### Question :

I need to pick out “x” number of non-repeating, random numbers out of a list. For example:

```
all_data = [1, 2, 2, 3, 4, 5, 6, 7, 8, 8, 9, 10, 11, 11, 12, 13, 14, 15, 15]
```

How do I pick out a list like `[2, 11, 15]`

and not `[3, 8, 8]`

?

##
Answer #1:

That’s exactly what `random.sample()`

does.

```
>>> random.sample(range(1, 16), 3)
[11, 10, 2]
```

**Edit**: I’m almost certain this is not what you asked, but I was pushed to include this comment: If the population you want to take samples from contains duplicates, you have to remove them first:

```
population = [1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1]
population = set(population)
samples = random.sample(population, 3)
```

##
Answer #2:

Something like this:

```
all_data = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
from random import shuffle
shuffle(all_data)
res = all_data[:3]# or any other number of items
```

OR:

```
from random import sample
number_of_items = 4
sample(all_data, number_of_items)
```

If all_data could contains duplicate entries than modify your code to remove duplicates first and then use shuffle or sample:

```
all_data = list(set(all_data))
shuffle(all_data)
res = all_data[:3]# or any other number of items
```

##
Answer #3:

Others have suggested that you use `random.sample`

. While this is a valid suggestion, there is one subtlety that everyone has ignored:

If the population contains repeats,

then each occurrence is a possible

selection in the sample.

Thus, you need to turn your list into a set, to avoid repeated values:

```
import random
L = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
random.sample(set(L), x) # where x is the number of samples that you want
```

##
Answer #4:

You can also generate a list of random choices, using `itertools.combinations`

and `random.shuffle`

.

```
all_data = [1,2,2,3,4,5,6,7,8,8,9,10,11,11,12,13,14,15,15]
# Remove duplicates
unique_data = set(all_data)
# Generate a list of combinations of three elements
list_of_three = list(itertools.combinations(unique_data, 3))
# Shuffle the list of combinations of three elements
random.shuffle(list_of_three)
```

Output:

```
[(2, 5, 15), (11, 13, 15), (3, 10, 15), (1, 6, 9), (1, 7, 8), ...]
```

##
Answer #5:

Another way, of course with all the solutions you have to be sure that there are at least 3 unique values in the original list.

```
all_data = [1,2,2,3,4,5,6,7,8,8,9,10,11,11,12,13,14,15,15]
choices = []
while len(choices) < 3:
selection = random.choice(all_data)
if selection not in choices:
choices.append(selection)
print choices
```

##
Answer #6:

```
import random
fruits_in_store = ['apple','mango','orange','pineapple','fig','grapes','guava','litchi','almond']
print('items available in store :')
print(fruits_in_store)
my_cart = []
for i in range(4):
#selecting a random index
temp = int(random.random()*len(fruits_in_store))
# adding element at random index to new list
my_cart.append(fruits_in_store[temp])
# removing the add element from original list
fruits_in_store.pop(temp)
print('items successfully added to cart:')
print(my_cart)
```

Output:

```
items available in store :
['apple', 'mango', 'orange', 'pineapple', 'fig', 'grapes', 'guava', 'litchi', 'almond']
items successfully added to cart:
['orange', 'pineapple', 'mango', 'almond']
```

##
Answer #7:

If the data being repeated implies that we are more likely to draw that particular data, we can’t turn it into a set right away (since we would loose that information by doing so). For this, we need to pick samples one by one and verify the size of the set that we generate has reached x (the number of samples that we want). Something like:

```
data=[0, 1, 2, 3, 4, 4, 4, 4, 5, 5, 6, 6]
x=3
res=set()
while(len(res)<x):
res.add(np.random.choice(data))
print(res)
```

some outputs :

```
{3, 4, 5}
{3, 5, 6}
{0, 4, 5}
{2, 4, 5}
```

As we can see 4 or 5 appear more frequently (I know 4 examples is not good enough statistics).