Python 3.6-使用计数器在列表中不包含重复值

最初，代码将查看列表A，然后计算是否有重复5次的数字。如果出现5次，代码将附加列表A中的编号和相应的列表B值

然后，如果列表a中有3个数字出现，将应用另一个条件，并打印列表B中的值和相应值

输出

a = ['1','1','1','1','1','3','3','3','1','1','1','1','1']
b = ['4','8','3','1','7','6','4','9','8','3','2','5','2']

from collections import Counter
counts = Counter(a)

c = []
d = []

for x, y in zip(a, b):
    if counts[x] == 5:
        c.append(x)
        d.append(y)
    if not counts[x] == 5:
       if counts[x] == 3:
           c.append(x)
           d.append(y)
    else:
        None

# c = ['3','3','3']

然而，在我的示例中，我相信如果以后重复，在搜索这5个事件时会感到困惑

所需输出

a = ['1','1','1','1','1','3','3','3','1','1','1','1','1']
b = ['4','8','3','1','7','6','4','9','8','3','2','5','2']

from collections import Counter
counts = Counter(a)

c = []
d = []

for x, y in zip(a, b):
    if counts[x] == 5:
        c.append(x)
        d.append(y)
    if not counts[x] == 5:
       if counts[x] == 3:
           c.append(x)
           d.append(y)
    else:
        None

# c = ['3','3','3']

我怎样才能让它指定一个实例中必须正好有5个实例

---

---

根据评论，这里有一个可能的解决方案。也许可以使用

itertools.groupby

或其他技巧来缩短它

a = ['1','1','1','1','1','3','3','3','1','1','1','1','1']
b = ['4','8','3','1','7','6','4','9','8','3','2','5','2']

counter, c, d = 1, [], []

a[-1] = ['-1']
b[-1] = ['-1']

for i,j in enumerate(a):
    if a[i-1]==a[i]:
        counter+=1

        if counter == 5 and a[i+1]!=a[i]:
            c += [j]*5
            d += b[i-4:i+1]
            counter = 1

        if counter == 3:
            c += [j]*3
            d += b[i-2:i+1]
            counter = 1   
    else:
        pass

def find_group（l，group_len）：
i=0
而我

我想你可以用一个简单的计数器，比如：

def find_group(l, group_len):
    i = 0
    while i < len(l):
        j = i + 1
        while j < len(l) and l[i] == l[j]:
            j += 1
            if j - i == group_len:
                return i
            i = j
    return None

a = ['1','1','1','1','1','3','3','3','1','1','1','1','1']
b = ['4','8','3','1','7','6','4','9','8','3','2','5','2']
start3 = find_group(a, 3)
start5 = find_group(a, 5)
c = []
d = []
if start5 is not None:
    c += a[start5:start5 + 5]
    d += b[start5:start5 + 5]

if start3 is not None:
    c += a[start3:start3 + 3]
    d += b[start3:start3 + 3]

print(c)
print(d)

这将产生以下结果：

#!/usr/bin/env python

a = ['1','1','1','1','1','3','3','3','1','1','1','1','1']
b = ['4','8','3','1','7','6','4','9','8','3','2','5','2']

counter, c, d = 1, [], []

# append non existing values in a,b such that not to count the 
# last consecutive series - in this case the '1's
# if you want to count the last 5 '1's then you can change the
# following lines to:
# a += ['-1'] and b += ['-1']

a[-1] = ['-1']
b[-1] = ['-1']

for i,j in enumerate(a):
    if a[i-1]==a[i]:
        counter+=1

        if counter == 3 and a[i+1]!=a[i]:
            c += [j]*3
            d += b[i-2:i+1]
            counter = 1

        elif counter == 5:
            c += [j]*5
            d += b[i-4:i+1]
            counter = 1   
    else:
        pass

print 'c list: {}'.format(c)
print 'd list: {}'.format(d)

更新

如果希望c列表不包含重复值，请将上述代码更改为：

c list: ['1', '1', '1', '1', '1', '3', '3', '3']
d list: ['4', '8', '3', '1', '7', '6', '4', '9']

结果:

#!/usr/bin/env python

a = ['1','1','1','1','1','3','3','3','1','1','1','1','1','4','4','4','2']
b = ['4','8','3','1','7','6','4','9','8','3','2','5','2','1','2','3','1']

counter, c, d = 1, [], []

a[-1] = ['-1']
b[-1] = ['-1']

for i,j in enumerate(a):
    if a[i-1]==a[i] and a[i] not in c:
        counter+=1

        if counter == 3 and a[i+1]!=a[i]:
            c += [j]*3
            d += b[i-2:i+1]
            counter = 1

        elif counter == 5:
            c += [j]*5
            d += b[i-4:i+1]
            counter = 1   
    else:
        pass

print 'c list: {}'.format(c)
print 'd list: {}'.format(d)

---

请记住，通过这种方式，您的代码会进行更多的检查，以查看该项是否在c-list上，因此比以前贵了一点。

您说过要过滤掉，但随后您要添加。@MosesKoledoye现在重新编写了它。请注意，

计数器会重复10次而不是5次。你是在要求连续重复吗？@MosesKoledoye是的，这是我的问题，第一次应该算5次，但我希望它忽略第二次instance@MosesKoledoye如果有两次发生5次，则只计算第一次。谢谢您的回复，然而，当我使用代码时，它似乎对c和d什么都没有给我？注释不是用于扩展讨论或调试；这段对话已经结束。如果从本次讨论中收集到任何相关见解，则应将其编辑成答案。