python - 计算列表与列表的 pandas 列的交集长度

Question

我有一个唯一随机整数列表和一个包含一列列表的数据框，如下所示:

>>> panel
    [1, 10, 9, 5, 6]

>>> df
       col1 
    0  [1, 5]
    1  [2, 3, 4]
    2  [9, 10, 6]

我想要的输出是 panel 和数据框中每个单独列表之间重叠的长度:

>>> result
       col1        res
    0  [1, 5]      2
    1  [2, 3, 4]   0
    2  [9, 10, 6]  3

目前，我正在使用 apply 函数，但我想知道是否有更快的方法，因为我需要创建很多面板并为每个面板循环执行此任务。

# My version right now
def cntOverlap(panel, series):
    # Typically the lists inside df will be much shorter than panel, 
    # so I think the fastest way would be converting the panel into a set 
    # and loop through the lists within the dataframe

    return sum(1 if x in panel for x in series)
    #return len(np.setxor1d(list(panel), series))
    #return len(panel.difference(series))


for i, panel in enumerate(list_of_panels):
    panel = set(panel)
    df[f"panel_{i}"] = df["col1"].apply(lambda x: cntOverlap(panel, x))

Answer 1

由于每行的可变长度数据，我们需要在 Python 中进行迭代(显式或隐式，即在幕后)。但是，我们可以优化到每次迭代计算最小化的水平。遵循这种理念，这里有一个带有数组分配和一些掩码的 -

# l is input list of unique random integers
s = df.col1
max_num = 10 # max number in df, if not known use : max(max(s))
map_ar = np.zeros(max_num+1, dtype=bool)
map_ar[l] = 1
df['res'] = [map_ar[v].sum() for v in s]

或者使用 2D 数组分配来进一步最小化每次迭代计算 -

map_ar = np.zeros((len(df),max_num+1), dtype=bool)
map_ar[:,l] = 1
for i,v in enumerate(s):
    map_ar[i,v] = 0
df['res'] = len(l)-map_ar.sum(1)