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Slicing tuples backwards - how to deal with the first element?

Let's say I need elements 1-3 of a tuple, back:

x = (0,1,2,3,4)
x[1:4]

Returns (1,2,3)

and

x[3:0:-1]

Returns (3,2,1)

But what if I want items 0-2?

I can do x[2::-1]

to get the correct answer (2,1,0)

, but it x[2:0:-1]

returns (2,1)

and x[2:-1:-1]

returns ()

.

Is there a way to cut off the last element without using an operator if

if I cut it at unknown intervals?

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4 answers


One slightly inefficient way:

x[0:3][::-1]

Which is equivalent:



tuple(reversed(x[0:3]))

I'm not sure how well the unneeded intermediate tuple will be optimized.

+3


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Instead of an empty element, you can use None

:



x[2:None:-1]
+2


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with x[-n:]

you can get the last index n

, and with with [::-1]

you can invert it!

>>> x[-2:][::-1]
(4, 3)
>>>
0


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Possibly merging a reverse-frequent tuple with your original tuple and an index that:

>>> x = (0, 1, 2, 3)
>>> y = (0,)

Then:

>>> (y + x)[3:0:-1]
(2, 1, 0)
>>> (y + x)[2:0:-1]
(1, 0)

Not as fast as Peter DeGlopper's solution:

#!/usr/bin/env python

import time

x = tuple(range(0,int(1e6)))
y = (0,)

start_time = time.time()
a = (y+x)[3:0:-1]
print("tuple add --- %s seconds ---" % (time.time() - start_time))

start_time = time.time()
b = x[0:3][::-1]
print("tuple rev --- %s seconds ---" % (time.time() - start_time))

Not even close, really:

$ ./test61.py
tuple add --- 0.0153260231018 seconds ---
tuple rev --- 6.19888305664e-06 seconds ---
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