i have a numpy array like the following
[[ 1 2 3 4 ]
[ 5 6 7 8 ]
......... ]
So basically I want to create 4 (can be different) lists where
list_1 = [1,5...], list_2 = [2,6....] and so on.
What is the pythonic way to do this?
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3 Answers
Given a numpy array
>>> numpy.array([[x for x in xrange(i,i+5)] for i in xrange(0,100,10)])
array([[ 0, 1, 2, 3, 4],
[10, 11, 12, 13, 14],
[20, 21, 22, 23, 24],
[30, 31, 32, 33, 34],
[40, 41, 42, 43, 44],
[50, 51, 52, 53, 54],
[60, 61, 62, 63, 64],
[70, 71, 72, 73, 74],
[80, 81, 82, 83, 84],
[90, 91, 92, 93, 94]])
You first have to transpose it
>>> narray=numpy.array([[x for x in xrange(i,i+5)] for i in xrange(0,100,10)])
>>> tarray=numpy.transpose(narray)
array([[ 0, 10, 20, 30, 40, 50, 60, 70, 80, 90],
[ 1, 11, 21, 31, 41, 51, 61, 71, 81, 91],
[ 2, 12, 22, 32, 42, 52, 62, 72, 82, 92],
[ 3, 13, 23, 33, 43, 53, 63, 73, 83, 93],
[ 4, 14, 24, 34, 44, 54, 64, 74, 84, 94]])
and then convert to list
>>> larray=tarray.tolist()
>>> larray
[[0, 10, 20, 30, 40, 50, 60, 70, 80, 90], [1, 11, 21, 31, 41, 51, 61, 71, 81, 91], [2, 12, 22, 32, 42, 52, 62, 72, 82, 92], [3, 13, 23, 33, 43, 53, 63, 73, 83, 93], [4, 14, 24, 34, 44, 54, 64, 74, 84, 94]]
now you can index larray as larray[0], larray[1] to get the individual lists.
2 Comments
Henry Gomersall
If you're happy with the elements stored in a 1D numpy array, you can just index the transposed array, like:
tarray[n] for the nth column. If you really want each element in a list, you can convert it as you need it: list(tarray[n])Henry Gomersall
Using a transpose to get a column is a bit opaque. Slicing (as described by @Anthony Kong) is a more transparent way to do it. As above, if you really want it as a list, something like
list_n = list(a[:,n]) for the nth column (where a is the numpy array).Given this
>>> a = array([[1,2,3,4], [5,6,7,8], [9, 10,11,12]])
>>> a
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
you can use slicing
>>> list1 = a[:,0]
>>> list1
array([1, 5, 9])
>>> list1 = a[:,1]
>>> list1
array([ 2, 6, 10])
Comments
a=[[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]
list1=[]
list2=[]
list3=[]
list4=[]
for x in a:
for i,y in enumerate(x):
if i==0:
list1.append(y)
elif i==1:
list2.append(y)
elif i==2:
list3.append(y)
elif i==3:
list4.append(y)
>>> print(list1)
[1, 5, 9, 13]
>>> print(list2)
[2, 6, 10, 14]
>>> print(list3)
[3, 7, 11, 15]
>>> print(list4)
[4, 8, 12, 16]
Dynamically :
>>>a=[[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]
>>>list=[[] for _ in range(len(a))]
>>>for x in a:
for i,y in enumerate(x):
list[i].append(y)
>>>print(list)
[[1, 5, 9, 13], [2, 6, 10, 14], [3, 7, 11, 15], [4, 8, 12, 16]]
answered Mar 25, 2012 at 8:59
Ashwini Chaudhary
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2 Comments
frazman
though this would fail when the number of list are more than 4.. not dynamic enough
Ashwini Chaudhary
actually I didn't read this part (want to create 4 (can be different)) of the question. Now I wrote a dynamic solution too.