Iterate over pandas dataframe columns containing nested arrays

user_13015189 0
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I hope you can help me with this issue,

I've this data below (Columns names whatever)

data=([['file0090',
    ([[ 84,  55, 189],
   [248, 100,  18],
   [ 68, 115,  88]])],
   ['file6565',
    ([[ 86,  58, 189],
   [24, 10,  118],
   [ 68, 11,  8]])
   ]])

I need to iterate over columns 0 and 1 into a list in sort I can transform into a Dataframe with this output:

col0          col1  col2   col3 
file0090      84     55     189
file0090      248    100      1
file0090      68     115    88
file6565      86     58    189
file6565      24    10     118
file6565      68    11      8

I've tested all dataframe iteration with iterrows, iteritems, items, and append into a list but the results always turn around the same output and I dont get how separate the items form these arrays

thank you in advance if you can help.

pandas dataframe python
4 Answers

You can try this:-

data_f = [[i[0]]+j for i in data for j in i[1]]
df = pd.DataFrame(data_f, columns =['col0','col1','col2','col3'])

Output:-

col0          col1  col2   col3 
file0090      84     55     189
file0090      248    100      1
file0090      68     115    88
file6565      86     58    189
file6565      24    10     118
file6565      68    11      8
user_87032840

You can do explode with a join after crreating another df from the series of lists:

df = pd.DataFrame(data).add_prefix('col')

out = df.explode('col1').reset_index(drop=True)
out = out.join(pd.DataFrame(out.pop('col1').tolist()).add_prefix('col_'))

Adding another solution if the list structure is similar:

l = [*itertools.chain.from_iterable(data)]
pd.DataFrame(np.vstack(l[1::2]),index = np.repeat(l[::2],len(l[1])))

      col0  col_0  col_1  col_2
0  file0090     84     55    189
1  file0090    248    100     18
2  file0090     68    115     88
3  file6565     86     58    189
4  file6565     24     10    118
5  file6565     68     11      8
user_98406370

We can do explode with row the do it explode with column again 

s = pd.DataFrame(data).set_index(0)[1].explode()
df = pd.DataFrame(s.tolist(), index = s.index.values)

df
Out[396]: 
            0    1    2
file0090   84   55  189
file0090  248  100   18
file0090   68  115   88
file6565   86   58  189
file6565   24   10  118
file6565   68   11    8
user_79645270

You can create a custom function to output the correct form of data. 

from itertools import chain
def transform(d):
    for l in d:
        *x, y = l
        yield list(map(lambda s: x+s, y))

df = pd.DataFrame(chain(*transform(data)))
df
          0    1    2    3
0  file0090   84   55  189
1  file0090  248  100   18
2  file0090   68  115   88
3  file6565   86   58  189
4  file6565   24   10  118
5  file6565   68   11    8

Timeit results of all the solutions:

# YOBEN_S's answer
In [275]: %%timeit
     ...: s = pd.DataFrame(data).set_index(0)[1].explode()
     ...: df = pd.DataFrame(s.tolist(), index = s.index.values)
     ...:
     ...:
1.52 ms ± 59.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#Anky's answer
In [276]: %%timeit
     ...: df = pd.DataFrame(data).add_prefix('col')
     ...: out = df.explode('col1').reset_index(drop=True)
     ...: out = out.join(pd.DataFrame(out.pop('col1').tolist()).add_prefix('col_'))
     ...:
     ...:
3.71 ms ± 606 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

#Dhaval's answer
In [277]: %%timeit
     ...: data_f = []
     ...: for i in data:
     ...:     for j in i[1]:
     ...:         data_f.append([i[0]]+j)
     ...: df = pd.DataFrame(data_f, columns =['col0','col1','col2','col3'])
     ...:
     ...:
712 µs ± 24.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#My answer
In [280]: %%timeit
     ...: pd.DataFrame(chain(*transform(data)))
     ...:
     ...:
489 µs ± 8.91 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#Using List comp of Dhaval's answer

In [306]: %%timeit
     ...: data_f = [[i[0]]+j for i in data for j in i[1]]
     ...: df = pd.DataFrame(data_f, columns =['col0','col1','col2','col3'])
     ...:
     ...:
586 µs ± 25 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

#Anky's 2nd solution

In [308]: %%timeit
     ...: l = [*chain.from_iterable(data)]
     ...: pd.DataFrame(np.vstack(l[1::2]),index = np.repeat(l[::2],len(l[1])))
     ...:
     ...:
221 µs ± 18.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
user_124164530
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