我有一长串的元组列表，其格式为
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，第二个元素对应于事件发生的那一天，第三个元素对应于事件的类别，第四个元素对应于事件的值。
任务是打印出前5天的最大值，例如在第6天，只考虑第1天到第5天的事件，每个单独的ID，即基本形式的东西
Day ID Max
7 A1 400
7 A2 350
8 A1 750
目前，在声明num_list之后，我有以下代码，其中ID是所有ID值的集合。使用14和18作为日期直到第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]}")
问题是这只返回第一个ID的最大结果，在这种情况下是A1。这是正确的，但我不确定为什么它没有为每个ID做这件事。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，对不起，如果以前发过类似的内容，但我找不到它
答案 0 :(得分：0)
通过您的示例，我可以轻松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 ]
<强>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)]
如果您想要一个范围，请使用range（）更改它：
print [item for item in num_list if item[1] in range(5,19) ]
<强>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，所选范围的最高值：
#!/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（最高值）
mortiz@alberta:~/Documents/projects/python$ python tuples.py
386.42
这就是你想要的吗？
答案 1 :(得分：0)
首先，此行中result[0]缺少左大括号
print(f"result[0]} {result[1]} {result[3]}")
但是你的主要问题是在你的while循环中，first_value和fifth_value继续增加而while循环运行第一个id，但是它们永远不会被重置，所以对于每个后续的id你永远不会输入while循环。
答案 2 :(得分：0)
我不确定你之后会怎样，但这是我的2美分。请注意，id是Python中的保留变量名称。 （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}')
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