Counting length of intersection of a list with pandas column of lists

Miguel Jiahao Wang :

I have a list of unique random integers and a dataframe with a column of lists, like below:

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

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

The output I would like to have is the length of the overlapping between panel and each individual list in the dataframe:

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

Currently, I am using the apply function, but I was wondering if there are faster ways, since I need to create a lot of panels and loop through this task for each panel.

# 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))

Divakar :

Owing to the variable length data per row, we need to iterate (explicitly or implicitly i.e. under the hoods) staying within Python. But, we can optimize to a level where per iteration compute is minimized. Going with that philosophy, here's one with array-assignment and some masking -

# 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]

Alternatively with 2D array-assignment to further minimize per-iteration-compute -

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)