<h2>用于检查重复序列模式中缺少元素的新代码</h2>
<p>现在我已经更清楚地理解了你的问题,下面是代码。此代码中的假设是列表将始终按从1到30的升序排列,然后从1开始再次重复。在1和30之间可能会缺少元素,但在1和30之间的顺序将始终是升序</p>
<p>如果源数据如列表<code>a1</code>所示,那么代码将导致<code>8</code>缺少元素</p>
<pre><code>a1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,
5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,
1,2]
a2 = a1.copy()
c = 1
missing = 0
while a2:
if a2[0] == c:
c+=1
a2.pop(0)
elif a2[0] > c:
missing +=1
c+=1
elif a2[0] < c:
missing += 31-c
c = 1
if c == 31: c=1
print (f'There are {missing} items missing in the list')
</code></pre>
<p>其输出将为:</p>
<pre><code>There are 8 items missing in the list
</code></pre>
<p>如果这能解决您的问题,请告诉我</p>
<h2>比较两个列表的早期代码</h2>
<p>不能使用<code>set</code>,因为这些项是重复的。因此,您需要<code>sort</code>它们,并找出每个元素在两个列表中的次数。差额将为您提供缺少的计数。您可以在<code>a1</code>中有一个元素,但在<code>a2</code>中没有,反之亦然。因此,找出丢失项目的绝对计数将给出结果</p>
<p>在下一次更新中,我将使用更好的变量更新响应</p>
<p>我是这样做的:</p>
<h2>带注释的代码:</h2>
<pre><code>a1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30]
a2 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,
5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,1,2]
#step 1: Find out which list is longer. We will use that as the master list
if len(a1) > len(a2):
master_list = a1.copy()
second_list = a2.copy()
else:
master_list = a2.copy()
second_list = a1.copy()
#step 2: We must sort both master and second list
# so we can compare against each other
master_list.sort()
second_list.sort()
#set the counter to zero
missing = 0
#iterate through the master list and find all values in master against second list
#for each iteration, remove the value in master[0] from both master and second list
#when you have iterated through the full list, you will get an empty master_list
#this will help you to use while statement to iterate until master_list is empty
while master_list:
#pick the first element of master list to search for
x = master_list[0]
#count the number of times master_list[0] is found in both master and second list
a_count = master_list.count(x)
b_count = second_list.count(x)
#absolute difference of both gives you how many are missing from each other
#master may have 4 occurrences and second may have 2 occurrences. abs diff is 2
#master may have 2 occurrences and second may have 5 occurrences. abs diff is 3
missing += abs(a_count - b_count)
#now remove all occurrences of master_list[0] from both master and second list
master_list = [i for i in master_list if i != x]
second_list = [i for i in second_list if i != x]
#iterate until master_list is empty
#you may end up with a few items in second_list that are not found in master list
#add them to the missing items list
#thats your absolute total of all missing items between lists a1 and a2
#if you want to know the difference between the bigger list and shorter one,
#then don't add the missing items from second list
missing += len(second_list)
#now print the count of missig elements between the two lists
print ('Total number of missing elements are:', missing)
</code></pre>
<p>此操作的输出为:</p>
<pre><code>Total number of missing elements are: 7
</code></pre>
<p>如果您想找出缺少哪些元素,那么需要再添加几行代码</p>
<p>在上面的示例中,元素<code>27,28,29,30, 4, 5</code>在<code>a2</code>中缺失,元素<code>31</code>在<code>a1</code>中缺失。所以缺失元素的总数是<code>7</code></p>
<h2>无注释代码:</h2>
<pre><code>a1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30]
a2 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,
5, 6, 7, 8, 9, 10, 11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,1,2]
if len(a1) > len(a2):
master_list = a1.copy()
second_list = a2.copy()
else:
master_list = a2.copy()
second_list = a1.copy()
master_list.sort()
second_list.sort()
missing = 0
while master_list:
x = master_list[0]
a_count = master_list.count(x)
b_count = second_list.count(x)
missing += abs(a_count - b_count)
master_list = [i for i in master_list if i != x]
second_list = [i for i in second_list if i != x]
missing += len(second_list)
print ('Total number of missing elements are:', missing)
</code></pre>