有关数字列表和间隔列表，请查找每个数字所在的时间间隔

Question

In a dictionary as follow:

example = {'Chr3' : [[1, 4], [5, 10], [13, 17]]}

How can I know in which list interval numbers 6 and 15 are located?

I can write a solution with a simple for loop:

for key in example.keys():
    for element in example[key]:
        if element[0]<= 6 <= element[1] or element[0] <= 15 <= element[1]:
            print element

But in my real problem, I don't have only two numbers but a list of numbers.

So, I was thinking if there is a more pythonic way of solving such a problem using special list or dictionary methods/functions?

EDIT: Because of down votes, I clarify that I know how to write the code using another loop on top for a list of numbers, I was wondering if there is a better solution to the same problem, a more efficient one!

Answer 1

Create a dictionary which maps any number from your list of numbers to the corresponding interval(s) in example['Chr3'] .

>>> {x:filter(lambda y: y[0] <= x <= y[1], example['Chr3']) for x in lst}
{6: [5, 10], 15: [13, 17]}

Answer 2

Use a function!

example = {'Chr3' : [[1, 4], [5, 10], [13, 17]]}
example_nums = [3, 6, 11, 15, 17]

def find_interval(dct, num):
    for key, values in dct.items():
        for left, right in values:
            if left <= num <= right:
                return [left, right]

print(find_interval(example, 6))
print(find_interval(example, 15))

# Now you can use the function to find the interval for each number
for num in example_nums:
    print(num, find_interval(example, num))

Output

[5, 10]
[13, 17]
3 [1, 4]
6 [5, 10]
11 None
15 [13, 17]
17 [13, 17]

https://repl.it/Bb78/4

Answer 3

Can you show more specific example of your list? I think your code though would work by incorporating the line for a loop over the list:

for x in my list:
    for key in example.keys():
         for element in example[key]:
            if element[0]<= x <= element[1] or element[0] <= x <= element[1]:
                   print element

Answer 4

Your code looks for any match not multiple matches, you can use a list comp with any

from itertools import chain

pot = [6, 15]
print([[a, b] for a, b in  chain(*example.values()) if any(a <= i <= b for i in pot)])
[[5, 10], [13, 17]]

If you just care about any match you only really need to check the min and max of the potential numbers:

pot = [6, 15,7,9,8]
from itertools import chain

mn,mx = min(pot), max(pot)
print([[a, b] for a, b in  chain(*example.values()) if mn <= a <= mx or mn <= b <= mx])

[[5, 10], [13, 17]]

So you are now doing constant work to check a and b instead of linear.

Or using python3 you could use range and use in which is an O(1) operation in python3:

pot = [6, 15,7,9,8]
mn ,mx = min(pot),  max(pot)
inter = range(mn, mx+1)
print([[a, b] for a, b in chain(*example.values()) if a in inter or b in inter])
[[5, 10], [13, 17]]