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问题描述

Python的新手,但我已经研究了几个小时了。如果我错过了一些明显的东西,请原谅我。

New to Python, but I have been researching this for a couple hours. Forgive me if I missed something obvious.

我有一个名为LineItem的类,它有一个属性_lineItems,属于给定LineItem的LineItem列表。一个子列表,基本上。

I have a class called LineItem, which has an attribute _lineItems, a list of LineItems that belong to the given LineItem. A sub-list, basically.

我想打印出一个LineItem及其所有子项(以及子项拥有的子项),但是我我遇到迭代问题。

I want to print out a LineItem and all of its sub-items (and the sub-items own sub-items), but I'm having trouble with the iteration.

from decimal import * 
class LineItem(object):
    """
    Instance attributes:
      amount: Decimal
      _lineItems: list of child internal lineitems (possibly an empty list)
      isInternal: bool
    """

    def __init__(self, **kw):
        self.amount = Decimal(0)
        self._lineItems = []
        self.isInternal = False
        for k, v in kw.items():
            setattr(self, k, v)

一个让我麻烦的LineItem示例在下面定义为ext2,有三个孩子。

An example LineItem that's giving me the trouble is defined below as ext2, with three children.

# External line item with one level of children
int1 = LineItem(amount=Decimal('1886.75'), description='State Dues',
         isInternal=True)
int2 = LineItem(amount=Decimal('232.50'), description='National Dues',
         isInternal=True)
int3 = LineItem(amount=Decimal('50'), description='Processing Fee',
         isInternal=True)
ext2 = LineItem(amount=Decimal('2169.25'), description='Dues',
         _lineItems=[int1, int2, int3])

我有这个递归函数来迭代所有子项(并打印它们编号,如1,2,2.1作为第二项的第一个子项,等等)

I have this recursive function for iterating all the sub-items (and printing them numbered, like 1, 2, 2.1 as the first subitem of the second item, etc)

def print_line_item(LineItems):
    count = 1
    for a in LineItems:
        print count, ' ', a.description, ' (', a.amount, ')'
        if a._lineItems != []:
            for b in a._lineItems:
                print count, '.', print_line_item(b),
        count+=1

但是当我尝试使用它时

def main():
    print_line_item([ext1, ext2, ext3]) #ext1 has no children, prints fine
if __name__=="__main__":
    main()

我得到

line 56, in print_line_item
    print count, '.', print_line_item(b),
line 51, in print_line_item
    for a in LineItems:
TypeError: 'LineItem' object is not iterable

好的,所以不知何故我搞砸了名单。

Okay, so somehow I'm screwing up lists.

如果我添加一个couple print:

if I add a couple print statements:

def print_line_item(LineItems):
    count = 1
    for a in LineItems:
        print count, ' ', a.description, ' (', a.amount, ')'
        if a._lineItems != []:
            print a._lineItems
            for b in a._lineItems:
                print b
                print count, '.', print_line_item(b),
        count+=1

我得到证明a._lineItems确实是一个列表,打印如下s:

I get proof that a._lineItems is indeed a list, printed as follows:

[<__main__.LineItem object at 0x0227C430>, <__main__.LineItem object at 0x0227C5F0>, <__main__.LineItem object at 0x0227C670>]

并且b我正试图传递给递归调用是一个LineItem的内存地址

and that the b I'm trying to pass to the recursing call is a memory address of a single LineItem

<__main__.LineItem object at 0x0227C430>

那我该怎么做才能做我想做的事情?我用.iter或 ___ iter ___ 尝试了几件事,但没有运气。

Sooo how am I actually supposed to do what I'm trying to do? I tried a couple things with .iter or ___iter___ but no luck.

另外,if a._lineItems! = []似乎也没有工作(也没有变化)。我得到打印行无

Also, the if a._lineItems != [] doesn't seem to be working either (nor variations on that). I get printed lines of "None"

推荐答案

def print_line_item(LineItems):
    count = 1
    for a in LineItems:
        print count, ' ', a.description, ' (', a.amount, ')'
        if a._lineItems != []:
            for b in a._lineItems:
                print count, '.', print_line_item(b),
        count+=1

它可能是正确的版本,未经过测试。

It might be correct version, not tested.

def print_line_item(LineItems, precedingNumber='1'):
    count = 1
    for a in LineItems:
        print precedingNumber, '.', count, ' ', a.description, ' (', a.amount, ')'
        print_line_item(a._lineItems, precedingNumber + '.' + count),
        count+=1

这篇关于对象列表的Python迭代“不可迭代”的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

09-16 07:58