I have a pandas data frame. One of the columns has a nested list. I would like to create new columns from the nested list
Example:
L = [[1,2,4],
[5,6,7,8],
[9,3,5]]
I want all the elements in the nested lists as columns. The value should be one if the list has the element and zero if it does not.
1 2 4 5 6 7 8 9 3
1 1 1 0 0 0 0 0 0
0 0 0 1 1 1 1 0 0
0 0 0 1 0 0 0 1 1
You can try the following:
df = pd.DataFrame({"A": L})
df
# A
#0 [1, 2, 4]
#1 [5, 6, 7, 8]
#2 [9, 3, 5]
# for each cell, use `pd.Series(1, x)` to create a Series object with the elements in the
# list as the index which will become the column headers in the result
df.A.apply(lambda x: pd.Series(1, x)).fillna(0).astype(int)
# 1 2 3 4 5 6 7 8 9
#0 1 1 0 1 0 0 0 0 0
#1 0 0 0 0 1 1 1 1 0
#2 0 0 1 0 1 0 0 0 1
pandasVery similar to @Psidom's answer. However, I use pd.value_counts and will handle repeats
Use @Psidom's df
df = pd.DataFrame({'A': L})
df.A.apply(pd.value_counts).fillna(0).astype(int)
numpyMore involved, but speedy
lst = df.A.values.tolist()
n = len(lst)
lengths = [len(sub) for sub in lst]
flat = np.concatenate(lst)
u, inv = np.unique(flat, return_inverse=True)
rng = np.arange(n)
slc = np.hstack([
rng.repeat(lengths)[:, None],
inv[:, None]
])
data = np.zeros((n, u.shape[0]), dtype=np.uint8)
data[slc[:, 0], slc[:, 1]] = 1
pd.DataFrame(data, df.index, u)
1 2 3 4 5 6 7 8 9
0 1 1 0 1 0 0 0 0 0
1 0 0 0 0 1 1 1 1 0
2 0 0 1 0 1 0 0 0 1