为什么我的清单不遍历每个清单？

Question

Currently working through tutorial "Python the hard way".

I'm learning about lists and loops (ex32).

At the end of the exercise Zed (Tutorial author) tells us to play around, which I have done.

# we can also build lists, first start with an empty one
elements = []
elements.append(range(0,6))

# then use the range function to do 0 to 5 counts
for element in elements:
    print "Adding %s to elements" % element


# now we can print them out too
for element in elements:
    print"Element was: %s" % element

This produces output like so:

Adding [0, 1, 2, 3, 4, 5] to elements
Element was: [0, 1, 2, 3, 4, 5]

I had expected to see something like this:

Adding 0 to elements
Adding 1 to elements
Adding 2 to elements
Adding 3 to elements
Adding 4 to elements
Adding 5 to elements
Element was: 0
Element was: 1
Element was: 2
Element was: 3
Element was: 4
Element was: 5

But instead Python wants to print out my script in a oner, rather than the concatenated string with each list component.

I know that I could change the script to reflect the authors script exactly

# we can also build lists, first start with an empty one
elements = []

# then use the range function to do 0 to 5 counts
for i in range(0, 6):
    print "Adding %d to the list." % i
    # append is a function that lists understand
    elements.append(i)

# now we can print them out too
for i in elements:
    print "Element was: %d" % i

but I'd just like to know why my piece does not work as expected?

Answer 1

you are appending a list to a list! you just want to create the list!

all you need to do is change the following:

elements = []
elements.append(range(0,6))

into

elements = range(0,6)

and you will get your expected results

why

when you created elements for the first time, it was a blank list. Then you appended range(0,6) to your empty list. Now elements looks like [[0,1,2,3,4,5]] (or [range(0,6)] ), which is a list with one element, a list.

Answer 2

This is because elements contains exactly one element , which is a list : [0, 1, 2, 3, 4, 5] . list.append() adds an item to the end of the list.

In [1]: elements = []

In [2]: elements.append(range(0,6))

In [3]: elements
Out[3]: [[0, 1, 2, 3, 4, 5]]

Perhaps you meant to extend the list:

In [1]: elements = []

In [2]: elements.extend(range(0, 6))

In [3]: elements
Out[3]: [0, 1, 2, 3, 4, 5]

Or replace it?

In [4]: elements = range(0,6)

In [5]: elements
Out[5]: [0, 1, 2, 3, 4, 5]

Or even:

In [6]: elements = [element for element in range(0,6)]

In [7]: elements
Out[7]: [0, 1, 2, 3, 4, 5]

The list comprehension is unnecessary in this example, but it demonstrates how it is easy to filter or map those elements.

Answer 3

.append adds a single element to the list. That single element is range(0, 6) , which is [0, 1, 2, 3, 4, 5] (Johnsyweb got it in before me). You could use .extend to append each one .

Answer 4

elements = []


elements.append(range(0,6)) 
# appends range(0,6) to elements. range(0,6) creates a list in Python 2.x but only in Python 2.x. thanks to adsmith for pointing this out.

print elements

[[0, 1, 2, 3, 4, 5]] # it's clear that elements has 1 element. A list.

that's why

for i in elements:
    print "adding %s to elements" % i

produces:

adding [0,1,2,3,4,5] to elements

To get the output you want:

elements = range(0,6)

OR

elements = [i for i in range(0,6)] # list comprehension

then

for i in elements:
    print "adding %s to elements" % i

outputs what you want