How to create a DataFrame from dict of unequal length lists, and truncating to a specific length?
pandas, python, python-3.x
Solution
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}
Problem
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 ```