从条目长度不同的字典创建数据框

Say I have a dictionary with 10 key-value pairs. Each entry holds a numpy array. However, the length of the array is not the same for all of them.

How can I create a dataframe where each column holds a different entry?

When I try:

pd.DataFrame(my_dict)

I get:

ValueError: arrays must all be the same length

Any way to overcome this? I am happy to have Pandas use NaN to pad those columns for the shorter entries.

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

You can also use pd.concat along axis=1 with a list of pd.Series objects:

import pandas as pd, numpy as np

d = {'A': np.array([1,2]), 'B': np.array([1,2,3,4])}

res = pd.concat([pd.Series(v, name=k) for k, v in d.items()], axis=1)

print(res)

     A  B
0  1.0  1
1  2.0  2
2  NaN  3
3  NaN  4

Both the following lines work perfectly :

pd.DataFrame.from_dict(df, orient='index').transpose() #A

pd.DataFrame(dict([ (k,pd.Series(v)) for k,v in df.items() ])) #B (Better)

But with %timeit on Jupyter, I’ve got a ratio of 4x speed for B vs A, which is quite impressive especially when working with a huge data set (mainly with a big number of columns/features).

If you don’t want it to show NaN and you have two particular lengths, adding a ‘space’ in each remaining cell would also work.

import pandas

long = [6, 4, 7, 3]
short = [5, 6]

for n in range(len(long) - len(short)):
    short.append(' ')

df = pd.DataFrame({'A':long, 'B':short}]
# Make sure Excel file exists in the working directory
datatoexcel = pd.ExcelWriter('example1.xlsx',engine = 'xlsxwriter')
df.to_excel(datatoexcel,sheet_name = 'Sheet1')
datatoexcel.save()

   A  B
0  6  5
1  4  6
2  7   
3  3   

If you have more than 2 lengths of entries, it is advisable to make a function which uses a similar method.

pd.DataFrame([my_dict]) will do!