我想从列表列表中找到最常见的符号序列

Question

我想从 [lists] 列表中找到最常用的符号序列

CATEGORIES = ["0","1","2","3","4","5","6","7","8","9",
              "A","B","C","D","E","F","G","H","I","J",
              "K","L","M","N","O","P","R","S","T","U",
              "V","W","X","Y","Z"]

KR8877J = [[0.002,0.006,0.004,0.045,0.002,0.017,0.006,0.077,0.001,0.035,0.042,0.005,0.004,0.039,0.001,0.002,0.001,0.008,0.058,0.352,0.002,0.007,0.017,0.004,0.007,0.007,0.007,0.004,0.005,0.009,0.089,0.036,0.053,0.041,0.004],[0.003,0.007,0.005,0.075,0.001,0.020,0.006,0.044,0.002,0.035,0.026,0.004,0.004,0.033,0.001,0.001,0.003,0.008,0.049,0.360,0.002,0.007,0.021,0.005,0.009,0.003,0.008,0.007,0.003,0.014,0.092,0.048,0.058,0.031,0.004],[0.002,0.000,0.025,0.012,0.006,0.002,0.001,0.627,0.006,0.021,0.022,0.008,0.004,0.006,0.004,0.033,0.000,0.006,0.011,0.009,0.002,0.002,0.009,0.000,0.002,0.040,0.007,0.005,0.015,0.000,0.035,0.001,0.008,0.015,0.053],[0.056,0.008,0.023,0.038,0.015,0.007,0.050,0.006,0.412,0.004,0.005,0.027,0.011,0.005,0.021,0.007,0.073,0.024,0.012,0.005,0.013,0.005,0.027,0.003,0.015,0.001,0.005,0.074,0.002,0.022,0.005,0.011,0.002,0.001,0.006],[0.025,0.011,0.025,0.034,0.018,0.027,0.090,0.008,0.258,0.006,0.007,0.026,0.016,0.008,0.026,0.011,0.079,0.030,0.026,0.008,0.018,0.011,0.033,0.003,0.016,0.001,0.003,0.106,0.004,0.021,0.012,0.013,0.003,0.005,0.014],[0.048,0.027,0.019,0.002,0.028,0.002,0.008,0.017,0.041,0.014,0.012,0.022,0.031,0.005,0.045,0.100,0.004,0.031,0.033,0.002,0.029,0.006,0.021,0.032,0.008,0.038,0.317,0.007,0.017,0.004,0.018,0.005,0.003,0.004,0.002],[0.013,0.002,0.002,0.000,0.164,0.001,0.060,0.004,0.006,0.002,0.018,0.003,0.035,0.002,0.008,0.008,0.001,0.008,0.028,0.005,0.383,0.013,0.063,0.010,0.004,0.002,0.014,0.016,0.002,0.005,0.048,0.011,0.028,0.017,0.012]]

KR8877J_1 = [[0.004,some data]]
KR8877J_2 = [[0.002,some data]]
KR8877J_3 = [[somedata]
KR8877J_4 = [[0.006,some data,0.008]]
KR8877J_5 = [[some data]]
KR8877J_6 = [[some data]]

def readable(x):
    tag = []
    for lst in x:
        index = max(enumerate(lst), key=lambda pair: pair[1])[0]
        tag.append(CATEGORIES[index])
    tag.reverse()
    str = tag 
    print(str)
    #print(tag)


for i in (KR8877J, KR8877J_1,KR8877J_2,KR8877J_3,KR8877J_4,KR8877J_5,KR8877J_6): 
    readable(i)
    

def compare_bitwise(a,b): 
    a_set = set(a) 
    b_set = set(b) 
    if (a_set & b_set): 
         return True 
    else: 
        return False


for i in (KR8877J, KR8877J_1,KR8877J_2,KR8877J_3,KR8877J_4,KR8877J_5,KR8877J_6):
    print(compare_bitwise(i, i+=1)) 

迭代的问题在这里： print(compare_bitwise(i, i+=1))我只在第一个列表中给出了一个数据示例，因为它们都是同一个专家，具有 output ['K', 'R', '8', '8', 'J', '7', 'J'] 代替 ['K', 'R', '8', '8', '7', 'J', 'J']

Answer 1

没有some data的例子，很难说出预期的结果。 在任何情况下，对于行print(compare_bitwise(i, i+=1)) - 如果您尝试将每个KR8877J_...序列与下一个序列进行比较，那么KR8877J_1与KR8877J_2 ， KR8877J_2与KR8877J_3等 - 然后最小的变化是：将该序列分配给列表或元组，然后进行相应的索引。

seq = (KR8877J_1,KR8877J_2,KR8877J_3,KR8877J_4,KR8877J_5,KR8877J_6)
for i in range(len(seq) - 1):
    print(compare_bitwise(seq[i], seq[i+1]))

但是在 Python 中，使用range()和len()进行索引并不好或不高效。 相反，最好将zip()与seq和seq[1:]一起使用：

seq = (KR8877J_1,KR8877J_2,KR8877J_3,KR8877J_4,KR8877J_5,KR8877J_6)
for i, j in zip(seq, seq[1:]):
    print(compare_bitwise(i, j))

---

编辑：我不完全理解您的意思，但您可以使用tag_tup = tuple(tag)将标签列表转换为元组并从readable() return值，而不仅仅是打印它。 或者只是return tag 。 然后通过执行counter = collections.Counter()将其作为collections.Counter的键，然后在循环中循环readable(i)然后在循环内使用counter.update({tag_tup: 1})以及tag_tup 。

def readable(x):
    tag = []
    for lst in x:
        index = max(enumerate(lst), key=lambda pair: pair[1])[0]
        tag.append(CATEGORIES[index])
    tag.reverse()
    print(tag)
    return tag


import collections
counter = collections.Counter()
for i in (KR8877J, KR8877J_1,KR8877J_2,KR8877J_3,KR8877J_4,KR8877J_5,KR8877J_6):
    tag_tup = tuple(readable(i))
    counter.update({tag_tup: 1})

print(counter)

请参阅collections.Counter文档（上面链接），然后获取most_common组合。