How to create a DataFrame from dict of unequal length lists, and truncating to a specific length?

Question

I have a dict of lists (which have variable lengths), I am looking forward to an efficient way of creating a Dataframe from it.

Assume I have minimum list length, so I can truncate size of bigger lists while creating Dataframe.

Here is my dummy code

data_dict = {'a': [1,2,3,4], 'b': [1,2,3], 'c': [2,45,67,93,82,92]}
min_length = 3

I can have a dictionary of 10k or 20k keys, so looking for an efficient way to create a DataFrame like bellow

>>> df
   a  b   c
0  1  1   2
1  2  2  45
2  3  3  67

Answer 1

You can filter values of dict in dict comprehension, then DataFrame works perfectly:

print ({k:v[:min_length] for k,v in data_dict.items()})
{'b': [1, 2, 3], 'c': [2, 45, 67], 'a': [1, 2, 3]}


df = pd.DataFrame({k:v[:min_length] for k,v in data_dict.items()})
print (df)
   a  b   c
0  1  1   2
1  2  2  45
2  3  3  67

If is possible some length can be less as min_length add Series:

data_dict = {'a': [1,2,3,4], 'b': [1,2], 'c': [2,45,67,93,82,92]}
min_length = 3

df = pd.DataFrame({k:pd.Series(v[:min_length]) for k,v in data_dict.items()})
print (df)
   a    b   c
0  1  1.0   2
1  2  2.0  45
2  3  NaN  67

Timings:

In [355]: %timeit (pd.DataFrame({k:v[:min_length] for k,v in data_dict.items()}))
The slowest run took 5.32 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 520 µs per loop

In [356]: %timeit (pd.DataFrame({k:pd.Series(v[:min_length]) for k,v in data_dict.items()}))
The slowest run took 4.50 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 937 µs per loop

#Allen's solution
In [357]: %timeit (pd.DataFrame.from_dict(data_dict,orient='index').T.dropna())
1 loop, best of 3: 16.7 s per loop

Code for timings:

np.random.seed(123)
L = list('ABCDEFGH')
N = 500000
min_length = 10000

data_dict = {k:np.random.randint(10, size=np.random.randint(N)) for k in L}