How do replace a list with another list in python?

You probably meant to do this:

gamewords = newgamewords[:]  # i.e. copy newgamewords

Another alternative would be

gamewords = list(newgamewords)

I find the latter more readable.

Note that when you 'copy' a list like both of these approaches do, changes to the new copied list will not effect the original. If you simply assigned newgamewords to gamewords (i.e. gamewords = newgamewords), then changes to gamewords would effect newgamewords.

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• list

I'm not sure what you exactly want. There are two options:

1. gamewords = newgamewords[:]
2. gamewords = newgamewords

The difference is that the first option copies the elements of newgamewords and assigns it to gamewords. The second option just assigns a reference of newgamewords to gamewords. Using the second version, you would change the original newgamewords-list if you changed gamewords.

Because you didn't give more of your source code I can't decide which will work properly for you, you have to figure it out yourself.

The obvious choice of functions to use to select one, or a set number of items randomly from a larger list would be random.choice() or random.choices().

>>> gamewords= ['blue','violet','red','orange','fuchsia','cyan','magenta','azure','black','turquoise','pink','scarlet']
>>> random.choice(gamewords)
'turquoise'
>>> random.choice(gamewords)
'orange'
>>> random.choice(gamewords)
'cyan'
>>> random.choices(gamewords, k=3)
['fuchsia', 'orange', 'red']
>>> random.choices(gamewords, k=2)
['turquoise', 'black']
>>> 

Not sure if you were able to find the answer to this but I had a similar issue and I resolved it using this method:

for x, y in zip(original_col, np.arange(0, len(original_col), 1)):
    df['Term Ldesc'] = df['Term Ldesc'].replace(x, y)

Hope this helps!