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How to convert a boolean array to an int array

I use Scilab, and want to convert an array of booleans into an array of integers:

>>> x = np.array([4, 3, 2, 1])
>>> y = 2 >= x
>>> y
array([False, False,  True,  True], dtype=bool)

In Scilab I can use:

>>> bool2s(y)
0.    0.    1.    1.  

or even just multiply it by 1:

>>> 1*y
0.    0.    1.    1.  

Is there a simple command for this in Python, or would I have to use a loop?

Are you asking for a way to convert a boolean array into an integer one without scipy, numpy and the like?
There's a separate way of formatting code. You don't have to use blockquote. It's done by indenting, and the curly braces button above the question editor will do it for you. Check it out.
Sukrit, I don't care if i have to use scipy, numpy or any other python module package.

t
the Tin Man

Numpy arrays have an astype method. Just do y.astype(int).

Note that it might not even be necessary to do this, depending on what you're using the array for. Bool will be autopromoted to int in many cases, so you can add it to int arrays without having to explicitly convert it:

>>> x
array([ True, False,  True], dtype=bool)
>>> x + [1, 2, 3]
array([2, 2, 4])

yes, i can also type x*1...and it does the same thing scilab does....*feels like dumbass now*.. thank you everyone for you help!....although the answer was right in my question, i really liked getting the variety of answers and seeing all the different ways to do it. Really opened my mind regarding python.
Re boolean arrays being autopromoted: unfortunately, numpy is not consistent with this. Try subtracting two boolean arrays, and you get a TypeError and a deprecation message.
t
the Tin Man

The 1*y method works in Numpy too:

>>> import numpy as np
>>> x = np.array([4, 3, 2, 1])
>>> y = 2 >= x
>>> y
array([False, False,  True,  True], dtype=bool)
>>> 1*y                      # Method 1
array([0, 0, 1, 1])
>>> y.astype(int)            # Method 2
array([0, 0, 1, 1]) 

If you are asking for a way to convert Python lists from Boolean to int, you can use map to do it:

>>> testList = [False, False,  True,  True]
>>> map(lambda x: 1 if x else 0, testList)
[0, 0, 1, 1]
>>> map(int, testList)
[0, 0, 1, 1]

Or using list comprehensions:

>>> testList
[False, False, True, True]
>>> [int(elem) for elem in testList]
[0, 0, 1, 1]

so, y = 1 if x else 0 is the same as y = 1 if x>0 else 0 and the same as if x: y = 1 ""NEXT LINE"" else: y = 0....how did you learn those tricks, i didn't see it in the if statement documentation?
No. y=1 if x else 0 is not the same as y=1 if x>0 else 0, since the latter doesn't take the negative numbers into consideration. This is just what Python defines as True or False, these are all in the documentation.
t
the Tin Man

Using numpy, you can do:

y = x.astype(int)

If you were using a non-numpy array, you could use a list comprehension:

y = [int(val) for val in x]

t
the Tin Man

Most of the time you don't need conversion:

>>>array([True,True,False,False]) + array([1,2,3,4])
array([2, 3, 3, 4])

The right way to do it is:

yourArray.astype(int)

or

yourArray.astype(float)

T
Tomas G.

A funny way to do this is

>>> np.array([True, False, False]) + 0 
np.array([1, 0, 0])

t
the Tin Man

I know you asked for non-looping solutions, but the only solutions I can come up with probably loop internally anyway:

map(int,y)

or:

[i*1 for i in y]

or:

import numpy
y=numpy.array(y)
y*1

yes, the looping is slow. from what i've read, if you need to do some time critical crunching you should call c from python. Do you know any references for doing this? also, thank you for your help. surprised how fast everyone responded!