[float(i) for i in lst]

prorgamming newbie--I was looking for answers to an exercise I was doing and got my answers from 开发者_如何学Pythonhere. 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!

---

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

---

Using map:

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

---

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

---

i only exists inside the list comprehension.

You want to say:

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

---

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

---

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