For a list x of lists, when I want to get the first member of all the lists in x, I type x[:][0]. But it gives me the first list in x. Can someone explain why this is so?
Here is an example.
x=[[1,2],[3,4]]
print x[0][:]
print x[:][0]
I get same answer for both x[:][0] and x[0][:] I get the same answer, namely [1,2].
I am using Python 2.6.6.
Thanks in advance.
x[:] just returns the contents of x. So x[:][0] is the same as x[0]. There is no built-in support for slicing a list of lists "vertically". You have to use a list comprehension as suggested by @squiguy's comment.
x[:] simply creates a shallow copy of x. For this reason, x[:][0] is the same as x[0][:] (both are the same as x[0]).
Numpy is perfect for what you're trying to do, though.
x = numpy.array([[1,2], [3,4]])
x[:,0]
x[0,:]
Try This,
x=[[1,2],[3,4]]
print x[0][0]
print x[1][0]
or
for child in x:
print child[0]
To complement @BrenBarn answer, to solve your problem you could use:
first_numbers = [l[0] for l in x]
You are almost using the numpy syntax
>>> import numpy as np
>>> x = np.array([[1,2],[3,4]])
>>> x[:,0]
array([1, 3])
If you want to do tricky things with arrays, consider whether you should be using numpy
Use one of the below statements
print x[:][1]
or
print x[1][:]
or
print x[1]
all produces the same output.
[x[0] for x in nested]?