观察计数器，计算总数并计算错过的计数
Watching a counter, tally total and counting missed counts
我正在尝试创建一段代码来监视一个计数器，其输出类似于：
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]
我希望代码能够计算总数并告诉我错过了多少计数，例如，如果发生这种情况：
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]
我总共还能得到 92 个，但得到的反馈是缺少 8 个。
我已经非常接近以下代码：
Blk_Tot = 0
CBN = 0
LBN = 0
x = 0
y = 0
z = 0
MissedBlocks = 0
for i in range(len(a1)):
CBN = a1[i]
if CBN - LBN <= 0:
if LBN == 30:
y = 30 - abs(CBN - LBN)
elif LBN < 30:
z = 30 - LBN
y = 30 - abs(CBN - LBN) + z
print(z)
Blk_Tot = Blk_Tot + y
else:
x = CBN - LBN
Blk_Tot = Blk_Tot + x
if x > 1:
MissedBlocks = MissedBlocks - 1 + x
LBN = CBN
print(Blk_Tot)
print(MissedBlocks)
如果我删除 1 到 30 之间的任何地方，它会完美地工作，但是如果我删除 30，比如 29,30,1,2 它 breaks.I 不要期望它能够错过 30连续，但仍然能够得出合适的总数。
有人对如何实现这一目标有任何想法吗？我觉得我错过了一个明显的答案 :D
抱歉，我想我不清楚，a1 是一个来自外部设备的计数器，它从 1 计数到 30，然后再次回到 1。每个计数实际上是消息的一部分，以表明消息已收到；所以说 1 2 4，我知道第三条消息丢失了。我要做的是找出应该收到的总数以及计数中遗漏了多少。
根据以下帖子的想法进行更新，另一种方法可能是：
输入：
123456
列表[1,2,3,4,5,6]
1.Check 首先输入以查看它在列表的哪一部分并从那里开始（以防我们不从零开始）
2.every 接收输入的时间检查是否与数组中的下一个值匹配
3.if 不是那么找到那个值需要多少步
用于检查重复序列模式中缺失元素的新代码
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 再次重复。1 到 30 之间可能缺少元素，但顺序将始终按 1 到 30 之间的升序排列。
如果源数据如列表a1所示，那么代码将导致8个缺失元素。
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')
这个输出将是：
There are 8 items missing in the list
如果这解决了您的问题，请告诉我
比较两个列表的较早代码
您不能使用 set，因为项目是重复的。所以你需要 sort 它们并找出每个元素在两个列表中的次数。差异将为您提供缺失的计数。您可能在 a1 中有一个元素，但在 a2 中没有，反之亦然。所以找出丢失项目的绝对数量会给你结果。
我会在下次更新时用更好的变量更新响应。
我是这样做的：
带注释的代码：
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)
这个输出是：
Total number of missing elements are: 7
如果要找出缺少哪些元素，则需要多添加几行代码。
在上面的示例中，a2 中缺少元素 27,28,29,30, 4, 5，a1 中缺少元素 31。所以缺失元素的总数是 7.
没有注释的代码：
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)
一般来说，如果你想统计两个列表之间的差异数，你可以很容易地使用字典。另一个答案也可以，但对于稍大的列表来说效率非常低。
def counter(lst):
# create a dictionary with count of each element
d = {}
for i in lst:
if d.get(i, None):
d[i] += 1
else:
d[i] = 1
return d
def compare(d1, d2):
# d1 and d2 are dictionaries
ans = 0
for i in d1.values():
if d2.get(i, None):
# comapares the common values in both lists
ans += abs(d1[i]-d2[i])
d2[i] = 0
else:
#for elements only in the first list
ans += d1[i]
for i in d2.values():
# for elements only in the second list
if d2[i]>0:
ans += d2[i]
return ans
l1 = [...]
l2 = [...]
print(compare(counter(l1), counter(l2)))
超过30线就不需要跟踪了。
只需与理想序列进行比较并计算缺失的数字即可。
不知道最后是否缺少部分。
不知道一个块中是否缺少超过 30 个零件。
from itertools import cycle
def idealSeqGen():
for i in cycle(range(1,31)):
yield(i)
def receivedSeqGen():
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]
for i in a1:
yield(i)
receivedSeq = receivedSeqGen()
idealSeq = idealSeqGen()
missing = 0
ideal = next(idealSeq)
try:
while True:
received = next(receivedSeq)
while received != ideal:
missing += 1
ideal = next(idealSeq)
ideal = next(idealSeq)
except StopIteration:
pass
print (f'There are {missing} items missing')
编辑
循环部分可以简单一点
missing = 0
try:
while True:
ideal = next(idealSeq)
received = next(receivedSeq)
while received != ideal:
missing += 1
ideal = next(idealSeq)
except StopIteration:
pass
print (f'There are {missing} items missing')
我正在尝试创建一段代码来监视一个计数器，其输出类似于：
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]
我希望代码能够计算总数并告诉我错过了多少计数，例如，如果发生这种情况：
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]
我总共还能得到 92 个，但得到的反馈是缺少 8 个。
我已经非常接近以下代码：
Blk_Tot = 0
CBN = 0
LBN = 0
x = 0
y = 0
z = 0
MissedBlocks = 0
for i in range(len(a1)):
CBN = a1[i]
if CBN - LBN <= 0:
if LBN == 30:
y = 30 - abs(CBN - LBN)
elif LBN < 30:
z = 30 - LBN
y = 30 - abs(CBN - LBN) + z
print(z)
Blk_Tot = Blk_Tot + y
else:
x = CBN - LBN
Blk_Tot = Blk_Tot + x
if x > 1:
MissedBlocks = MissedBlocks - 1 + x
LBN = CBN
print(Blk_Tot)
print(MissedBlocks)
如果我删除 1 到 30 之间的任何地方，它会完美地工作，但是如果我删除 30，比如 29,30,1,2 它 breaks.I 不要期望它能够错过 30连续，但仍然能够得出合适的总数。
有人对如何实现这一目标有任何想法吗？我觉得我错过了一个明显的答案 :D
抱歉，我想我不清楚，a1 是一个来自外部设备的计数器，它从 1 计数到 30，然后再次回到 1。每个计数实际上是消息的一部分，以表明消息已收到；所以说 1 2 4，我知道第三条消息丢失了。我要做的是找出应该收到的总数以及计数中遗漏了多少。
根据以下帖子的想法进行更新，另一种方法可能是：
输入：
123456
列表[1,2,3,4,5,6]
1.Check 首先输入以查看它在列表的哪一部分并从那里开始（以防我们不从零开始）
2.every 接收输入的时间检查是否与数组中的下一个值匹配
3.if 不是那么找到那个值需要多少步
用于检查重复序列模式中缺失元素的新代码
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 再次重复。1 到 30 之间可能缺少元素，但顺序将始终按 1 到 30 之间的升序排列。
如果源数据如列表a1所示，那么代码将导致8个缺失元素。
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')
这个输出将是：
There are 8 items missing in the list
如果这解决了您的问题，请告诉我
比较两个列表的较早代码
您不能使用 set，因为项目是重复的。所以你需要 sort 它们并找出每个元素在两个列表中的次数。差异将为您提供缺失的计数。您可能在 a1 中有一个元素，但在 a2 中没有，反之亦然。所以找出丢失项目的绝对数量会给你结果。
我会在下次更新时用更好的变量更新响应。
我是这样做的：
带注释的代码：
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)
这个输出是：
Total number of missing elements are: 7
如果要找出缺少哪些元素，则需要多添加几行代码。
在上面的示例中，a2 中缺少元素 27,28,29,30, 4, 5，a1 中缺少元素 31。所以缺失元素的总数是 7.
没有注释的代码：
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)
一般来说，如果你想统计两个列表之间的差异数，你可以很容易地使用字典。另一个答案也可以，但对于稍大的列表来说效率非常低。
def counter(lst):
# create a dictionary with count of each element
d = {}
for i in lst:
if d.get(i, None):
d[i] += 1
else:
d[i] = 1
return d
def compare(d1, d2):
# d1 and d2 are dictionaries
ans = 0
for i in d1.values():
if d2.get(i, None):
# comapares the common values in both lists
ans += abs(d1[i]-d2[i])
d2[i] = 0
else:
#for elements only in the first list
ans += d1[i]
for i in d2.values():
# for elements only in the second list
if d2[i]>0:
ans += d2[i]
return ans
l1 = [...]
l2 = [...]
print(compare(counter(l1), counter(l2)))
超过30线就不需要跟踪了。
只需与理想序列进行比较并计算缺失的数字即可。
不知道最后是否缺少部分。
不知道一个块中是否缺少超过 30 个零件。
from itertools import cycle
def idealSeqGen():
for i in cycle(range(1,31)):
yield(i)
def receivedSeqGen():
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]
for i in a1:
yield(i)
receivedSeq = receivedSeqGen()
idealSeq = idealSeqGen()
missing = 0
ideal = next(idealSeq)
try:
while True:
received = next(receivedSeq)
while received != ideal:
missing += 1
ideal = next(idealSeq)
ideal = next(idealSeq)
except StopIteration:
pass
print (f'There are {missing} items missing')
编辑
循环部分可以简单一点
missing = 0
try:
while True:
ideal = next(idealSeq)
received = next(receivedSeq)
while received != ideal:
missing += 1
ideal = next(idealSeq)
except StopIteration:
pass
print (f'There are {missing} items missing')