迭代列表列表并检查条件 Python

Question

我有一个数据 object 是我通过操作一个更大的数据集创建的。 它是一个列表列表，其中最高级别是星期几（1-7，但为简洁起见，我只包括第一天），第二级别是每天的时间。 object 看起来像这样：

act_all_fun = [[[datetime.time(0, 0), 254, 'F37', 'd1'],
                [datetime.time(2, 0), 254, 'F7', 'd1'],
                [datetime.time(5, 0), 254, 'F17', 'd1'],
                [datetime.time(6, 30), 254, 'F37', 'd1'],
                [datetime.time(7, 0), 15, 'F37', 'd1'],
                [datetime.time(10, 0), 17, 'F37', 'd0'],
                [datetime.time(12, 0), 19, 'F37', 'd0'],
                [datetime.time(21, 0), 15, 'F7', 'd1'],
                [datetime.time(23, 0), 254, 'F37', 'd1']],
               [[datetime.time(0, 0), 254, 'F37', 'd1'],
                [datetime.time(1, 30), 254, 'F', 'd1'],
                [datetime.time(5, 0), 2, 'F7', 'd1'],
                [datetime.time(6, 30), 2, 'F37', 'd0'],
                [datetime.time(8, 0), 7, 'F37', 'd0'],
                [datetime.time(9, 15), 11, 'F', 'd0'],
                [datetime.time(10, 15), 15, 'F', 'd0'],
                [datetime.time(15, 30), 5, 'F37', 'd0'],
                [datetime.time(16, 15), 6, 'F', 'd0'],
                [datetime.time(19, 30), 9, 'F37', 'd0'],
                [datetime.time(23, 0), 254, 'F37', 'd1']]]

我想遍历 act_all_fun 中的每个特定列表（在时间级别）并且：

1. 检查每个的第二个元素是否等于 254： 如果为真，请查看有多少个独特的第三个元素（例如“F”或“F7”或“F17”等）：对于每个独特的第三个元素，从 254 中减去 1
2. 如果第二个元素不等于 254： a． 例如，如果第二个元素等于 15：检查任何其他为 15 的元素，并查看当第三个元素为 15 作为第二个元素时有多少个唯一值。对于每个唯一元素，将 30 添加到第二个元素.

例如，所需的 output 将是：

    act_all_fun = [[[datetime.time(0, 0), 254, 'F37', 'd1'],
                [datetime.time(2, 0), 253, 'F7', 'd1'],
                [datetime.time(5, 0), 252, 'F17', 'd1'],
                [datetime.time(6, 30), 254, 'F37', 'd1'],
                [datetime.time(7, 0), 15, 'F37', 'd1'],
                [datetime.time(10, 0), 17, 'F37', 'd0'],
                [datetime.time(12, 0), 19, 'F37', 'd0'],
                [datetime.time(21, 0), 45, 'F7', 'd1'],
                [datetime.time(23, 0), 254, 'F37', 'd1']],
               [[datetime.time(0, 0), 254, 'F37', 'd1'],
                [datetime.time(1, 30), 251, 'F', 'd1'],
                [datetime.time(5, 0), 2, 'F7', 'd1'],
                [datetime.time(6, 30), 2, 'F37', 'd0'],
                [datetime.time(8, 0), 7, 'F37', 'd0'],
                [datetime.time(9, 15), 11, 'F', 'd0'],
                [datetime.time(10, 15), 75, 'F', 'd0'],
                [datetime.time(15, 30), 5, 'F37', 'd0'],
                [datetime.time(16, 15), 6, 'F', 'd0'],
                [datetime.time(19, 30), 9, 'F37', 'd0'],
                [datetime.time(23, 0), 254, 'F37', 'd1']]]

为便于参考，查output at: a-day 1次：(0,0)/(2,0), (5,0)/(6,30)/(21,0) b-day 2次: (1,30)/(10,15)

我试过这样的事情：

for day in act_all_fun:
for act in day:
    if act[1] in range(240,255) and act[2] not in [x[2] for x in day if x != act]:
        act[1] = act[1] - 1

但这并没有产生我预期的结果。

我尝试了多种不同的方法，从将列表展平到将其放入 dataframe 但无济于事。 我也试图尽可能清楚，但意识到我可能没有做到。 如果您需要任何说明，请告诉我。

Answer 1

好吧，经过大量工作并尝试多次迭代，从概念上讲，这个问题可以通过以下方式解决：a- 扁平化列表 b- 创建我们想要比较的项目对 c- 索引我们得到的匹配项 d- 处理（即操纵我们需要的值）满足既定条件的项目。

这是代码，希望它能帮助其他人解决这个问题，或者这个问题的任何其他衍生问题：

#flattening the list for ease of operation
def flatten(input_list):
  flat_list = []
  for sublist in input_list:
    for item in sublist:
      flat_list.append(item)
  return flat_list
act_flt = flatten(flatten(act_all_fun))

# Creating unique pattern-based actions
unq_act = []
idx_act = []
for i in range(len(act_flt)):
    if isinstance(act_flt[i], int) and act_flt[i] != 254:
        unq_act.append([act_flt[i], act_flt[i+1]])
        idx_act.append(i)
unique_second_elements = []
for item in new_list:
    if item[1] not in unique_second_elements:
        unique_second_elements.append(item[1])
        
for second_element in unique_second_elements:
    for item in unq_act:
        if item[1] == second_element:
            item[0] += (unique_second_elements.index(second_element) * 30)
            
# Creating unique free pattern-based actions
unq_free = []
idx_free = []
for i in range(len(act_flt)):
    if isinstance(act_flt[i], int) and act_flt[i] == 254:
        unq_free.append([act_flt[i], act_flt[i+1]])
        idx_free.append(i)
unique_second_elements = []
for item in unq_free:
    if item[1] not in unique_second_elements:
        unique_second_elements.append(item[1])
        
for second_element in unique_second_elements:
    for item in unq_free:
        if item[1] == second_element:
            item[0] += (unique_second_elements.index(second_element) - 2)

现在处理之前的输出：

dow1 = len(act_all_fun[0])*4 -1
dow2 = len(act_all_fun[1])*4 
dow3 = len(act_all_fun[2])*4 
dow4 = len(act_all_fun[3])*4 
dow5 = len(act_all_fun[4])*4 
dow6 = len(act_all_fun[5])*4 
dow7 = len(act_all_fun[6])*4 

for i in idx_act:
    if 0<=i<=dow1:
        day = 0
        act_tmp = i//4 #finding the action entry
        act_all_fun[day][act_tmp][1] = unq_act[idx_act.index(i)][0]
        print(i)
    elif dow1<i<=(dow1+dow2):
        day = 1
        act_tmp = (i-dow1)//4 #finding the action entry
        act_all_fun[day][act_tmp][1] = unq_act[idx_act.index(i)][0]
        print(i)
    elif (dow1+dow2)<i<=(dow1+dow2+dow3):
        day = 2
        act_tmp = (i-(dow1+dow2))//4
        act_all_fun[day][act_tmp][1] = unq_act[idx_act.index(i)][0]
        print(i)
    elif dow1+dow2+dow3<i<=(dow1+dow2+dow3+dow4):
        day = 3
        act_tmp = (i-(dow1+dow2+dow3))//4
        act_all_fun[day][act_tmp][1] = unq_act[idx_act.index(i)][0]
        print(i)
    elif dow1+dow2+dow3+dow4<i<=(dow1+dow2+dow3+dow4+dow5):
        day = 4
        act_tmp = (i-(dow1+dow2+dow3+dow4))//4
        act_all_fun[day][act_tmp][1] = unq_act[idx_act.index(i)][0]
        print(i)
    elif dow1+dow2+dow3+dow4+dow5<i<=(dow1+dow2+dow3+dow4+dow5+dow6):
        day = 5
        act_tmp = (i-(dow1+dow2+dow3+dow4+dow5))//4
        act_all_fun[day][act_tmp][1] = unq_act[idx_act.index(i)][0]
        print(i)
    else:
        day = 6
        act_tmp = (i-(dow1+dow2+dow3+dow4+dow5+dow6))//4
        act_all_fun[day][act_tmp][1] = unq_act[idx_act.index(i)][0]
        print(i)

for j in idx_free:
    if 0<=j<=dow1:
        day = 0
        act_tmp = j//4
        act_all_fun[day][act_tmp][1] = unq_free[idx_free.index(j)][0]
        print(j)
    elif dow1<j<=(dow1+dow2):
        day = 1
        act_tmp = (j-dow1)//4
        act_all_fun[day][act_tmp][1] = unq_free[idx_free.index(j)][0]
        print(j)
    elif dow1+dow2<j<=(dow1+dow2+dow3):
        day = 2
        act_tmp = (j-(dow1+dow2))//4
        act_all_fun[day][act_tmp][1] = unq_free[idx_free.index(j)][0]
        print(j)
    elif dow1+dow2+dow3<j<=(dow1+dow2+dow3+dow4):
        day = 3
        act_tmp = (j-(dow1+dow2+dow3))//4
        act_all_fun[day][act_tmp][1] = unq_free[idx_free.index(j)][0]
        print(j)
    elif dow1+dow2+dow3+dow4<j<=(dow1+dow2+dow3+dow4+dow5):
        day = 4
        act_tmp = (j-(dow1+dow2+dow3+dow4))//4
        act_all_fun[day][act_tmp][1] = unq_free[idx_free.index(j)][0]
        print(j)
    elif dow1+dow2+dow3+dow4+dow5<j<=(dow1+dow2+dow3+dow4+dow5+dow6):
        day = 5
        act_tmp = (j-(dow1+dow2+dow3+dow4+dow5))//4
        act_all_fun[day][act_tmp][1] = unq_free[idx_free.index(j)][0]
        print(j)
    else:
        day = 6
        act_tmp = (j-(dow1+dow2+dow3+dow4+dow5+dow6))//4
        act_all_fun[day][act_tmp][1] = unq_free[idx_free.index(j)][0]
        print(j)

随意从中提取您需要的内容。