Question

我有一个列表 [T20，T5，T10，T1，T2，T8，T16，T17，T9，T4，T12，T13，T18]

我已经去除了T，将其转换为整数类型，并对列表进行排序以获取此信息：

sorted_ids = [1、2、4、5、8、9、10、12、13、16、17、18、20]

我正在遍历列表，并检查当前数字的下一个数字是否按数字顺序。如果不是，我想在其位置插入“ V” 。

所以最终列表应如下所示： [1、2，V，4、5，V，V，8、9、10，V，12、13，V，V，16、17、18 ，V，20]

但是，我无法在正确的位置插入 V的确切编号。

def arrange_tickets(tickets_list):

    ids=[]
    for item in tickets_list:
        new_str=item.strip("T")
        ids.append(int(new_str))
    sorted_ids = sorted(ids)
    temp_ids = []
    print("Sorted: ",sorted_ids)
    #size = len(sorted_ids)
    for i in range(len(sorted_ids)-1):
        temp_ids.append(sorted_ids[i])

        if sorted_ids[i]+1 != sorted_ids[i+1] :
            temp_ids.insert(i+1,"V")
    print(temp_ids)
    #print(sorted_ids)


tickets_list = ['T20', 'T5', 'T10', 'T1', 'T2', 'T8', 'T16', 'T17', 'T9', 'T4', 'T12', 'T13', 'T18']
print("Ticket ids of all the available students :")
print(tickets_list)
result=arrange_tickets(tickets_list)

实际结果： [1、2，'V'，4，'V'，5、8，'V'，9，'V'，10、12，'V'，13、16 17，18]

预期结果： [T1，T2，V，T4，T5，V，V，T8，T9，T10，V，T12，T13，V，V，T16，T17，T18，V，T20]

Answer 1

这是一个解决方案：

sorted_ids=[1, 2, 4, 5, 8, 9, 10, 12, 13, 16, 17, 18, 20]

def arrange(inList):
    newList = []
    newList.append('T'+str(inList[0]))
    for i in range(1,len(inList)):
        diff = inList[i] - inList[i-1]
        if diff > 1:
            for d in range(diff-1):
                newList.append('V')
            newList.append('T'+str(inList[i]))
        else:
            newList.append('T'+str(inList[i]))
    return newList

print(arrange(sorted_ids))

输出：

['T1', 'T2', 'V', 'T4', 'T5', 'V', 'V', 'T8', 'T9', 'T10', 'V', 'T12', 'T13', 'V', 'V', 'T16', 'T17', 'T18', 'V', 'T20']

Answer 2

这是值得考虑的另一个解决方案：

TableA(B_NUM)

结果：

sorted_ids=[1, 2, 4, 5, 8, 9, 10, 12, 13, 16, 17, 18, 20]

for i in range(min(sorted_ids), max(sorted_ids)): 
     if sorted_ids[i] != i + 1: 
         sorted_ids.insert(i, 'V')

final_list = [ "T" + str(x) if isinstance(x, int) else x for x in sorted_ids]

Answer 3

以下是列表理解功能，可让您获得所需的内容：

sorted_ids=[1, 2, 4, 5, 8, 9, 10, 12, 13, 16, 17, 18, 20]
a = sorted_ids[0]
b = sorted_ids[-1]
nums = set(sorted_ids)
expected = ["T" + str(i) if i in nums else 'V' for i in range(a,b+1)]
print(expected)

输出：

['T1', 'T2', 'V', 'T4', 'T5', 'V', 'V', 'T8', 'T9', 'T10', 'V', 'T12', 'T13', 'V', 'V', 'T16', 'T17', 'T18', 'V', 'T20']

Answer 4

temp_ids.insert(i+1,"V")

这是麻烦的声明。通过以下方式更新代码

temp_ids=[]

for i in range(len(sorted_ids)-1):
    temp_ids.append(sorted_ids[i])

    if sorted_ids[i]+1 != sorted_ids[i+1] :
        for i in range(sorted_ids[i+1]-sorted_ids[i]-1):
            temp_ids.append("V") # appends as many V's as required

temp_ids.append(sorted_ids[-1]) # appends last element

这应该有效假设排序数组是[1,2,6]

因此，我们的期望输出应为[1,2，'V'，'V'，'V'，6]。所以每次

sorted_ids[i]+1 != sorted_ids[i+1]

条件成立，我们将不得不追加几个V。现在确定要附加的V数，看到在2和6之间，将附加3个V。因此，通常我们会附加（sorted_ids [i + 1]-sorted [i] -1）V。

现在看到此行

for i in range(len(sorted_ids)-1):

由于这一行，我们的列表仅在[1,2,6]中以[1,2]运行，并且我们从不在For循环中附加6，因此在退出For Loop之后将其附加。

Answer 5

首先考虑列表中应该包含哪些ID，假设它们从1开始并以最大的ID结束。然后检查每个期望的id是否确实存在，如果没有，请在其中放置“ V”。作为副作用，这也会对列表进行排序。

def arrange_tickets(tickets_list):
    ids = [int(ticket[1:]) for ticket in tickets_list]
    expected_ids = range(1, max(ids) + 1)
    return ["T%d" % n if n in ids else "V" for n in expected_ids]

tickets_list = ['T20', 'T5', 'T10', 'T1', 'T2', 'T8', 'T16', 'T17', 'T9', 'T4', 'T12', 'T13', 'T18']
print("Ticket ids of all the available students :")
print(tickets_list)
result=arrange_tickets(tickets_list)
print(result)

结果：

Ticket ids of all the available students :
['T20', 'T5', 'T10', 'T1', 'T2', 'T8', 'T16', 'T17', 'T9', 'T4', 'T12', 'T13', 'T18']

['T1', 'T2', 'V', 'T4', 'T5', 'V', 'V', 'T8', 'T9', 'T10', 'V', 'T12', 'T13', 'V', 'V', 'T16', 'T17', 'T18', 'V', 'T20']

Answer 6

您可以使用以下itertools recipe来对连续的数字进行分组：

from itertools import groupby
from operator import itemgetter

def groupby_consecutive(lst):
    for _, g in groupby(enumerate(lst), lambda x: x[0] - x[1]):
        yield list(map(itemgetter(1), g))

sorted_ids = [1, 2, 4, 5, 8, 9, 10, 12, 13, 16, 17, 18, 20]
print(list(groupby_consecutive(lst=sorted_ids)))
# [[1, 2], [4, 5], [8, 9, 10], [12, 13], [16, 17, 18], [20]]

然后，您可以创建一个函数，以从先前的分组中获取散布的V值：

def interperse(lst):
    for x, y in zip(lst, lst[1:]):
        yield ["V"] * (y[0] - x[-1] - 1)

groups = list(groupby_consecutive(lst))
print(list(interperse(groups)))
# [['V'], ['V', 'V'], ['V'], ['V', 'V'], ['V']]

然后，您最终可以将上述结果压缩在一起：

def add_prefix(lst, prefix):
    return [prefix + str(x) for x in lst]

def create_sequences(lst, prefix='T'):
    groups = list(groupby_consecutive(lst))
    between = list(interperse(groups))

    result = add_prefix(groups[0], prefix)
    for x, y in zip(between, groups[1:]):
        result.extend(x + add_prefix(y, prefix))

    return result

sorted_ids = [1, 2, 4, 5, 8, 9, 10, 12, 13, 16, 17, 18, 20]
print(create_sequences(lst=sorted_ids))
# ['T1', 'T2', 'V', 'T4', 'T5', 'V', 'V', 'T8', 'T9', 'T10', 'V', 'T12', 'T13', 'V', 'V', 'T16', 'T17', 'T18', 'V', 'T20']

Answer 7

一枪直接形成原始阵列

array = ['T20', 'T5', 'T10', 'T1', 'T2', 'T8', 'T16', 'T17', 'T9', 'T4', 'T12', 'T13', 'T18']

您可以定义一个完成所有工作的方法：

def add_vs_between_not_cons(array):
  iterable = sorted(array, key= lambda x: int(x[1:]))
  i, size = 0, len(iterable)
  while i < size:
    delta = int(iterable[i][1:]) - int(iterable[i-1][1:])
    for _ in range(delta-1):
      yield "V"
    yield iterable[i]
    i += 1

因此，您可以致电：

print(list(add_vs_between_not_cons(array)))

#=> ['T1', 'T2', 'V', 'T4', 'T5', 'V', 'V', 'T8', 'T9', 'T10', 'V', 'T12', 'T13', 'V', 'V', 'T16', 'T17', 'T18', 'V', 'T20']

Answer 8

我对infytq查询的解决方案：

def arrange_tickets(tickets_list):
    ids = [int(ticket[1:]) for ticket in tickets_list]
    expected_ids = range(1, max(ids) + 1)
    listt=["T%d" % n if n in ids else "V" for n in expected_ids]
    list1=listt[0:10]
    list2=listt[11:]
    for i in range(10):
        if 'V' in list2:
            list2.remove('V')
    for j in range(0,len(list2)):
        for n, i in enumerate(list1):
            if i == 'V':
                list1[n] = list2[j]
                j+=1
    return list1
tickets_list = ['T5','T7','T1','T2','T8','T15','T17','T19','T6','T12','T13']
print("Ticket ids of all the available students :")
print(tickets_list)
result=arrange_tickets(tickets_list)
print()
print("Ticket ids of the ten students in Group-1:")
print(result[0:10])

检查python列表中的序列

8 个答案: