在集合列表中查找重复项的计数

发布于 2025-02-05 03:38:33 字数 668 浏览 3 评论 0原文

嗨，我必须从数据中检查字符串的重叠和重复，我可以这样做，任何人都可以帮助我找到字符串的副本。我有此数据s = [（100，350，“ a”），（125，145，“ a”），（200，400，“ d”），（0，10，“ a”）]，我完成了重叠部分，但是请检查ODF字符串，我需要帮助。

def overlap(a, b) -> bool:
  a_start, a_end, _ = a
  b_start, b_end, _ = b
  return a_start < b_end and b_start < a_end
ls = [(100, 350,"a"), (125, 145,"a"), (200, 400, "d"), (0, 10, "a")]
overlaps = set()
for idx_a in range(len(ls)):
    for idx_b in range(len(ls)):
        if idx_a != idx_b:  
            if overlap(ls[idx_a], ls[idx_b]):
                overlaps.add(ls[idx_a])
                overlaps.add(ls[idx_b])

print(f"Number of overlaps: {len(overlaps)}")

原文

Hi i have to check the overlapping and duplicate of string from the data , i could do it can anyone help me to find the duplicate of string .I have this data s = [(100, 350,"a"), (125, 145,"a"), (200, 400, "d"), (0, 10, "a")] and i done the overlap part but theduplicate check odf string i need help .

def overlap(a, b) -> bool:
  a_start, a_end, _ = a
  b_start, b_end, _ = b
  return a_start < b_end and b_start < a_end
ls = [(100, 350,"a"), (125, 145,"a"), (200, 400, "d"), (0, 10, "a")]
overlaps = set()
for idx_a in range(len(ls)):
    for idx_b in range(len(ls)):
        if idx_a != idx_b:  
            if overlap(ls[idx_a], ls[idx_b]):
                overlaps.add(ls[idx_a])
                overlaps.add(ls[idx_b])

print(f"Number of overlaps: {len(overlaps)}")

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入画浅相思 2025-02-12 03:38:33

似乎您不需要使用设置，以防您只需要重叠数量。

我会这样解决您的问题：

def is_overlapped(a, b) -> bool:  # changed the name for readability
    a_start, a_end, _ = a
    b_start, b_end, _ = b
    return a_start < b_end and b_start < a_end
ls = [(100, 350,"a"), (125, 145,"a"), (200, 400, "d"), (0, 10, "a")]
overlaps = 0 # int instead of set
for idx_a, value_a in enumerate(ls):  # this is more pythonic way to access the index and item at the same time
    for idx_b, value_b in enumerate(ls):
        if idx_a != idx_b:
            if is_overlapped(value_a, value_b):
                overlaps += 1
print(f"Number of overlaps: {overlaps}")

It seems like you don't need to use set in case you need only number of overlaps.

I would solve your problem like this:

def is_overlapped(a, b) -> bool:  # changed the name for readability
    a_start, a_end, _ = a
    b_start, b_end, _ = b
    return a_start < b_end and b_start < a_end
ls = [(100, 350,"a"), (125, 145,"a"), (200, 400, "d"), (0, 10, "a")]
overlaps = 0 # int instead of set
for idx_a, value_a in enumerate(ls):  # this is more pythonic way to access the index and item at the same time
    for idx_b, value_b in enumerate(ls):
        if idx_a != idx_b:
            if is_overlapped(value_a, value_b):
                overlaps += 1
print(f"Number of overlaps: {overlaps}")

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