Python：如何进行循环列表并将其追加到新列表

Question

Practicing my python.

Task: Loop through list A and create a new list with only items form list A that's between 0-5.

What am I doing wrong here

a = [100, 1, 10, 2, 3, 5, 8, 13, 21, 34, 55, 98]


def new_list(x):

    for item in range(len(x)):
        new = []

        if x[item] < 5 and x[item] > 0:
            (new.append(item))
            return new


print(new_list(a))

I'm just getting [1] as an answer.

Answer 1

You return command is inside the loop so as soon as it goes through the first case it returns the value exiting the function.

Here is an example of what your code should look like

a = [100, 1, 10, 2, 3, 5, 8, 13, 21, 34, 55, 98]


def new_list(x):
    new = []
    for item in range(len(x)):            

        if x[item] < 5 and x[item] > 0:
            new.append(x[item])
    return new


print new_list(a)

You can achieve the same result by using a list comprehension

def new_list(x):
    return [item for item in x if 0 < item < 5]

Answer 2

You're resetting new to a brand new empty list each time through the loop, which discards any work done in prior iterations.

Also, in the if statement you're calling return , which exits your function immediately, so you never process the remainder of the list.

You probably wanted something like this instead:

def new_list(x):
    new = []
    for item in x:
        if 0 < item < 5:
            new.append(item)
    return new

Answer 3

Just my recommendation. You could use filter() here instead of a making your own loop.

a = [100, 1, 10, 2, 3, 5, 8, 13, 21, 34, 55, 98]

def new_list(x, low=0, high=5):
    return filter(lambda f: f in range(low, high), x)

Filter returns a new list with elements passing a given predicate and it's equivalent to

[item for item in iterable if function(item)]

as per the documentation.

Therefore

print new_list(a)

Results in:

[1, 2, 3, 5]

This way you can check any values such as:

print new_list(a, 5, 10)
[5, 8]

Answer 4

Three errors:

1. you are reinstantiating new with each iteration of the for loop.
2. you should return new when the list is finished building, at the end of the function.
3. You are appending item , but this is your index. In your code, you would have to append x[item] .

Code with corrections:

a = [100, 1, 10, 2, 3, 5, 8, 13, 21, 34, 55, 98]

def new_list(x):
    new = []

    for item in range(len(x)):
        if x[item] < 5 and x[item] > 0:
            new.append(x[item])
    return new

print(new_list(a))

Output:

[1, 2, 3]

Suggestions:

1. Don't index, loop over the items of x directly ( for item in x: ... ).
2. Use chained comparisons, eg 0 < item < 5 .
3. Consider a list comprehension.

Code with all three suggestions:

>>> [item for item in a if 0 < item < 5]
>>> [1, 2, 3]

Answer 5

Just a suggestion!

1. The empty list is inside the For Loop meaning that a new empty list is created every iteration

2. The 'return' is also inside the for loop which is less than ideal, you want it to be returned after the loop has been exhausted and all suitable elements have been appended.

    a = [100, 1, 10, 2, 3, 5, 8, 13, 21, 34, 55, 98] def new_list(x): new = [] for item in range(len(x)): if x[item] < 5 and x[item] > 0: new.append(item) return new print(new_list(a))