Python: How do I randomly mix two lists?

If I generate two random lists for example:

N=5

X=20



parent1 = np.random.choice(X, N, replace=True)

parent2 = np.random.choice(X, N, replace=True)

Would give two lists e.g:

[2,5,1,1,12]

[3,18,4,5,1]

How could I make a new list which is a random mix of the two lists with the same amount of numbers?

e.g.

[2,18,1,5,12]

[18,5,1,12,5]

It doesn't matter about order.

asked Nov 22 '18 at 2:00

P erry

455

Is it allowed to pick same number multiple times?

– Krishna
Nov 22 '18 at 2:03

random.sample(a+b, N) would work?

– bro-grammer
Nov 22 '18 at 2:06

In a loop; randomly generate a one or a zero; if it is a one pick a random choice from the first list; else pick a random choice from the second list; stop when the item limit is reached,

– wwii
Nov 22 '18 at 2:07

Are the parents necessarily the same length?

– Mark_Anderson
Nov 22 '18 at 2:25

add a comment |

If I generate two random lists for example:

N=5

X=20



parent1 = np.random.choice(X, N, replace=True)

parent2 = np.random.choice(X, N, replace=True)

Would give two lists e.g:

[2,5,1,1,12]

[3,18,4,5,1]

How could I make a new list which is a random mix of the two lists with the same amount of numbers?

e.g.

[2,18,1,5,12]

[18,5,1,12,5]

It doesn't matter about order.

asked Nov 22 '18 at 2:00

P erry

455

Is it allowed to pick same number multiple times?

– Krishna
Nov 22 '18 at 2:03

random.sample(a+b, N) would work?

– bro-grammer
Nov 22 '18 at 2:06

In a loop; randomly generate a one or a zero; if it is a one pick a random choice from the first list; else pick a random choice from the second list; stop when the item limit is reached,

– wwii
Nov 22 '18 at 2:07

Are the parents necessarily the same length?

– Mark_Anderson
Nov 22 '18 at 2:25

add a comment |

If I generate two random lists for example:

N=5

X=20



parent1 = np.random.choice(X, N, replace=True)

parent2 = np.random.choice(X, N, replace=True)

Would give two lists e.g:

[2,5,1,1,12]

[3,18,4,5,1]

How could I make a new list which is a random mix of the two lists with the same amount of numbers?

e.g.

[2,18,1,5,12]

[18,5,1,12,5]

It doesn't matter about order.

asked Nov 22 '18 at 2:00

P erry

455

If I generate two random lists for example:

N=5

X=20



parent1 = np.random.choice(X, N, replace=True)

parent2 = np.random.choice(X, N, replace=True)

Would give two lists e.g:

[2,5,1,1,12]

[3,18,4,5,1]

How could I make a new list which is a random mix of the two lists with the same amount of numbers?

e.g.

[2,18,1,5,12]

[18,5,1,12,5]

It doesn't matter about order.

python python-3.x list random genetic-algorithm

asked Nov 22 '18 at 2:00

P erry

455

asked Nov 22 '18 at 2:00

P erry

455

asked Nov 22 '18 at 2:00

P erry

455

asked Nov 22 '18 at 2:00

P erry

455

asked Nov 22 '18 at 2:00

P erry

455

Is it allowed to pick same number multiple times?

– Krishna
Nov 22 '18 at 2:03

random.sample(a+b, N) would work?

– bro-grammer
Nov 22 '18 at 2:06

In a loop; randomly generate a one or a zero; if it is a one pick a random choice from the first list; else pick a random choice from the second list; stop when the item limit is reached,

– wwii
Nov 22 '18 at 2:07

Are the parents necessarily the same length?

– Mark_Anderson
Nov 22 '18 at 2:25

add a comment |

Is it allowed to pick same number multiple times?

– Krishna
Nov 22 '18 at 2:03

random.sample(a+b, N) would work?

– bro-grammer
Nov 22 '18 at 2:06

In a loop; randomly generate a one or a zero; if it is a one pick a random choice from the first list; else pick a random choice from the second list; stop when the item limit is reached,

– wwii
Nov 22 '18 at 2:07

Are the parents necessarily the same length?

– Mark_Anderson
Nov 22 '18 at 2:25

Is it allowed to pick same number multiple times?

– Krishna
Nov 22 '18 at 2:03

random.sample(a+b, N) would work?

– bro-grammer
Nov 22 '18 at 2:06

In a loop; randomly generate a one or a zero; if it is a one pick a random choice from the first list; else pick a random choice from the second list; stop when the item limit is reached,

– wwii
Nov 22 '18 at 2:07

Are the parents necessarily the same length?

– Mark_Anderson
Nov 22 '18 at 2:25

add a comment |

4 Answers
4

active

oldest

votes

Continuing your example, you could simply try this:

result = np.random.choice(np.concatenate([parent1,parent2]), N, replace=False)

Wether to sample with replacement or not is your choice (argument replace).

edited Nov 22 '18 at 2:31

answered Nov 22 '18 at 2:06

normanius

1,4861129

parent1+parent2 will add corresponding elements of two arrays!

– bro-grammer
Nov 22 '18 at 2:11

@bro-grammer You're right. Forgot that parent1 and parent2 were arrays and not lists. Fixed it by concatenating them.

– normanius
Nov 22 '18 at 2:16

Brill thanks very much!

– P erry
Nov 22 '18 at 2:29

add a comment |

if you are given parent1 and parent2, you can use np.choose:

which_one = np.random.int(2, size=N).astype(np.bool)

out_array = np.choose(which_one, [parent1, parent2])

however, keep in mind that if parent1 and parent2 are both random as per in your code, then you could simply do

out_array = np.random.choice(X, N, replace=True)

answered Nov 22 '18 at 2:14

ealfie

add a comment |

A few ways to get a uniform random selection from 2 lists.

Random 0 or 1 to pick list, then chose randomly in list.

Random number up to the sum of the lists, then use logic to chose from the first or second list according to that number.

import random

#### Option 1

for i in range(len(parent1)):

   list_number = random.randint(1,3)

   output(i) = exec('np.random.choice(parent'+str(list_number)+')')



#### Option 2

output_list = np.random.choice(np.concatenate([parent1,parent2]), N, replace=False)

The former represents each list equally in the output regardless of their relative length, the latter samples according to the relative length of the lists.

edited Nov 22 '18 at 2:18

answered Nov 22 '18 at 2:06

Mark_Anderson

421216

Option 2 is so simple and exactly what I was looking for. Thanks! :)

– P erry
Nov 22 '18 at 2:28

What does "replace=False" actually mean?

– P erry
Nov 22 '18 at 2:28

When repeatedly picking from the list [parent1, parent2], do you "replace" the numbers you've picked (such that they can be picked again), or not?

– Mark_Anderson
Nov 22 '18 at 2:34

No same number can be used twice unless there is a duplicate of the number

– P erry
Nov 22 '18 at 11:01

Sorry, that was my explanation of Replace=False. The code above does as you require.

– Mark_Anderson
Nov 22 '18 at 12:55

add a comment |

One way would be:

N=5

select_one = np.random.randint(0,2,N) # Outputs an array of 1 or 0s

mixed_list = select_one*parent1 + (1 - select_one)*parent2

If you also want to randomly permute the list elements, you can use:

mixed_list = np.random.permutation(rand_fact*parent1 + (1 - rand_fact)*parent2)

answered Nov 22 '18 at 2:20

RR_28023

134

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

Continuing your example, you could simply try this:

result = np.random.choice(np.concatenate([parent1,parent2]), N, replace=False)

Wether to sample with replacement or not is your choice (argument replace).

edited Nov 22 '18 at 2:31

answered Nov 22 '18 at 2:06

normanius

1,4861129

parent1+parent2 will add corresponding elements of two arrays!

– bro-grammer
Nov 22 '18 at 2:11

@bro-grammer You're right. Forgot that parent1 and parent2 were arrays and not lists. Fixed it by concatenating them.

– normanius
Nov 22 '18 at 2:16

Brill thanks very much!

– P erry
Nov 22 '18 at 2:29

add a comment |

Continuing your example, you could simply try this:

result = np.random.choice(np.concatenate([parent1,parent2]), N, replace=False)

Wether to sample with replacement or not is your choice (argument replace).

edited Nov 22 '18 at 2:31

answered Nov 22 '18 at 2:06

normanius

1,4861129

parent1+parent2 will add corresponding elements of two arrays!

– bro-grammer
Nov 22 '18 at 2:11

@bro-grammer You're right. Forgot that parent1 and parent2 were arrays and not lists. Fixed it by concatenating them.

– normanius
Nov 22 '18 at 2:16

Brill thanks very much!

– P erry
Nov 22 '18 at 2:29

add a comment |

Continuing your example, you could simply try this:

result = np.random.choice(np.concatenate([parent1,parent2]), N, replace=False)

Wether to sample with replacement or not is your choice (argument replace).

edited Nov 22 '18 at 2:31

answered Nov 22 '18 at 2:06

normanius

1,4861129

Continuing your example, you could simply try this:

result = np.random.choice(np.concatenate([parent1,parent2]), N, replace=False)

Wether to sample with replacement or not is your choice (argument replace).

edited Nov 22 '18 at 2:31

answered Nov 22 '18 at 2:06

normanius

1,4861129

edited Nov 22 '18 at 2:31

answered Nov 22 '18 at 2:06

normanius

1,4861129

answered Nov 22 '18 at 2:06

normanius

1,4861129

answered Nov 22 '18 at 2:06

normanius

1,4861129

parent1+parent2 will add corresponding elements of two arrays!

– bro-grammer
Nov 22 '18 at 2:11

@bro-grammer You're right. Forgot that parent1 and parent2 were arrays and not lists. Fixed it by concatenating them.

– normanius
Nov 22 '18 at 2:16

Brill thanks very much!

– P erry
Nov 22 '18 at 2:29

add a comment |

parent1+parent2 will add corresponding elements of two arrays!

– bro-grammer
Nov 22 '18 at 2:11

@bro-grammer You're right. Forgot that parent1 and parent2 were arrays and not lists. Fixed it by concatenating them.

– normanius
Nov 22 '18 at 2:16

Brill thanks very much!

– P erry
Nov 22 '18 at 2:29

parent1+parent2 will add corresponding elements of two arrays!

– bro-grammer
Nov 22 '18 at 2:11

@bro-grammer You're right. Forgot that parent1 and parent2 were arrays and not lists. Fixed it by concatenating them.

– normanius
Nov 22 '18 at 2:16

Brill thanks very much!

– P erry
Nov 22 '18 at 2:29

add a comment |

if you are given parent1 and parent2, you can use np.choose:

which_one = np.random.int(2, size=N).astype(np.bool)

out_array = np.choose(which_one, [parent1, parent2])

however, keep in mind that if parent1 and parent2 are both random as per in your code, then you could simply do

out_array = np.random.choice(X, N, replace=True)

answered Nov 22 '18 at 2:14

ealfie

add a comment |

if you are given parent1 and parent2, you can use np.choose:

which_one = np.random.int(2, size=N).astype(np.bool)

out_array = np.choose(which_one, [parent1, parent2])

however, keep in mind that if parent1 and parent2 are both random as per in your code, then you could simply do

out_array = np.random.choice(X, N, replace=True)

answered Nov 22 '18 at 2:14

ealfie

add a comment |

if you are given parent1 and parent2, you can use np.choose:

which_one = np.random.int(2, size=N).astype(np.bool)

out_array = np.choose(which_one, [parent1, parent2])

however, keep in mind that if parent1 and parent2 are both random as per in your code, then you could simply do

out_array = np.random.choice(X, N, replace=True)

answered Nov 22 '18 at 2:14

ealfie

if you are given parent1 and parent2, you can use np.choose:

which_one = np.random.int(2, size=N).astype(np.bool)

out_array = np.choose(which_one, [parent1, parent2])

however, keep in mind that if parent1 and parent2 are both random as per in your code, then you could simply do

out_array = np.random.choice(X, N, replace=True)

answered Nov 22 '18 at 2:14

ealfie

answered Nov 22 '18 at 2:14

ealfie

answered Nov 22 '18 at 2:14

ealfie

answered Nov 22 '18 at 2:14

ealfie

add a comment |

A few ways to get a uniform random selection from 2 lists.

Random 0 or 1 to pick list, then chose randomly in list.

Random number up to the sum of the lists, then use logic to chose from the first or second list according to that number.

import random

#### Option 1

for i in range(len(parent1)):

   list_number = random.randint(1,3)

   output(i) = exec('np.random.choice(parent'+str(list_number)+')')



#### Option 2

output_list = np.random.choice(np.concatenate([parent1,parent2]), N, replace=False)

The former represents each list equally in the output regardless of their relative length, the latter samples according to the relative length of the lists.

edited Nov 22 '18 at 2:18

answered Nov 22 '18 at 2:06

Mark_Anderson

421216

Option 2 is so simple and exactly what I was looking for. Thanks! :)

– P erry
Nov 22 '18 at 2:28

What does "replace=False" actually mean?

– P erry
Nov 22 '18 at 2:28

When repeatedly picking from the list [parent1, parent2], do you "replace" the numbers you've picked (such that they can be picked again), or not?

– Mark_Anderson
Nov 22 '18 at 2:34

No same number can be used twice unless there is a duplicate of the number

– P erry
Nov 22 '18 at 11:01

Sorry, that was my explanation of Replace=False. The code above does as you require.

– Mark_Anderson
Nov 22 '18 at 12:55

add a comment |

A few ways to get a uniform random selection from 2 lists.

Random 0 or 1 to pick list, then chose randomly in list.

Random number up to the sum of the lists, then use logic to chose from the first or second list according to that number.

import random

#### Option 1

for i in range(len(parent1)):

   list_number = random.randint(1,3)

   output(i) = exec('np.random.choice(parent'+str(list_number)+')')



#### Option 2

output_list = np.random.choice(np.concatenate([parent1,parent2]), N, replace=False)

The former represents each list equally in the output regardless of their relative length, the latter samples according to the relative length of the lists.

edited Nov 22 '18 at 2:18

answered Nov 22 '18 at 2:06

Mark_Anderson

421216

Option 2 is so simple and exactly what I was looking for. Thanks! :)

– P erry
Nov 22 '18 at 2:28

What does "replace=False" actually mean?

– P erry
Nov 22 '18 at 2:28

When repeatedly picking from the list [parent1, parent2], do you "replace" the numbers you've picked (such that they can be picked again), or not?

– Mark_Anderson
Nov 22 '18 at 2:34

No same number can be used twice unless there is a duplicate of the number

– P erry
Nov 22 '18 at 11:01

Sorry, that was my explanation of Replace=False. The code above does as you require.

– Mark_Anderson
Nov 22 '18 at 12:55

add a comment |

A few ways to get a uniform random selection from 2 lists.

Random 0 or 1 to pick list, then chose randomly in list.

Random number up to the sum of the lists, then use logic to chose from the first or second list according to that number.

import random

#### Option 1

for i in range(len(parent1)):

   list_number = random.randint(1,3)

   output(i) = exec('np.random.choice(parent'+str(list_number)+')')



#### Option 2

output_list = np.random.choice(np.concatenate([parent1,parent2]), N, replace=False)

The former represents each list equally in the output regardless of their relative length, the latter samples according to the relative length of the lists.

edited Nov 22 '18 at 2:18

answered Nov 22 '18 at 2:06

Mark_Anderson

421216

A few ways to get a uniform random selection from 2 lists.

Random 0 or 1 to pick list, then chose randomly in list.

Random number up to the sum of the lists, then use logic to chose from the first or second list according to that number.

import random

#### Option 1

for i in range(len(parent1)):

   list_number = random.randint(1,3)

   output(i) = exec('np.random.choice(parent'+str(list_number)+')')



#### Option 2

output_list = np.random.choice(np.concatenate([parent1,parent2]), N, replace=False)

The former represents each list equally in the output regardless of their relative length, the latter samples according to the relative length of the lists.

edited Nov 22 '18 at 2:18

answered Nov 22 '18 at 2:06

Mark_Anderson

421216

edited Nov 22 '18 at 2:18

answered Nov 22 '18 at 2:06

Mark_Anderson

421216

answered Nov 22 '18 at 2:06

Mark_Anderson

421216

answered Nov 22 '18 at 2:06

Mark_Anderson

421216

Option 2 is so simple and exactly what I was looking for. Thanks! :)

– P erry
Nov 22 '18 at 2:28

What does "replace=False" actually mean?

– P erry
Nov 22 '18 at 2:28

When repeatedly picking from the list [parent1, parent2], do you "replace" the numbers you've picked (such that they can be picked again), or not?

– Mark_Anderson
Nov 22 '18 at 2:34

No same number can be used twice unless there is a duplicate of the number

– P erry
Nov 22 '18 at 11:01

Sorry, that was my explanation of Replace=False. The code above does as you require.

– Mark_Anderson
Nov 22 '18 at 12:55

add a comment |

Option 2 is so simple and exactly what I was looking for. Thanks! :)

– P erry
Nov 22 '18 at 2:28

What does "replace=False" actually mean?

– P erry
Nov 22 '18 at 2:28

When repeatedly picking from the list [parent1, parent2], do you "replace" the numbers you've picked (such that they can be picked again), or not?

– Mark_Anderson
Nov 22 '18 at 2:34

No same number can be used twice unless there is a duplicate of the number

– P erry
Nov 22 '18 at 11:01

Sorry, that was my explanation of Replace=False. The code above does as you require.

– Mark_Anderson
Nov 22 '18 at 12:55

Option 2 is so simple and exactly what I was looking for. Thanks! :)

– P erry
Nov 22 '18 at 2:28

What does "replace=False" actually mean?

– P erry
Nov 22 '18 at 2:28

When repeatedly picking from the list [parent1, parent2], do you "replace" the numbers you've picked (such that they can be picked again), or not?

– Mark_Anderson
Nov 22 '18 at 2:34

No same number can be used twice unless there is a duplicate of the number

– P erry
Nov 22 '18 at 11:01

Sorry, that was my explanation of Replace=False. The code above does as you require.

– Mark_Anderson
Nov 22 '18 at 12:55

add a comment |

One way would be:

N=5

select_one = np.random.randint(0,2,N) # Outputs an array of 1 or 0s

mixed_list = select_one*parent1 + (1 - select_one)*parent2

If you also want to randomly permute the list elements, you can use:

mixed_list = np.random.permutation(rand_fact*parent1 + (1 - rand_fact)*parent2)

answered Nov 22 '18 at 2:20

RR_28023

134

add a comment |

One way would be:

N=5

select_one = np.random.randint(0,2,N) # Outputs an array of 1 or 0s

mixed_list = select_one*parent1 + (1 - select_one)*parent2

If you also want to randomly permute the list elements, you can use:

mixed_list = np.random.permutation(rand_fact*parent1 + (1 - rand_fact)*parent2)

answered Nov 22 '18 at 2:20

RR_28023

134

add a comment |

One way would be:

N=5

select_one = np.random.randint(0,2,N) # Outputs an array of 1 or 0s

mixed_list = select_one*parent1 + (1 - select_one)*parent2

If you also want to randomly permute the list elements, you can use:

mixed_list = np.random.permutation(rand_fact*parent1 + (1 - rand_fact)*parent2)

answered Nov 22 '18 at 2:20

RR_28023

134

One way would be:

N=5

select_one = np.random.randint(0,2,N) # Outputs an array of 1 or 0s

mixed_list = select_one*parent1 + (1 - select_one)*parent2

If you also want to randomly permute the list elements, you can use:

mixed_list = np.random.permutation(rand_fact*parent1 + (1 - rand_fact)*parent2)

answered Nov 22 '18 at 2:20

RR_28023

134

answered Nov 22 '18 at 2:20

RR_28023

134

answered Nov 22 '18 at 2:20

RR_28023

134

answered Nov 22 '18 at 2:20

RR_28023

134

add a comment |

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