将字符串拆分为列表并将项目转换为 int

Question

I have a pandas dataframe where I have a column values like this:

0       16 0
1    7 1 2 0
2          5
3          1
4         18

What I want is to create another column, modified_values , that contains a list of all the different numbers that I will get after splitting each value. The new column will be like this:

0       [16, 0]
1    [7, 1, 2, 0]
2          [5]
3          [1]
4         [18]

Beware the values in this list should be int and not strings .

Things that I am aware of:

1) I can split the column in a vectorized way like this df.values.str.split(" ") . This will give me the list but the objects inside the list will be strings. I can add another operation on top of that like this df.values.str.split(" ").apply(func to convert values to int) but that wouldn't be vectorized

2) I can directly do this df['modified_values']= df['values'].apply(func that splits as well as converts to int)

The second one will be much slower than the first for sure but I am wondering if the same thing can be achieved in a vectorized way.

Answer 1

No native "vectorised" solution is possible

I'm highlighting this because it's a common mistake to assume pd.Series.str methods are vectorised. They aren't. They offer convenience and error-handling at the cost of efficiency. For clean data only , eg no NaN values, a list comprehension is likely your best option:

df = pd.DataFrame({'A': ['16 0', '7 1 2 0', '5', '1', '18']})

df['B'] = [list(map(int, i.split())) for i in df['A']]

print(df)

         A             B
0     16 0       [16, 0]
1  7 1 2 0  [7, 1, 2, 0]
2        5           [5]
3        1           [1]
4       18          [18]

Performance benchmarking

To illustrate performance issues with pd.Series.str , you can see for larger dataframes how the more operations you pass to Pandas, the more performance deteriorates:

df = pd.concat([df]*10000)

%timeit [list(map(int, i.split())) for i in df['A']]            # 55.6 ms
%timeit [list(map(int, i)) for i in df['A'].str.split()]        # 80.2 ms
%timeit df['A'].str.split().apply(lambda x: list(map(int, x)))  # 93.6 ms

list as elements in pd.Series is also anti-Pandas

As described here , holding lists in series gives 2 layers of pointers and is not recommended:

Don't do this . Pandas was never designed to hold lists in series / columns. You can concoct expensive workarounds, but these are not recommended.

The main reason holding lists in series is not recommended is you lose the vectorised functionality which goes with using NumPy arrays held in contiguous memory blocks. Your series will be of object dtype, which represents a sequence of pointers, much like list . You will lose benefits in terms of memory and performance, as well as access to optimized Pandas methods.

See also What are the advantages of NumPy over regular Python lists? The arguments in favour of Pandas are the same as for NumPy.

Answer 2

The double for comprehension is 33% faster than the map comprehension from the jpp's answer. Numba trick is 250 times faster than the map comprehension from jpp's answer, but you get a pandas DataFrame with floats and nan 's and not a series of lists. Numba is included in Anaconda.

Benchmarks:

%timeit pd.DataFrame(nb_calc(df.A))            # numba trick       0.144 ms
%timeit [int(x) for i in df['A'] for x in i.split()]            # 23.6   ms
%timeit [list(map(int, i.split())) for i in df['A']]            # 35.6   ms
%timeit [list(map(int, i)) for i in df['A'].str.split()]        # 50.9   ms
%timeit df['A'].str.split().apply(lambda x: list(map(int, x)))  # 56.6   ms

Code for Numba function:

@numba.jit(nopython=True, nogil=True)
def str2int_nb(nb_a):
    n1 = nb_a.shape[0]
    n2 = nb_a.shape[1]
    res = np.empty(nb_a.shape)
    res[:] = np.nan
    j_res_max = 0
    for i in range(n1):
        j_res = 0
        s = 0
        for j in range(n2):
            x = nb_a[i,j]
            if x == 32:
                res[i,j_res]=np.float64(s)
                s=0
                j_res+=1
            elif x == 0:
                break
            else:
                s=s*10+x-48
        res[i,j_res]=np.float64(s)
        if j_res>j_res_max:
            j_res_max = j_res

    return res[:,:j_res_max+1]

def nb_calc(s):
    a_temp = s_a.values.astype("U")
    nb_a = a_temp.view("uint32").reshape(len(s_a),-1).astype(np.int8)
    str2int_nb(nb_a)

Numba does not support strings. So I first convert to array of int8 and only then work with it. Conversion to int8 actually takes 3/4 of the execution time.

The output of my numba function looks like this:

      0    1    2    3
-----------------------
0  16.0  0.0  NaN  NaN
1   7.0  1.0  2.0  0.0
2   5.0  NaN  NaN  NaN
3   1.0  NaN  NaN  NaN
4  18.0  NaN  NaN  NaN