I have this df:
l1 = ['a', 'b', 'c']
l2 = ['x', ['y1', 'y2', 'y3'], 'z']
df = pd.DataFrame(list(zip(l1, l2)), columns = ['l1', 'l2'])
result:
l1 l2
0 a x
1 b [y1, y2, y3]
2 c z
What i need is to unpack the inner list in l2 and spread the corresponding value in l1 like this:
l1 l2
0 a x
1 b y1
2 b y2
3 b y3
4 c z
What is the proper way to do this? Thanks.
I think you can use numpy.repeat for repeat values by legths by str.len and flat values of nested lists by chain:
from itertools import chain
df1 = pd.DataFrame({
"l1": np.repeat(df.l1.values, df.l2.str.len()),
"l2": list(chain.from_iterable(df.l2))})
print (df1)
l1 l2
0 a x
1 b y1
2 b y2
3 b y3
4 c z
Timings:
#[100000 rows x 2 columns]
np.random.seed(10)
N = 100000
l1 = ['a', 'b', 'c']
l1 = np.random.choice(l1, N)
l2 = [list(tuple(string.ascii_letters[:np.random.randint(1, 10)])) for _ in np.arange(N)]
df = pd.DataFrame({"l1":l1, "l2":l2})
df.l2 = df.l2.apply(lambda x: x if len(x) !=1 else x[0])
#print (df)
In [91]: %timeit (pd.DataFrame([(left, right) for outer in zip(l1, l2) for left, right in zip_longest(*outer, fillvalue=outer[0])]))
1 loop, best of 3: 242 ms per loop
In [92]: %timeit (pd.DataFrame({ "l1": np.repeat(df.l1.values, df.l2.str.len()), "l2": list(chain.from_iterable(df.l2))}))
10 loops, best of 3: 84.6 ms per loop
Conclusion:
numpy.repeat is 3 times faster as zip_longest solution in larger df.
EDIT:
For compare with loop version is necessery smaller df, because very slow:
#[1000 rows x 2 columns]
np.random.seed(10)
N = 1000
l1 = ['a', 'b', 'c']
l1 = np.random.choice(l1, N)
l2 = [list(tuple(string.ascii_letters[:np.random.randint(1, 10)])) for _ in np.arange(N)]
df = pd.DataFrame({"l1":l1, "l2":l2})
df.l2 = df.l2.apply(lambda x: x if len(x) !=1 else x[0])
#print (df)
def alexey(df):
df2 = pd.DataFrame(columns=df.columns,index=df.index)[0:0]
for idx in df.index:
new_row = df.loc[idx, :].copy()
for res in df.ix[idx, 'l2']:
new_row.set_value('l2', res)
df2.loc[len(df2)] = new_row
return df2
print (alexey(df))
In [20]: %timeit (alexey(df))
1 loop, best of 3: 11.4 s per loop
In [21]: %timeit pd.DataFrame([(left, right) for outer in zip(l1, l2) for left, right in zip_longest(*outer, fillvalue=outer[0])])
100 loops, best of 3: 2.57 ms per loop
In [22]: %timeit pd.DataFrame({ "l1": np.repeat(df.l1.values, df.l2.str.len()), "l2": list(chain.from_iterable(df.l2))})
The slowest run took 4.42 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 1.41 ms per loop