Vous pouvez d'abord créer tuples de toutes les valeurs et les séparer pour les sous-listes, si la performance est importante:
from itertools import groupby
L = list(zip(df['words'], df['tags']))
print (L)
[('I', 'WW'), ('am', 'XX'), ('newbie', 'YY'),
('.', 'ZZ'), ('You', 'WW'), ('are', 'XX'),
('cool', 'YY'), ('.', 'ZZ')]
sep = ('.','ZZ')
new_L = [list(g) + [sep] for k, g in groupby(L, lambda x: x==sep) if not k]
print (new_L)
[[('I', 'WW'), ('am', 'XX'), ('newbie', 'YY'), ('.', 'ZZ')],
[('You', 'WW'), ('are', 'XX'), ('cool', 'YY'), ('.', 'ZZ')]]
minutage:
df = pd.concat([df]*1000).reset_index(drop=True)
def zero(df):
dft = df.apply(tuple, 1)
return ([x.values.tolist() for _, x in dft.groupby((dft == ('.', 'ZZ')).shift().cumsum().bfill())])
In [55]: %timeit ([list(g) + [('.','ZZ')] for k, g in groupby(list(zip(df['words'], df['tags'])), lambda x: x==('.','ZZ')) if not k])
100 loops, best of 3: 4.14 ms per loop
def pir(df):
v = df.values
return ([list(map(tuple, x)) for x in np.split(v, np.where((v == ['.', 'ZZ']).all(1)[:-1])[0] + 1)])
In [68]: %timeit (pir(df))
10 loops, best of 3: 21.9 ms per loop
In [56]: %timeit (zero(df))
1 loop, best of 3: 328 ms per loop
In [57]: %timeit (df.groupby((df.shift().values == ['.', 'ZZ']).all(axis=1).cumsum()).apply(lambda group: list(zip(group['words'], group['tags']))).values.tolist())
1 loop, best of 3: 286 ms per loop
In [58]: %timeit (list(filter(None,[i.apply(tuple,1).values.tolist() for i in np.array_split(df,df[(df['words'] == '.') & (df['tags'] == 'ZZ')].index+1)])))
1 loop, best of 3: 1.31 s per loop
Pour séparés à des listes secondaires que je crée question, vous pouvez vérifier la solution here:
def jez_coldspeed(df):
L = list(zip(df['words'], df['tags']))
L2 = []
for i in L[::-1]:
if i == ('.','ZZ'):
L2.append([])
L2[-1].append(i)
return [x[::-1] for x in L2[::-1]]
def jez_coldspeed1(df):
L = list(zip(df['words'], df['tags']))
L2 = []
sep = ('.','ZZ')
for i in reversed(L):
if i == sep:
L2.append([])
L2[-1].append(i)
return [x[::-1] for x in reversed(L2)]
In [74]: %timeit (jez_coldspeed(df))
100 loops, best of 3: 2.96 ms per loop
In [75]: %timeit (jez_coldspeed1(df))
100 loops, best of 3: 2.95 ms per loop
def jez_theBuzzyCoder(df):
L = list(zip(df['words'], df['tags']))
a = list()
start = 0
sep = ('.', 'ZZ')
while start < len(L) and (L.index(sep, start) != -1):
end = L.index(sep, start) + 1
a.append(L[start:end])
start = end
return a
print (jez_theBuzzyCoder(df))
In [81]: %timeit (jez_theBuzzyCoder(df))
100 loops, best of 3: 3.16 ms per loop
Cette méthode est certainement la plus rapide. – Dark
Très rapide en effet. – Alexander
Aha! Rapide en effet! –