这里有一个（不那么简单）的方法。它的速度很快，因为它使用矢量化函数，但它的时间和内存复杂度都是O(len(df1) * len(df2))。根据数据集的规模，内存需求可能会超过计算机硬件。
其思想是使用numpy广播将df1中的每一行与df2中的每一行进行比较，搜索满足以下条件的对：

• 具有相同的id
• 具有重叠持续时间（start - end）。

...然后对它们执行计算：

# Extract the columns to numpy array
# For the columns of df1, raise each by one dimension to prepare
# for numpy broadcasting
id1, start1, end1 = [col[:, None] for col in df1.to_numpy().T]
id2, start2, end2, quantity2 = df2.to_numpy().T

# Match each row in df1 to each row in df2
# `is_match` is a matrix where if cell (i, j) is True, row i of
# df1 matches row j of df2
is_match = (id1 == id2) & (start1 <= end2) & (start2 <= end1)

# `start` is a matrix where cell (i, j) is the maximum start time
# between row i of df1 and row j of df2
start = np.maximum(
    np.tile(start1, len(df2)),
    np.tile(start2, (len(df1), 1))
)

# Likewise, `end` is a matrix where cell (i, j) is the minium end
# time between row i of df1 and row j of df2
end = np.minimum(
    np.tile(end1, len(df2)),
    np.tile(end2, (len(df1), 1))
)

# This assumes that every row in df1 has a duration of 15
df1["quantity"] = (is_match * (end - start) * quantity2).sum(axis=1) / 15

# This allow each row in df1 to have a different duration
df1["quantity"] = (is_match * (end - start) * quantity2).sum(axis=1) / (end1 - start1)[:, 0]