NumPy first and last element from array
Question:
I am trying to dynamically get the first and last element from an array.
So, let us suppose the array has 6 elements.
test = [1,23,4,6,7,8]
If I am trying to get the first and last = 1,8, 23,7 and 4,6. Is there a way to get elements in this order?
I looked at a couple of questions Link Link2. I took help of these links and I came up with this prototype..
#!/usr/bin/env python
import numpy
test = [1,23,4,6,7,8]
test1 = numpy.array([1,23,4,6,7,8])
len_test = len(test)
first_list = [0,1,2]
len_first = len(first_list)
second_list = [-1,-2,-3]
len_second = len(second_list)
for a in range(len_first):
print numpy.array(test)[[first_list[a] , second_list[a]]]
print test1[[first_list[a], second_list[a]]]
But this prototype won’t scale for if you have more than 6 elements. So, I was wondering if there is way to dynamically get the pair of elements.
Thanks!
Answers:
>>> test = [1,23,4,6,7,8]
>>> from itertools import izip_longest
>>> for e in izip_longest(test, reversed(test)):
print e
(1, 8)
(23, 7)
(4, 6)
(6, 4)
(7, 23)
(8, 1)
Another option
>>> test = [1,23,4,6,7,8]
>>> start, end = iter(test), reversed(test)
>>> try:
while True:
print map(next, [start, end])
except StopIteration:
pass
[1, 8]
[23, 7]
[4, 6]
[6, 4]
[7, 23]
[8, 1]
How about:
In [10]: arr = numpy.array([1,23,4,6,7,8])
In [11]: [(arr[i], arr[-i-1]) for i in range(len(arr) // 2)]
Out[11]: [(1, 8), (23, 7), (4, 6)]
Depending on the size of arr, writing the entire thing in NumPy may be more performant:
In [41]: arr = numpy.array([1,23,4,6,7,8]*100)
In [42]: %timeit [(arr[i], arr[-i-1]) for i in range(len(arr) // 2)]
10000 loops, best of 3: 167 us per loop
In [43]: %timeit numpy.vstack((arr, arr[::-1]))[:,:len(arr)//2]
100000 loops, best of 3: 16.4 us per loop
How about this?
>>> import numpy
>>> test1 = numpy.array([1,23,4,6,7,8])
>>> forward = iter(test1)
>>> backward = reversed(test1)
>>> for a in range((len(test1)+1)//2):
... print forward.next(), backward.next()
...
1 8
23 7
4 6
The (len(test1)+1)//2 ensures that the middle element of odd length arrays is also returned:
>>> test1 = numpy.array([1,23,4,9,6,7,8]) # additional element '9' in the middle
>>> forward = iter(test1)
>>> backward = reversed(test1)
>>> for a in range((len(test1)+1)//2):
... print forward.next(), backward.next()
1 8
23 7
4 6
9 9
Using just len(test1)//2 will drop the middle elemen of odd length arrays.
Assuming the list has a even number of elements, you could do:
test = [1,23,4,6,7,8]
test_rest = reversed(test[:len(test)/2])
for n in len(test_rest):
print [test[n], test_test[n]]
This does it. Note that with an odd number of elements the one in the middle won’t be included.
test = [1, 23, 4, 6, 7, 8, 5]
for i in range(len(test)/2):
print (test[i], test[-1-i])
Output:
(1, 5)
(23, 8)
(4, 7)
Using Numpy’s fancy indexing:
>>> test
array([ 1, 23, 4, 6, 7, 8])
>>> test[::-1] # test, reversed
array([ 8, 7, 6, 4, 23, 1])
>>> numpy.vstack([test, test[::-1]]) # stack test and its reverse
array([[ 1, 23, 4, 6, 7, 8],
[ 8, 7, 6, 4, 23, 1]])
>>> # transpose, then take the first half;
>>> # +1 to cater to odd-length arrays
>>> numpy.vstack([test, test[::-1]]).T[:(len(test) + 1) // 2]
array([[ 1, 8],
[23, 7],
[ 4, 6]])
vstack copies the array, but all the other operations are constant-time pointer tricks (including reversal) and hence are very fast.
I ended here, because I googled for “python first and last element of array”, and found everything else but this. So here’s the answer to the title question:
a = [1,2,3]
a[0] # first element (returns 1)
a[-1] # last element (returns 3)
You can simply use take method and index of element (Last index can be -1).
arr = np.array([1,2,3])
last = arr.take(-1)
# 3
arr = np.array([1,2,3,4])
arr[-1] # last element
I am trying to dynamically get the first and last element from an array.
So, let us suppose the array has 6 elements.
test = [1,23,4,6,7,8]
If I am trying to get the first and last = 1,8, 23,7 and 4,6. Is there a way to get elements in this order?
I looked at a couple of questions Link Link2. I took help of these links and I came up with this prototype..
#!/usr/bin/env python
import numpy
test = [1,23,4,6,7,8]
test1 = numpy.array([1,23,4,6,7,8])
len_test = len(test)
first_list = [0,1,2]
len_first = len(first_list)
second_list = [-1,-2,-3]
len_second = len(second_list)
for a in range(len_first):
print numpy.array(test)[[first_list[a] , second_list[a]]]
print test1[[first_list[a], second_list[a]]]
But this prototype won’t scale for if you have more than 6 elements. So, I was wondering if there is way to dynamically get the pair of elements.
Thanks!
>>> test = [1,23,4,6,7,8]
>>> from itertools import izip_longest
>>> for e in izip_longest(test, reversed(test)):
print e
(1, 8)
(23, 7)
(4, 6)
(6, 4)
(7, 23)
(8, 1)
Another option
>>> test = [1,23,4,6,7,8]
>>> start, end = iter(test), reversed(test)
>>> try:
while True:
print map(next, [start, end])
except StopIteration:
pass
[1, 8]
[23, 7]
[4, 6]
[6, 4]
[7, 23]
[8, 1]
How about:
In [10]: arr = numpy.array([1,23,4,6,7,8])
In [11]: [(arr[i], arr[-i-1]) for i in range(len(arr) // 2)]
Out[11]: [(1, 8), (23, 7), (4, 6)]
Depending on the size of arr, writing the entire thing in NumPy may be more performant:
In [41]: arr = numpy.array([1,23,4,6,7,8]*100)
In [42]: %timeit [(arr[i], arr[-i-1]) for i in range(len(arr) // 2)]
10000 loops, best of 3: 167 us per loop
In [43]: %timeit numpy.vstack((arr, arr[::-1]))[:,:len(arr)//2]
100000 loops, best of 3: 16.4 us per loop
How about this?
>>> import numpy
>>> test1 = numpy.array([1,23,4,6,7,8])
>>> forward = iter(test1)
>>> backward = reversed(test1)
>>> for a in range((len(test1)+1)//2):
... print forward.next(), backward.next()
...
1 8
23 7
4 6
The (len(test1)+1)//2 ensures that the middle element of odd length arrays is also returned:
>>> test1 = numpy.array([1,23,4,9,6,7,8]) # additional element '9' in the middle
>>> forward = iter(test1)
>>> backward = reversed(test1)
>>> for a in range((len(test1)+1)//2):
... print forward.next(), backward.next()
1 8
23 7
4 6
9 9
Using just len(test1)//2 will drop the middle elemen of odd length arrays.
Assuming the list has a even number of elements, you could do:
test = [1,23,4,6,7,8]
test_rest = reversed(test[:len(test)/2])
for n in len(test_rest):
print [test[n], test_test[n]]
This does it. Note that with an odd number of elements the one in the middle won’t be included.
test = [1, 23, 4, 6, 7, 8, 5]
for i in range(len(test)/2):
print (test[i], test[-1-i])
Output:
(1, 5)
(23, 8)
(4, 7)
Using Numpy’s fancy indexing:
>>> test
array([ 1, 23, 4, 6, 7, 8])
>>> test[::-1] # test, reversed
array([ 8, 7, 6, 4, 23, 1])
>>> numpy.vstack([test, test[::-1]]) # stack test and its reverse
array([[ 1, 23, 4, 6, 7, 8],
[ 8, 7, 6, 4, 23, 1]])
>>> # transpose, then take the first half;
>>> # +1 to cater to odd-length arrays
>>> numpy.vstack([test, test[::-1]]).T[:(len(test) + 1) // 2]
array([[ 1, 8],
[23, 7],
[ 4, 6]])
vstack copies the array, but all the other operations are constant-time pointer tricks (including reversal) and hence are very fast.
I ended here, because I googled for “python first and last element of array”, and found everything else but this. So here’s the answer to the title question:
a = [1,2,3]
a[0] # first element (returns 1)
a[-1] # last element (returns 3)
You can simply use take method and index of element (Last index can be -1).
arr = np.array([1,2,3])
last = arr.take(-1)
# 3
arr = np.array([1,2,3,4])
arr[-1] # last element