose[rc, listz], First -> Last], #, ]& /@ Range[k]

"C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"

Update: To get k non-empty sublists:

SeedRandom[1]
rc = Module[r, While[Length[Counts[r = RandomChoice[Range@k, Length @ listz]]] < k]; r];
Pick[listz , rc, # ] & /@ Range[k ]

"A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"

Lookup[GroupBy[Transpose[rc , listz], First -> Last], #, ] & /@ Range[k]

"A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"

edited 8 mins ago

answered 57 mins ago

kglr

169k8192396

listz = CharacterRange["A", "K"];
k = 4;

SeedRandom[1]
rc = RandomChoice[Range @ k, Length @ listz];
m, n, p, r = Pick[listz, rc, #]& /@ Range[k];

"C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"

Also

Lookup[GroupBy[Transpose[rc, listz], First -> Last], #, ]& /@ Range[k]

"C", "F", "G", "H", "K", "B", "D", "E", "I", , "A", "J"

Update: To get k non-empty sublists:

SeedRandom[1]
rc = Module[r, While[Length[Counts[r = RandomChoice[Range@k, Length @ listz]]] < k]; r];
Pick[listz , rc, # ] & /@ Range[k ]

"A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"

Lookup[GroupBy[Transpose[rc , listz], First -> Last], #, ] & /@ Range[k]

"A", "B", "C", "E", "H", "I", "F", "D", "G", "J", "K"

edited 8 mins ago

answered 57 mins ago

kglr

169k8192396

edited 8 mins ago

answered 57 mins ago

kglr

169k8192396

answered 57 mins ago

kglr

169k8192396

answered 57 mins ago

kglr

169k8192396

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