Python 从元组列表创建一个列表

我有这样一份清单：

mylist = [(20, 'Start', '2008-10-10', 'TBS'),...,(20, 'End', '2008-11-09', 'NG'), 
          (21, 'Start', '2008-12-10', 'TBS'),...,(21, 'End', '2008-12-15', 'G'), 
          (22, 'Start', '2009-01-10', 'TBS'),...,(22, 'End', '2009-12-10', 'B'),..]

[[20, 'Start', '2008-10-10', 'End', '2008-11-09', 'NG'] ,
 [21, 'Start', '2008-12-10', 'End', '2008-12-15', 'G'], 
 [22, 'Start', '2009-01-10', 'End', '2009-12-10', 'B']]

[[[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[21, 'Start', '2008-12-10', 'End'], ['2008-12-15', 'G']], 
 [[21, 'Start', '2008-12-10', 'End'], ['2008-12-15', 'G']],
 [[22, 'Start', '2009-01-10', 'End'], ['2009-12-10', 'B']]]

code = 0
brr = []
for row in mylist:

    if row[1] == "Start":
        arr = []
        code = row[0]
        arr.extend([row[0], row[1], row[2]])
        # continue not needed here

    if row[0] == code and row[1] == "End":
        arr.extend([row[1], row[2], row[3]])

    if arr not in brr:
        brr.append(arr)

for k in brr:
    print(k)

我在上面的例子中加上“…”表示每个id都有其他项，比如列表中的20、21和22，但我不想要它们。我想要的唯一项目是包含“开始”或“结束”的项目。（其他项目的单词与这两个词不同。）

我想创建一个嵌套列表，如下所示：

mylist = [(20, 'Start', '2008-10-10', 'TBS'),...,(20, 'End', '2008-11-09', 'NG'), 
          (21, 'Start', '2008-12-10', 'TBS'),...,(21, 'End', '2008-12-15', 'G'), 
          (22, 'Start', '2009-01-10', 'TBS'),...,(22, 'End', '2009-12-10', 'B'),..]

[[20, 'Start', '2008-10-10', 'End', '2008-11-09', 'NG'] ,
 [21, 'Start', '2008-12-10', 'End', '2008-12-15', 'G'], 
 [22, 'Start', '2009-01-10', 'End', '2009-12-10', 'B']]

[[[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[21, 'Start', '2008-12-10', 'End'], ['2008-12-15', 'G']], 
 [[21, 'Start', '2008-12-10', 'End'], ['2008-12-15', 'G']],
 [[22, 'Start', '2009-01-10', 'End'], ['2009-12-10', 'B']]]

code = 0
brr = []
for row in mylist:

    if row[1] == "Start":
        arr = []
        code = row[0]
        arr.extend([row[0], row[1], row[2]])
        # continue not needed here

    if row[0] == code and row[1] == "End":
        arr.extend([row[1], row[2], row[3]])

    if arr not in brr:
        brr.append(arr)

for k in brr:
    print(k)

这是我的密码：

code = 0
brr = []
for row in myList:
    if row[1] == "Start":
        arr = []
        code = row[0]
        arr.append([row[0], row[1], row[2]])
        continue

    if row[0] == code and row[1] == "End":
        arr.append([row[1], row[2], row[3]])
    brr.append(arr)
for k in brr:
    print(k)

但问题是它产生了这样的东西：

mylist = [(20, 'Start', '2008-10-10', 'TBS'),...,(20, 'End', '2008-11-09', 'NG'), 
          (21, 'Start', '2008-12-10', 'TBS'),...,(21, 'End', '2008-12-15', 'G'), 
          (22, 'Start', '2009-01-10', 'TBS'),...,(22, 'End', '2009-12-10', 'B'),..]

[[20, 'Start', '2008-10-10', 'End', '2008-11-09', 'NG'] ,
 [21, 'Start', '2008-12-10', 'End', '2008-12-15', 'G'], 
 [22, 'Start', '2009-01-10', 'End', '2009-12-10', 'B']]

[[[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[21, 'Start', '2008-12-10', 'End'], ['2008-12-15', 'G']], 
 [[21, 'Start', '2008-12-10', 'End'], ['2008-12-15', 'G']],
 [[22, 'Start', '2009-01-10', 'End'], ['2009-12-10', 'B']]]

code = 0
brr = []
for row in mylist:

    if row[1] == "Start":
        arr = []
        code = row[0]
        arr.extend([row[0], row[1], row[2]])
        # continue not needed here

    if row[0] == code and row[1] == "End":
        arr.extend([row[1], row[2], row[3]])

    if arr not in brr:
        brr.append(arr)

for k in brr:
    print(k)

对于每个项目，我在列表中有多行。我不知道为什么？ 如果我的问题太长，很抱歉。

您需要使用arr.extend（）函数

您的

brr.append（arr）

总是为每行添加一个数组，这就是为什么输出中有6个元素。将

brr.append（arr）

更改为：

if arr not in brr:
    brr.append(arr)

至于格式，

arr.append（[row[0]，row[1]，row[2]]）

添加了一个包含3个元素的列表，而不是3个单独的元素。改用

您的最终代码应该如下所示：

mylist = [(20, 'Start', '2008-10-10', 'TBS'),...,(20, 'End', '2008-11-09', 'NG'), 
          (21, 'Start', '2008-12-10', 'TBS'),...,(21, 'End', '2008-12-15', 'G'), 
          (22, 'Start', '2009-01-10', 'TBS'),...,(22, 'End', '2009-12-10', 'B'),..]

[[20, 'Start', '2008-10-10', 'End', '2008-11-09', 'NG'] ,
 [21, 'Start', '2008-12-10', 'End', '2008-12-15', 'G'], 
 [22, 'Start', '2009-01-10', 'End', '2009-12-10', 'B']]

[[[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[20, 'Start', '2008-10-10', 'End'], ['2008-11-09', 'NG']] ,
 [[21, 'Start', '2008-12-10', 'End'], ['2008-12-15', 'G']], 
 [[21, 'Start', '2008-12-10', 'End'], ['2008-12-15', 'G']],
 [[22, 'Start', '2009-01-10', 'End'], ['2009-12-10', 'B']]]

code = 0
brr = []
for row in mylist:

    if row[1] == "Start":
        arr = []
        code = row[0]
        arr.extend([row[0], row[1], row[2]])
        # continue not needed here

    if row[0] == code and row[1] == "End":
        arr.extend([row[1], row[2], row[3]])

    if arr not in brr:
        brr.append(arr)

for k in brr:
    print(k)

---

您也可以通过以下方式轻松实现这一目标：

输出：

[[20, 'Start', '2008-10-10', 'End', '2008-11-09', 'NG'],
 [21, 'Start', '2008-12-10', 'End', '2008-12-15', 'G'],
 [22, 'Start', '2009-01-10', 'End', '2009-12-10', 'B']]

请试试这个

startlist=[]
endlist=[]
for item in mylist:
    if 'Start' in list(item):
        startlist.append(list(item))
    elif 'End' in list(item):
        endlist.append(list(item))
outlist=[i+j for i,j in zip(startlist,endlist)]     

---

开始和结束是否始终是行的第一个和最后一个元素？不，myList是许多元组的列表。但开始总是在结束之前@用户4343502在每个元组中，开始和结束是否出现一次且仅出现一次？此外，您删除了

TBS

，这是有意的吗？对于每个元组，只有一个单词。我想说，如果它是start，那么将它存储到一个列表中，然后忽略其他单词，我应该搜索特定代码的end，并将其保存在与start相同的列表中@user4343502@TemporalWolf：

arr.append

只接受一个参数<代码>arr.append（1，2）无效。完美！我在这里工作。但我仍然有多行的每一个项目，我不知道为什么@卡佩尔·弗洛里安斯基只是说。。。

brr.append（arr）

将

arr

添加到

brr

每个循环迭代中。您迭代了6个元组，因此您将在输出中添加6个元素。太棒了！谢谢