[float(i) for i in lst]

Question

prorgamming newbie--I was looking for answers to an exercise I was doing and got my answers from here. My question is this--from that thread, the one chosen as best answer, was this code

[float(i) for i in lst]

The code did what it was supposed to do, but when I tried to get to that new list, I am getting errors

>>> xs = '12 10 32 3 66 17 42 99 20'.split()
>>> [float(i) for i in xs]
[12.0, 10.0, 32.0, 3.0, 66.0, 17.0, 42.0, 99.0, 20.0]
>>> i
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'i' is not defined

How should I do it?

Thanks!

Whenever the python prompt prints something out, it's because whatever you just typed returned a value. You can keep that value by assigning a variable to the expression you typed, but then nothing will be printed out. — SingleNegationElimination, May 15 '11 at 22:57

score 7 · Accepted Answer · answered May 15 '11 at 22:52

7

You have to assign [float(i) for i in xs] to something:

>>> new_list = [float(i) for i in xs]
>>> new_list
[12.0, 10.0, 32.0, 3.0, 66.0, 17.0, 42.0, 99.0, 20.0]
>>> new_list[0]
12.0
>>> new_list[5]
17.0

answered May 15 '11 at 22:52

hyperslug

3,473
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score 3 · Answer 2 · answered May 16 '11 at 00:36

3

Using map:

xs = '12 10 32 3 66 17 42 99 20'.split()
new_xs = map(float, xs)

answered May 16 '11 at 00:36

riza

16,274
7
29
29

2

Using [`map()`](http://docs.python.org/library/functions.html#map) is certainly the neatest way to achieve this. – johnsyweb May 16 '11 at 02:09

score 1 · Answer 3 · answered May 15 '11 at 22:57

1

Unlike for loops that leave the last iteration bound to the variable the var inside a list comp stops existing after evaluation.

answered May 15 '11 at 22:57

Jakob Bowyer

33,878
8
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score 0 · Answer 4 · answered May 15 '11 at 22:52

0

i only exists inside the list comprehension.

You want to say:

i = [float(i) for i in lst]

answered May 15 '11 at 22:52

mdonoughe

515
5
12

2

Using `i` as a variable inside and outside of the list comprehension could lead to unnecessary confusion. – johnsyweb May 15 '11 at 22:56
or rather, maybe what you really *want* to say is `floatlist = [....`, since naming the list after the iterator is a bit confusing. – SingleNegationElimination May 15 '11 at 22:58
1

for completeness, `i` only exists inside the list comprehension in Python 3. In Python 2, you can access `i` outside. – carl May 15 '11 at 23:03

score 0 · Answer 5 · answered May 16 '11 at 03:15

This is something related to Python versions. Look: On Python 2, if you do:

>>> i = 1234
>>> j = [i for i in xrange(10)]
>>> print i
9

But this was fixed in Python 3:

>>> i = 1234
>>> j = [i for i in range(10)]
>>> print(i)
1234

If you have Python 3.0 or superior, the variable will only be available in the comprehension. It will not affect the rest of the environment

score 0 · Answer 6 · answered May 16 '11 at 07:44

As an alternative to assigning a name to your list comprehension, in Python interactive mode, the symbol _ is automatically assigned to the last expression evaluated by the interpreter. So you could do the following:

>>> xs = '12 10 32 3 66 17 42 99 20'.split()
>>> [float(i) for i in xs]
[12.0, 10.0, 32.0, 3.0, 66.0, 17.0, 42.0, 99.0, 20.0]
>>> _
[12.0, 10.0, 32.0, 3.0, 66.0, 17.0, 42.0, 99.0, 20.0]
>>> _[2]
32.0
>>> _
32.0

[float(i) for i in lst]

6 Answers6