Starting iteration from a nth row, and finish all the rows-CodePudding

So, I have this array:

a = [[ 1,  -1 ]
 [ 0,   1 ]
 [-1.5, -1 ]]

I want to start an iteration from the second row, and continue iteration until I have passed through all the array (thus, iterate in this index order: 1, 2, 0).

How do I do this in Python/numpy?

CodePudding user response：

With numpy you can use numpy.roll:

a = [[ 1, -1 ], [ 0, 1 ], [-1.5, -1 ]]

np.roll(a, -1, axis=0)

Output:

array([[ 0. ,  1. ],
       [-1.5, -1. ],
       [ 1. , -1. ]])

Or, rolling the indices:

for i in np.roll(range(len(a)), -1):
   # do something

CodePudding user response：

You can use % operator. n is your input

for i in range(len(a)):
    a[(i n)%len(a)] ....

CodePudding user response：

Using a list or tuple is possible just by using slice indexing.

>>> a = [1,2,3]
>>> a[1:] a[:1]
[2, 3, 1]
>>> [*a[1:],*a[:1]] #or if you prefer..
[2, 3, 1]

using the extend method would be something like this:

>>> a = [1,2,3]
>>> b = a[1:]
>>> b.extend(a[1:])
>>> b
[2,3,1]

as the other answer suggested you could use np.roll(x,-1,axis=0) with numpy

to iterate you simply do a for loop and it's done:

a = [1,2,3]
b = a[:1]
b.extend(a[1:])
for x in b:
   print(x,end=' ')
# 2 3 1

or you could use the modulo operator:

from functools import partial
mod = lambda x,y:x%y
a = [1,2,3]
l = len(a)
for x in map(partial(mod,y=l),range(l 1)):
    print(a[x],end=' ')
>>> 2 3 1

CodePudding user response：

Try this code

start_idx=1
for i in range(a.size):
    print(a[(i//a.shape[1]   start_idx)%a.shape[0], i%a.shape[1]])

0
1
-1.5
-1
1
-1