How to change values in a tuple?
I have a tuple called values
which contains the following:
('275', '54000', '0.0', '5000.0', '0.0')
I want to change the first value (i.e., 275
) in this tuple but I understand that tuples are immutable so values[0] = 200
will not work. How can I achieve this?
It's possible via:
t = ('275', '54000', '0.0', '5000.0', '0.0')
lst = list(t)
lst[0] = '300'
t = tuple(lst)
But if you're going to need to change things, you probably are better off keeping it as a list
Depending on your problem slicing can be a really neat solution:
>>> b = (1, 2, 3, 4, 5)
>>> b[:2] + (8,9) + b[3:]
(1, 2, 8, 9, 4, 5)
>>> b[:2] + (8,) + b[3:]
(1, 2, 8, 4, 5)
This allows you to add multiple elements or also to replace a few elements (especially if they are "neighbours". In the above case casting to a list is probably more appropriate and readable (even though the slicing notation is much shorter).
Well, as Trufa has already shown, there are basically two ways of replacing a tuple's element at a given index. Either convert the tuple to a list, replace the element and convert back, or construct a new tuple by concatenation.
In [1]: def replace_at_index1(tup, ix, val):
...: lst = list(tup)
...: lst[ix] = val
...: return tuple(lst)
...:
In [2]: def replace_at_index2(tup, ix, val):
...: return tup[:ix] + (val,) + tup[ix+1:]
...:
So, which method is better, that is, faster?
It turns out that for short tuples (on Python 3.3), concatenation is actually faster!
In [3]: d = tuple(range(10))
In [4]: %timeit replace_at_index1(d, 5, 99)
1000000 loops, best of 3: 872 ns per loop
In [5]: %timeit replace_at_index2(d, 5, 99)
1000000 loops, best of 3: 642 ns per loop
Yet if we look at longer tuples, list conversion is the way to go:
In [6]: k = tuple(range(1000))
In [7]: %timeit replace_at_index1(k, 500, 99)
100000 loops, best of 3: 9.08 µs per loop
In [8]: %timeit replace_at_index2(k, 500, 99)
100000 loops, best of 3: 10.1 µs per loop
For very long tuples, list conversion is substantially better!
In [9]: m = tuple(range(1000000))
In [10]: %timeit replace_at_index1(m, 500000, 99)
10 loops, best of 3: 26.6 ms per loop
In [11]: %timeit replace_at_index2(m, 500000, 99)
10 loops, best of 3: 35.9 ms per loop
Also, performance of the concatenation method depends on the index at which we replace the element. For the list method, the index is irrelevant.
In [12]: %timeit replace_at_index1(m, 900000, 99)
10 loops, best of 3: 26.6 ms per loop
In [13]: %timeit replace_at_index2(m, 900000, 99)
10 loops, best of 3: 49.2 ms per loop
So: If your tuple is short, slice and concatenate. If it's long, do the list conversion!
It is possible with a one liner:
values = ('275', '54000', '0.0', '5000.0', '0.0')
values = ('300', *values[1:])
As Hunter McMillen mentioned, tuples are immutable, you need to create a new tuple in order to achieve this. For instance:
>>> tpl = ('275', '54000', '0.0', '5000.0', '0.0')
>>> change_value = 200
>>> tpl = (change_value,) + tpl[1:]
>>> tpl
(200, '54000', '0.0', '5000.0', '0.0')