python 列表解析示例

Question

我想知道如何解析（或拆分）列表的元素？

我有一个（字符串的）列表列表，例如：

resultList =  [['TWP-883 PASS'], ['TWP-1080 PASS'], ['TWP-1081 PASS']]

其中：

resultList[0] = ['TWP-883 PASS']
resultList[1] = ['TWP-1080 PASS']

本质上，我需要一个变量来表示列表中每个元素中的两个条目。例如：

issueId = 'TWP-883'
status = 'PASS'

什么允许遍历此列表并像上面那样进行解析？

Answer 1

好吧，就这么简单：

# You can also assign as you iterate as suggested in the comments.
for issue, status in resultList:
    print issue, status

这输出

TWP-883 PASS
TWP-1080  PASS
TWP-1081  PASS
TWP-1082  PASS
TWP-884  FAIL
TWP-885  PASS

这是另一个例子：

>>> x = [1, 2] # or (1, 2), or '12' works with collections
>>> y, z = x
>>> y
1
>>> z
2
>>>

顺带一提，在Python3.x中，您还可以：

In [1]: x = [1, 2, 3, 4]
In [2]: y, z, *rest = x
In [3]: y
Out[3]: 1
In [4]: z
Out[4]: 2
In [5]: rest
Out[5]: [3, 4]

Answer 2

您只需要一个简单的 for 循环来利用 Python 的元组解包机制。

for issueId, status in resultList:
    # Do stuff with issueId and status

Answer 3

如果你以后想做更多的转换，使用genexp，

g = (func(issueid, status) for issueid, status in resultList) # func returns non-None objects

如果你只想使用可迭代对象，

for issueid, status in resultList:
    # print(issueid, status)
    # whatever

Answer 4

注意：我更改了这个答案以反映对问题的编辑。具体来说，我添加了一个 split() 将嵌套列表中的字符串分隔为两个字符串（issueId 和 status）。

我会使用列表和字典理解将您的列表列表转换为具有键 issueId 和 status:

的字典列表

resultList =  [['TWP-883 PASS'], ['TWP-1080 PASS'], ['TWP-1081 PASS']]

result_dicts = [{("issueId","status")[x[0]]:x[1] for x in enumerate(lst[0].split())} for lst in resultList]

现在可以通过以下方式进行查找：

>>> result_dicts[0]["status"]
'PASS'
>>> result_dicts[0]["issueId"]
'TWP-883'
>>> result_dicts[1]
{'status': 'PASS', 'issueId': 'TWP-1080'}
>>>

要为列表中每个字典中的每个值声明变量并打印它们，请使用以下代码：

for entry in result_dicts:
    issueId = entry["issueId"]
    status = entry["status"]
    print("The status of {0: <10} is {1}".format(issueId, status))

输出：

The status of TWP-883    is PASS
The status of TWP-1080   is PASS
The status of TWP-1081   is PASS

Answer 5

您可以通过

获取字符串列表

issueId = [y for x in resultList for (i,y) in enumerate(x.split()) if(i%2==0)]
status = [y for x in resultList for (i,y) in enumerate(x.split()) if(i%2==1)]

查看每个问题 ID 和您可以使用的相应状态

for id,st in zip(issueId,status):
    print(id, " : ", st)

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