Creating dataframe from a dictionary where entries have different lengths

In Python 3.x:

import pandas as pd
import numpy as np

d = dict( A = np.array([1,2]), B = np.array([1,2,3,4]) )
    
pd.DataFrame(dict([ (k,pd.Series(v)) for k,v in d.items() ]))

Out[7]: 
    A  B
0   1  1
1   2  2
2 NaN  3
3 NaN  4

In Python 2.x:

replace d.items() with d.iteritems().

Here's a simple way to do that:

In[20]: my_dict = dict( A = np.array([1,2]), B = np.array([1,2,3,4]) )
In[21]: df = pd.DataFrame.from_dict(my_dict, orient='index')
In[22]: df
Out[22]: 
   0  1   2   3
A  1  2 NaN NaN
B  1  2   3   4
In[23]: df.transpose()
Out[23]: 
    A  B
0   1  1
1   2  2
2 NaN  3
3 NaN  4

A way of tidying up your syntax, but still do essentially the same thing as these other answers, is below:

>>> mydict = {'one': [1,2,3], 2: [4,5,6,7], 3: 8}

>>> dict_df = pd.DataFrame({ key:pd.Series(value) for key, value in mydict.items() })

>>> dict_df

   one  2    3
0  1.0  4  8.0
1  2.0  5  NaN
2  3.0  6  NaN
3  NaN  7  NaN

A similar syntax exists for lists, too:

>>> mylist = [ [1,2,3], [4,5], 6 ]

>>> list_df = pd.DataFrame([ pd.Series(value) for value in mylist ])

>>> list_df

     0    1    2
0  1.0  2.0  3.0
1  4.0  5.0  NaN
2  6.0  NaN  NaN

Another syntax for lists is:

>>> mylist = [ [1,2,3], [4,5], 6 ]

>>> list_df = pd.DataFrame({ i:pd.Series(value) for i, value in enumerate(mylist) })

>>> list_df

   0    1    2
0  1  4.0  6.0
1  2  5.0  NaN
2  3  NaN  NaN

You may additionally have to transpose the result and/or change the column data types (float, integer, etc).

While this does not directly answer the OP's question. I found this to be an excellent solution for my case when I had unequal arrays and I'd like to share:

from pandas documentation

In [31]: d = {'one' : Series([1., 2., 3.], index=['a', 'b', 'c']),
   ....:      'two' : Series([1., 2., 3., 4.], index=['a', 'b', 'c', 'd'])}
   ....: 

In [32]: df = DataFrame(d)

In [33]: df
Out[33]: 
   one  two
a    1    1
b    2    2
c    3    3
d  NaN    4