Question

我有以下代码片段，我想以这样的方式更改它，我想为每个changeInfo ['changeInfo'] [1..x]进行循环，我该怎么做

  for changeInfo in MainchangeInfo:
    if (changeInfo['CRStatus'] == 'Fix' and (('Not Provided' in changeInfo['changeInfo'][0]['Url'] or 'Wrong change Provided' in changeInfo['changeInfo'][0]['Url']) or 'NEW' in changeInfo['changeInfo'][0]['Status'] or 'ABANDONED' in changeInfo['changeInfo'][0]['Status'] or 'Yes' not in changeInfo['RNotesStatus'] or 'Provided' not in changeInfo['RCAInfo'] or 'False' in str(changeInfo['IsDevComplete']))):
      if 'Wrong change Provided' in changeInfo['changeInfo'][0]['Url'] or changeInfo['changeInfo'][0]['Info'] != 'Available' ://want to loop over for changeInfo['changeInfo'][0]
        changeMailBody = changeMailBody + "<tr bgcolor=\"Red\">"

Answer 1

[1..x]是什么意思？

如果要循环遍历changeInfo['changeInfo']中的“前x个”元素（在这种情况下，您的意思是[0..x]？），您可以使用切片表示法：

for item in changeInfo['changeInfo'][:x]:
    # do stuff

如果你真的想从1到x，即跳过你可以做的第一个元素：

for item in changeInfo['changeInfo'][1:x]:
    # do stuff

请参阅此处的文档：
http://docs.python.org/2.3/whatsnew/section-slices.html

编辑：

在评论中澄清之后，您甚至不需要切片表示法，只需要for item in changeInfo['changeInfo'] ...我认为您的完整示例如下：

for changeInfo in MainchangeInfo:
    for item in changeInfo['changeInfo']:
        if (changeInfo['CRStatus'] == 'Fix' and (('Not Provided' in item['Url'] or 'Wrong change Provided' in item['Url']) or 'NEW' in item['Status'] or 'ABANDONED' in item['Status'] or 'Yes' not in changeInfo['RNotesStatus'] or 'Provided' not in changeInfo['RCAInfo'] or 'False' in str(changeInfo['IsDevComplete']))):
            if 'Wrong change Provided' in item['Url'] or item['Info'] != 'Available' ://want to loop over for changeInfo['changeInfo'][0]
                changeMailBody = changeMailBody + "<tr bgcolor=\"Red\">"

Answer 2

您可以使用显式循环执行此操作：

for change in changeInfo['changeInfo']:
    info = change['info']
    # do stuff with it

或者，如果你想把它写成一个理解：

infos = [change['info'] for change in changeInfo['changeInfo']]

或者，如果你只想要一个迭代器，而不是一个列表：

infos = (change['info'] for change in changeInfo['changeInfo'])

在任何情况下，您都会获得changeInfo['changeInfo'][0]['info']，然后是changeInfo['changeInfo'][1]['info']，依此类推。

我在这里假设changeInfo['changeInfo']是一个列表，而不是一个dict，它的键恰好是小整数。如果该假设有误，请在您选择的任何解决方案中将changeInfo['changeInfo']替换为changeInfo['changeInfo'].values()。（当然，在这种情况下，订单将是任意的，但这正是您对dict的期望。）

因此，在您现有的代码中，而不是：

  if ('Wrong change Provided' in changeInfo['changeInfo'][0]['Url'] or
      changeInfo['changeInfo'][0]['Info'] != 'Available'):

...我将使用生成器表达式和any函数，如下所示：

  if any('Wrong change Provided' in change['Url'] or change['Info'] != 'Available'
         for change in changeInfo['changeInfo']):

现在，不是检查changeInfo['changeInfo']中的第一个值是否为真，即changeInfo['changeInfo'][0] - 您正在测试是否属实changeInfo['changeInfo']中任何的值。据我所知，这正是你想要的。

如何遍历每个changeInfo ['changeInfo'] [1 ... x] ['Url']

2 个答案: