Generating set of semi-random numbers with some guaranteed present

I am attempting to create a Python 3 script which prints out all possible sets of 6 numbers, 4 of which are fixed (must appear), but can appear anywhere in the 6. The numbers also must all be between 0 and 9.

For instance, if the "fixed" numbers are 1,2,3,4, then 1,2,4,3,5,7 is acceptable, 1,2,3,3,4,4 is also acceptable, but 0,4,3,7,8,2 or 8,0,5,6,7,8 are not and should not be printed.

I have attempted so solve this so far, but have run into a mental roadblock when it comes to conceptually understanding where exactly to get started on this.

So far I have attempted to use random.sample, just displaying if the numbers picked are within the range, but ultimately determined that it would not be suitable and that I am most likely barking up the wrong tree here. (I realize that the below is not optimal, just trying to get a working solution. This is just for my personal use for generating 6 digit arma map coordinates, so long as it works I don't overly care what it looks like)

import random

count = 0

while count < 4:

    res = random.sample([0, 1, 2, 3, 4, 5, 6, 7, 8, 9],  6)

    if [0, 0, 2, 8] in res:

        print(res);

        count += 1

Any help is greatly appreciated.

edited Jan 1 at 17:22

Dmytro Chasovskyi

829828

asked Jan 1 at 8:35

Markyroson

347

1

Not sure, if this will work for you but you can consider random.choice?

– Dmytro Chasovskyi
Jan 1 at 8:41

3

How about: fix your 4 numbers, then generate the last 2 using your favourite RNG.

– Hong Ooi
Jan 1 at 8:58

1

Use your favourite shuffle to randomise the order

– Hong Ooi
Jan 1 at 9:16

1

@DmytroChasovskyi Why use acceptance rejection when it can be done quite directly?

– pjs
Jan 1 at 20:51

1

Your question says that you want a script that "prints out all possible sets of 6 numbers [...]". Is that what you want, or do you just want to generate some of the possible sets at random? If you want all of them, it's a bit strange to be using the random module.

– Mark Dickinson
Jan 2 at 17:39

|
show 5 more comments

For instance, if the "fixed" numbers are 1,2,3,4, then 1,2,4,3,5,7 is acceptable, 1,2,3,3,4,4 is also acceptable, but 0,4,3,7,8,2 or 8,0,5,6,7,8 are not and should not be printed.

I have attempted so solve this so far, but have run into a mental roadblock when it comes to conceptually understanding where exactly to get started on this.

import random

count = 0

while count < 4:

    res = random.sample([0, 1, 2, 3, 4, 5, 6, 7, 8, 9],  6)

    if [0, 0, 2, 8] in res:

        print(res);

        count += 1

Any help is greatly appreciated.

edited Jan 1 at 17:22

Dmytro Chasovskyi

829828

asked Jan 1 at 8:35

Markyroson

347

1

Not sure, if this will work for you but you can consider random.choice?

– Dmytro Chasovskyi
Jan 1 at 8:41

3

How about: fix your 4 numbers, then generate the last 2 using your favourite RNG.

– Hong Ooi
Jan 1 at 8:58

1

Use your favourite shuffle to randomise the order

– Hong Ooi
Jan 1 at 9:16

1

@DmytroChasovskyi Why use acceptance rejection when it can be done quite directly?

– pjs
Jan 1 at 20:51

1

Your question says that you want a script that "prints out all possible sets of 6 numbers [...]". Is that what you want, or do you just want to generate some of the possible sets at random? If you want all of them, it's a bit strange to be using the random module.

– Mark Dickinson
Jan 2 at 17:39

|
show 5 more comments

For instance, if the "fixed" numbers are 1,2,3,4, then 1,2,4,3,5,7 is acceptable, 1,2,3,3,4,4 is also acceptable, but 0,4,3,7,8,2 or 8,0,5,6,7,8 are not and should not be printed.

I have attempted so solve this so far, but have run into a mental roadblock when it comes to conceptually understanding where exactly to get started on this.

import random

count = 0

while count < 4:

    res = random.sample([0, 1, 2, 3, 4, 5, 6, 7, 8, 9],  6)

    if [0, 0, 2, 8] in res:

        print(res);

        count += 1

Any help is greatly appreciated.

edited Jan 1 at 17:22

Dmytro Chasovskyi

829828

asked Jan 1 at 8:35

Markyroson

347

For instance, if the "fixed" numbers are 1,2,3,4, then 1,2,4,3,5,7 is acceptable, 1,2,3,3,4,4 is also acceptable, but 0,4,3,7,8,2 or 8,0,5,6,7,8 are not and should not be printed.

I have attempted so solve this so far, but have run into a mental roadblock when it comes to conceptually understanding where exactly to get started on this.

import random

count = 0

while count < 4:

    res = random.sample([0, 1, 2, 3, 4, 5, 6, 7, 8, 9],  6)

    if [0, 0, 2, 8] in res:

        print(res);

        count += 1

Any help is greatly appreciated.

python-3.x random

edited Jan 1 at 17:22

Dmytro Chasovskyi

829828

asked Jan 1 at 8:35

Markyroson

347

edited Jan 1 at 17:22

Dmytro Chasovskyi

829828

asked Jan 1 at 8:35

Markyroson

347

edited Jan 1 at 17:22

Dmytro Chasovskyi

829828

edited Jan 1 at 17:22

Dmytro Chasovskyi

829828

edited Jan 1 at 17:22

Dmytro Chasovskyi

829828

asked Jan 1 at 8:35

Markyroson

347

asked Jan 1 at 8:35

Markyroson

347

asked Jan 1 at 8:35

Markyroson

347

1

Not sure, if this will work for you but you can consider random.choice?

– Dmytro Chasovskyi
Jan 1 at 8:41

3

How about: fix your 4 numbers, then generate the last 2 using your favourite RNG.

– Hong Ooi
Jan 1 at 8:58

1

Use your favourite shuffle to randomise the order

– Hong Ooi
Jan 1 at 9:16

1

@DmytroChasovskyi Why use acceptance rejection when it can be done quite directly?

– pjs
Jan 1 at 20:51

1

Your question says that you want a script that "prints out all possible sets of 6 numbers [...]". Is that what you want, or do you just want to generate some of the possible sets at random? If you want all of them, it's a bit strange to be using the random module.

– Mark Dickinson
Jan 2 at 17:39

|
show 5 more comments

1

Not sure, if this will work for you but you can consider random.choice?

– Dmytro Chasovskyi
Jan 1 at 8:41

3

How about: fix your 4 numbers, then generate the last 2 using your favourite RNG.

– Hong Ooi
Jan 1 at 8:58

1

Use your favourite shuffle to randomise the order

– Hong Ooi
Jan 1 at 9:16

1

@DmytroChasovskyi Why use acceptance rejection when it can be done quite directly?

– pjs
Jan 1 at 20:51

1

Your question says that you want a script that "prints out all possible sets of 6 numbers [...]". Is that what you want, or do you just want to generate some of the possible sets at random? If you want all of them, it's a bit strange to be using the random module.

– Mark Dickinson
Jan 2 at 17:39

Not sure, if this will work for you but you can consider random.choice?

– Dmytro Chasovskyi
Jan 1 at 8:41

How about: fix your 4 numbers, then generate the last 2 using your favourite RNG.

– Hong Ooi
Jan 1 at 8:58

Use your favourite shuffle to randomise the order

– Hong Ooi
Jan 1 at 9:16

@DmytroChasovskyi Why use acceptance rejection when it can be done quite directly?

– pjs
Jan 1 at 20:51

Your question says that you want a script that "prints out all possible sets of 6 numbers [...]". Is that what you want, or do you just want to generate some of the possible sets at random? If you want all of them, it's a bit strange to be using the random module.

– Mark Dickinson
Jan 2 at 17:39

|
show 5 more comments

2 Answers
2

active

oldest

votes

The following takes a specified list of fixed values, determines how many additional random numbers are needed and appends them, then shuffles the list and returns it.

import random



def generate_set(fixed_vals = [1, 2, 3, 4], length = 6, min = 0, max = 9):

    for _ in range(length - len(fixed_vals)):

        fixed_vals.append(random.randint(min, max))

    random.shuffle(fixed_vals)

    return fixed_vals



print(generate_set())  # produces, e.g., [0, 3, 4, 8, 2, 1]

edited Jan 1 at 22:05

answered Jan 1 at 20:47

pjs

13.2k41541

clearly, superior answer

– Severin Pappadeux
Jan 7 at 18:51

add a comment |

My solution works out for the sequence that should contain 0, 0, 2, 8 (weights specified for this sequence). The solution could be further generalized.

def generate_sequence(weights):

    return random.choices([0,1,2,3,4,5,6,7,8,9], weights=weights, k=6)



def check_sequence(sequence, check):

    return check(sequence)



def check():

   # TODO: Add your own implementation of sequence checker

   pass



# You could play with weights to generate sequence with higher probability

weights = [4,1,2,1,1,1,1,1,2,1]  

sequence = generate_sequence(weights)



while check_sequence(sequence, check):

    sequence = generate_sequence(weights)

answered Jan 1 at 9:28

Dmytro Chasovskyi

829828

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

The following takes a specified list of fixed values, determines how many additional random numbers are needed and appends them, then shuffles the list and returns it.

import random



def generate_set(fixed_vals = [1, 2, 3, 4], length = 6, min = 0, max = 9):

    for _ in range(length - len(fixed_vals)):

        fixed_vals.append(random.randint(min, max))

    random.shuffle(fixed_vals)

    return fixed_vals



print(generate_set())  # produces, e.g., [0, 3, 4, 8, 2, 1]

edited Jan 1 at 22:05

answered Jan 1 at 20:47

pjs

13.2k41541

clearly, superior answer

– Severin Pappadeux
Jan 7 at 18:51

add a comment |

The following takes a specified list of fixed values, determines how many additional random numbers are needed and appends them, then shuffles the list and returns it.

import random



def generate_set(fixed_vals = [1, 2, 3, 4], length = 6, min = 0, max = 9):

    for _ in range(length - len(fixed_vals)):

        fixed_vals.append(random.randint(min, max))

    random.shuffle(fixed_vals)

    return fixed_vals



print(generate_set())  # produces, e.g., [0, 3, 4, 8, 2, 1]

edited Jan 1 at 22:05

answered Jan 1 at 20:47

pjs

13.2k41541

clearly, superior answer

– Severin Pappadeux
Jan 7 at 18:51

add a comment |

The following takes a specified list of fixed values, determines how many additional random numbers are needed and appends them, then shuffles the list and returns it.

import random



def generate_set(fixed_vals = [1, 2, 3, 4], length = 6, min = 0, max = 9):

    for _ in range(length - len(fixed_vals)):

        fixed_vals.append(random.randint(min, max))

    random.shuffle(fixed_vals)

    return fixed_vals



print(generate_set())  # produces, e.g., [0, 3, 4, 8, 2, 1]

edited Jan 1 at 22:05

answered Jan 1 at 20:47

pjs

13.2k41541

The following takes a specified list of fixed values, determines how many additional random numbers are needed and appends them, then shuffles the list and returns it.

import random



def generate_set(fixed_vals = [1, 2, 3, 4], length = 6, min = 0, max = 9):

    for _ in range(length - len(fixed_vals)):

        fixed_vals.append(random.randint(min, max))

    random.shuffle(fixed_vals)

    return fixed_vals



print(generate_set())  # produces, e.g., [0, 3, 4, 8, 2, 1]

edited Jan 1 at 22:05

answered Jan 1 at 20:47

pjs

13.2k41541

edited Jan 1 at 22:05

answered Jan 1 at 20:47

pjs

13.2k41541

answered Jan 1 at 20:47

pjs

13.2k41541

answered Jan 1 at 20:47

pjs

13.2k41541

clearly, superior answer

– Severin Pappadeux
Jan 7 at 18:51

add a comment |

clearly, superior answer

– Severin Pappadeux
Jan 7 at 18:51

clearly, superior answer

– Severin Pappadeux
Jan 7 at 18:51

add a comment |

My solution works out for the sequence that should contain 0, 0, 2, 8 (weights specified for this sequence). The solution could be further generalized.

def generate_sequence(weights):

    return random.choices([0,1,2,3,4,5,6,7,8,9], weights=weights, k=6)



def check_sequence(sequence, check):

    return check(sequence)



def check():

   # TODO: Add your own implementation of sequence checker

   pass



# You could play with weights to generate sequence with higher probability

weights = [4,1,2,1,1,1,1,1,2,1]  

sequence = generate_sequence(weights)



while check_sequence(sequence, check):

    sequence = generate_sequence(weights)

answered Jan 1 at 9:28

Dmytro Chasovskyi

829828

add a comment |

My solution works out for the sequence that should contain 0, 0, 2, 8 (weights specified for this sequence). The solution could be further generalized.

def generate_sequence(weights):

    return random.choices([0,1,2,3,4,5,6,7,8,9], weights=weights, k=6)



def check_sequence(sequence, check):

    return check(sequence)



def check():

   # TODO: Add your own implementation of sequence checker

   pass



# You could play with weights to generate sequence with higher probability

weights = [4,1,2,1,1,1,1,1,2,1]  

sequence = generate_sequence(weights)



while check_sequence(sequence, check):

    sequence = generate_sequence(weights)

answered Jan 1 at 9:28

Dmytro Chasovskyi

829828

add a comment |

My solution works out for the sequence that should contain 0, 0, 2, 8 (weights specified for this sequence). The solution could be further generalized.

def generate_sequence(weights):

    return random.choices([0,1,2,3,4,5,6,7,8,9], weights=weights, k=6)



def check_sequence(sequence, check):

    return check(sequence)



def check():

   # TODO: Add your own implementation of sequence checker

   pass



# You could play with weights to generate sequence with higher probability

weights = [4,1,2,1,1,1,1,1,2,1]  

sequence = generate_sequence(weights)



while check_sequence(sequence, check):

    sequence = generate_sequence(weights)

answered Jan 1 at 9:28

Dmytro Chasovskyi

829828

My solution works out for the sequence that should contain 0, 0, 2, 8 (weights specified for this sequence). The solution could be further generalized.

def generate_sequence(weights):

    return random.choices([0,1,2,3,4,5,6,7,8,9], weights=weights, k=6)



def check_sequence(sequence, check):

    return check(sequence)



def check():

   # TODO: Add your own implementation of sequence checker

   pass



# You could play with weights to generate sequence with higher probability

weights = [4,1,2,1,1,1,1,1,2,1]  

sequence = generate_sequence(weights)



while check_sequence(sequence, check):

    sequence = generate_sequence(weights)

answered Jan 1 at 9:28

Dmytro Chasovskyi

829828

answered Jan 1 at 9:28

Dmytro Chasovskyi

829828

answered Jan 1 at 9:28

Dmytro Chasovskyi

829828

answered Jan 1 at 9:28

Dmytro Chasovskyi

829828

add a comment |

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