I have a list of objects and I want to shuffle them. I thought I could use the `random.shuffle`

method, but this seems to fail when the list is of objects. Is there a method for shuffling objects or another way around this?

```
import random
class A:
foo = "bar"
a1 = a()
a2 = a()
b = [a1, a2]
print(random.shuffle(b))
```

This will fail.

As you learned the in-place shuffling was the problem. I also have problem frequently, and often seem to forget how to copy a list, too. Using `sample(a, len(a))`

is the solution, using `len(a)`

as the sample size. See https://docs.python.org/3.6/library/random.html#random.sample for the Python documentation.

Here's a simple version using `random.sample()`

that returns the shuffled result as a new list.

```
import random
a = range(5)
b = random.sample(a, len(a))
print a, b, "two list same:", a == b
# print: [0, 1, 2, 3, 4] [2, 1, 3, 4, 0] two list same: False
# The function sample allows no duplicates.
# Result can be smaller but not larger than the input.
a = range(555)
b = random.sample(a, len(a))
print "no duplicates:", a == list(set(b))
try:
random.sample(a, len(a) + 1)
except ValueError as e:
print "Nope!", e
# print: no duplicates: True
# print: Nope! sample larger than population
```

For one-liners, use`random.sample(list_to_be_shuffled, length_of_the_list)`

with an example:

```
import random
random.sample(list(range(10)), 10)
```

outputs: [2, 9, 7, 8, 3, 0, 4, 1, 6, 5]

You can go for this:

```
>>> A = ['r','a','n','d','o','m']
>>> B = [1,2,3,4,5,6]
>>> import random
>>> random.sample(A+B, len(A+B))
[3, 'r', 4, 'n', 6, 5, 'm', 2, 1, 'a', 'o', 'd']
```

if you want to go back to two lists, you then split this long list into two.

```
def shuffle(_list):
if not _list == []:
import random
list2 = []
while _list != []:
card = random.choice(_list)
_list.remove(card)
list2.append(card)
while list2 != []:
card1 = list2[0]
list2.remove(card1)
_list.append(card1)
return _list
```

you can either use shuffle or sample . both of which come from random module.

```
import random
def shuffle(arr1):
n=len(arr1)
b=random.sample(arr1,n)
return b
```

OR

```
import random
def shuffle(arr1):
random.shuffle(arr1)
return arr1
```

It took me some time to get that too. But the documentation for shuffle is very clear:

shuffle list x

in place; return None.

So you shouldn't `print(random.shuffle(b))`

. Instead do `random.shuffle(b)`

and then `print(b)`

.

One can define a function called `shuffled`

(in the same sense of `sort`

vs `sorted`

)

```
def shuffled(x):
import random
y = x[:]
random.shuffle(y)
return y
x = shuffled([1, 2, 3, 4])
print x
```

'print func(foo)' will print the return value of 'func' when called with 'foo'. 'shuffle' however has None as its return type, as the list will be modified in place, hence it prints nothing. Workaround:

```
# shuffle the list in place
random.shuffle(b)
# print it
print(b)
```

If you're more into functional programming style you might want to make the following wrapper function:

```
def myshuffle(ls):
random.shuffle(ls)
return ls
```

If you have multiple lists, you might want to define the permutation (the way you shuffle the list / rearrange the items in the list) first and then apply it to all lists:

```
import random
perm = list(range(len(list_one)))
random.shuffle(perm)
list_one = [list_one[index] for index in perm]
list_two = [list_two[index] for index in perm]
```

If your lists are numpy arrays, it is simpler:

```
import numpy as np
perm = np.random.permutation(len(list_one))
list_one = list_one[perm]
list_two = list_two[perm]
```

I've created the small utility package `mpu`

which has the `consistent_shuffle`

function:

```
import mpu
# Necessary if you want consistent results
import random
random.seed(8)
# Define example lists
list_one = [1,2,3]
list_two = ['a', 'b', 'c']
# Call the function
list_one, list_two = mpu.consistent_shuffle(list_one, list_two)
```

Note that `mpu.consistent_shuffle`

takes an arbitrary number of arguments. So you can also shuffle three or more lists with it.

```
""" to shuffle random, set random= True """
def shuffle(x,random=False):
shuffled = []
ma = x
if random == True:
rando = [ma[i] for i in np.random.randint(0,len(ma),len(ma))]
return rando
if random == False:
for i in range(len(ma)):
ave = len(ma)//3
if i < ave:
shuffled.append(ma[i+ave])
else:
shuffled.append(ma[i-ave])
return shuffled
```

In some cases when using numpy arrays, using `random.shuffle`

created duplicate data in the array.

An alternative is to use `numpy.random.shuffle`

. If you're working with numpy already, this is the preferred method over the generic `random.shuffle`

.

**Example**

```
>>> import numpy as np
>>> import random
```

Using `random.shuffle`

:

```
>>> foo = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> foo
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
>>> random.shuffle(foo)
>>> foo
array([[1, 2, 3],
[1, 2, 3],
[4, 5, 6]])
```

Using `numpy.random.shuffle`

:

```
>>> foo = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> foo
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
>>> np.random.shuffle(foo)
>>> foo
array([[1, 2, 3],
[7, 8, 9],
[4, 5, 6]])
```

```
import random
class a:
foo = "bar"
a1 = a()
a2 = a()
b = [a1.foo,a2.foo]
random.shuffle(b)
```

If you happen to be using numpy already (very popular for scientific and financial applications) you can save yourself an import.

```
import numpy as np
np.random.shuffle(b)
print(b)
```

https://numpy.org/doc/stable/reference/random/generated/numpy.random.shuffle.html

```
import random
class a:
foo = "bar"
a1 = a()
a2 = a()
a3 = a()
a4 = a()
b = [a1,a2,a3,a4]
random.shuffle(b)
print(b)
```

`shuffle`

is in place, so do not print result, which is `None`

, but the list.

Plan: Write out the shuffle without relying on a library to do the heavy lifting. Example: Go through the list from the beginning starting with element 0; find a new random position for it, say 6, put 0’s value in 6 and 6’s value in 0. Move on to element 1 and repeat this process, and so on through the rest of the list

```
import random
iteration = random.randint(2, 100)
temp_var = 0
while iteration > 0:
for i in range(1, len(my_list)): # have to use range with len()
for j in range(1, len(my_list) - i):
# Using temp_var as my place holder so I don't lose values
temp_var = my_list[i]
my_list[i] = my_list[j]
my_list[j] = temp_var
iteration -= 1
```

It works fine. I am trying it here with functions as list objects:

```
from random import shuffle
def foo1():
print "foo1",
def foo2():
print "foo2",
def foo3():
print "foo3",
A=[foo1,foo2,foo3]
for x in A:
x()
print "\r"
shuffle(A)
for y in A:
y()
```

It prints out: foo1 foo2 foo3 foo2 foo3 foo1 (the foos in the last row have a random order)

```
from random import random
my_list = range(10)
shuffled_list = sorted(my_list, key=lambda x: random())
```

This alternative may be useful for some applications where you want to swap the ordering function.

```
#!/usr/bin/python3
import random
s=list(range(5))
random.shuffle(s) # << shuffle before print or assignment
print(s)
# print: [2, 4, 1, 3, 0]
```

you could build a function that takes a list as a parameter and returns a shuffled version of the list:

```
from random import *
def listshuffler(inputlist):
for i in range(len(inputlist)):
swap = randint(0,len(inputlist)-1)
temp = inputlist[swap]
inputlist[swap] = inputlist[i]
inputlist[i] = temp
return inputlist
```

**The shuffling process is "with replacement"**, so the occurrence of each item may change! At least when when items in your list is also list.

E.g.,

```
ml = [[0], [1]] * 10
```

After,

```
random.shuffle(ml)
```

The number of [0] may be 9 or 8, but not exactly 10.

```
>>> import random
>>> a = ['hi','world','cat','dog']
>>> random.shuffle(a,random.random)
>>> a
['hi', 'cat', 'dog', 'world']
```

It works fine for me. Make sure to set the random method.

Source: Stackoverflow.com