Python for 循环未正确迭代第二个条件

Question

我有一长串采用这种形式的元组

num_list = [('A1', 4, 'FF', 977.98), ('A4', 14, 'CC', 249.12), ('A1', 14, 'EE', 386.42), ('A9', 19, 'BB', 919.21), ('A4', 14, 'CC', 109.80)]

每个元组的第一个元素对应一个ID，第二个是事件发生的日期，第三个是事件的类别，第四个是事件的值。

任务是打印前 5 天的最大值，例如在第 6 天，对于每个单独的 ID，即基本形式的东西，只应考虑第 1 天到第 5 天的事件

Day   ID    Max
7     A1    400
7     A2    350
8     A1    750

目前我在声明 num_list 后有以下代码，其中 ID 是所有 ID 值的集合。 14 和 18 用于天数到第 19 天

first_value = 1
fifth_value = 5

for id in ID:
    while first_value <= 14 and fifth_value <= 18:
        result = max([i for i in num_list if i[1] <= fifth_value and i[1] >= first_value and i[0] == id], key = lambda  x:x[3])
        first_value += 1
        fifth_value += 1
        print(f"result[0]} {result[1]} {result[3]}")   

问题在于这仅返回第一个 ID 的最大结果，在本例中为 A1。这是正确的，但我不确定为什么它不是为每个 ID 都这样做。我已经检查过，在 while 循环之前它会返回每个 ID，所以我不确定这里的问题是什么

先谢谢了，如果以前发布过类似的东西，我很抱歉，但我找不到它

Answer 1

通过您的示例，我可以很容易地访问任何范围的元组：

#!/usr/bin/python

num_list = [('A1', 4, 'FF', 977.98), ('A4', 14, 'CC', 249.12), ('A1', 14, 'EE', 386.42), ('A9', 19, 'BB', 919.21), ('A4', 14, 'CC', 109.80)]

print [item for item in num_list if item[1] < 19 ]

输出：

mortiz@alberta:~/Documents/projects/python$ python tuples.py 
[('A1', 4, 'FF', 977.98), ('A4', 14, 'CC', 249.12), ('A1', 14, 'EE', 386.42), ('A4', 14, 'CC', 109.8)]

如果你想要一个范围，然后使用 range() 更改它：

print [item for item in num_list if item[1] in range(5,19) ]

输出：

mortiz@alberta:~/Documents/projects/python$ python tuples.py 
[('A4', 14, 'CC', 249.12), ('A1', 14, 'EE', 386.42), ('A4', 14, 'CC', 109.8)]

最后是所选范围的最大值：

#!/usr/bin/python

num_list = [('A1', 4, 'FF', 977.98), ('A4', 14, 'CC', 249.12), ('A1', 14, 'EE', 386.42), ('A9', 19, 'BB', 919.21), ('A4', 14, 'CC', 109.80)]
result=[item for item in num_list if item[1] in range(5,19)]
highest=[item[3] for item in result]

print max(highest)

输出（最大值）

mortiz@alberta:~/Documents/projects/python$ python tuples.py
386.42

这是你想要的吗？

Answer 2

首先，您在此行中缺少 result[0] 的左大括号 print(f"result[0]} {result[1]} {result[3]}") 但是您的主要问题是，在您的 while 循环中，first_value 和 fifth_value 在 while 循环针对第一个 id 运行时不断增加，但是它们永远不会重置，因此对于每个后续 id，您永远不会进入 while 循环。

Answer 3

我不确定你想要什么，但这是我的 2 美分。请注意，id 是 Python 中的保留变量名。 （因此使用id_）

#Let's say we have the data shown in the `num_list`.

num_list = [('A1', 4, 'FF', 977.98), ('A4', 14, 'CC', 249.12), ('A1', 14, 'EE', 386.42), ('A9', 19, 'BB', 919.21), ('A4', 14, 'CC', 109.80), ('A7', 6, 'DD', 243.12), ('A3', 11, 'GG', 612.21)]

#First, we want to sort it by the `day` of occurence. Then we want to list all the possible IDs.

sorted_list = sorted(num_list, key=lambda x:x[1])
ID = sorted(list(set(x[0] for x in num_list)))

# Then, let's define a function, which will get the entry/row with the maximum value
# We give the function two arguments. The `day`, which is the day of observation.
#    If the day of observation is 6, then, only look at the entries which have day of the event 1-5
# The another argument is the id_ we are looking at ('A1', 'A4', ..etc.)

def get_max_within_time_window(day, id_):

    # Get all the entries with the wanted id
    entries_id = [x for x in sorted_list if x[0] == id_]

    # Get all the entries within the 5-day time window. The day of observation not included.
    entries_filtered = [x for x in entries_id if 0 < day-x[1] <= 5]

    # In case of zero matches, return empty list
    if not entries_filtered:
        return []

    # Return the event with highest `value`
    return max(entries_filtered, key=lambda x:x[3])

# Now, let's put this in action.
# Day*: The day of observation
# Day: The day of event occurence. If day of observation is 6, then day of occurence can be 1, 2, 3, 4 or 5.
print('Day*\tDay\tID\tMax')
for day in range(20):
    for id_ in ID:
        found = get_max_within_time_window(day, id_)
        if not found:
            continue
        max_val = found[3]
        day_occured = found[1]
        print(f'{day}\t{day_occured}\t{id_}\t{max_val}')

结果

Day*    Day     ID      Max
5       4       A1      977.98
6       4       A1      977.98
7       4       A1      977.98
7       6       A7      243.12
8       4       A1      977.98
8       6       A7      243.12
9       4       A1      977.98
9       6       A7      243.12
10      6       A7      243.12
11      6       A7      243.12
12      11      A3      612.21
13      11      A3      612.21
14      11      A3      612.21
15      14      A1      386.42
15      11      A3      612.21
15      14      A4      249.12
16      14      A1      386.42
16      11      A3      612.21
16      14      A4      249.12
17      14      A1      386.42
17      14      A4      249.12
18      14      A1      386.42
18      14      A4      249.12
19      14      A1      386.42
19      14      A4      249.12