How to sort an array based on two conditions in python

x[0] and x[1] == 1

This is a logical expression which will either evaluate either as True (if x[0] != 0 and x[1] == 1) or False. So your sort key can only take two possible values.

By my understanding, what you want is:

• all cases where x[1] == 1 to appear after all the cases where x[1] != 1
• subject to above, outputs to be sorted by x[0]

You can't do this easily with a one-dimensional key. To understand why, think about the range of possible inputs in between [-Inf,0] and [Inf,0] and the range between [-Inf,1] and [Inf,1].

Each of these ranges is infinitely large in both directions. If you want to sort them with a one-dimensional key, you somehow need to map two double-endedly infinite ranges onto one number line.

This isn't impossible - if you really had to do it that way, there are tricks you could use to make that happen. But it's a very roundabout way to solve the problem.

It's much easier just to use a two-dimensional key. The sort function checks the zero'th position of a tuple first, and only goes to the first position as a tiebreaker. So, because the value of x[1] takes priority in the sort order you want, the zeroth entry of the tuple is based on x[1] and the first is based on x[0], like this:

x=[[8, 1], [1, 1], [3, 1], [8, 1], [4, 0], [2, 0]]
sorted(x,key=lambda x:(x[1]==1,x[0]))

Output:

[[2, 0], [4, 0], [1, 1], [3, 1], [8, 1], [8, 1]]