Create a pandas dataframe from a dict of uneven length

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For a dictionary:

d = {
    "a": [1],
    "b": 2,
    "c": [[7, 8, 9], ["a", "b", "c"], [9, 10, 11]],
    "d": None,
}

I want to achieve this dataframe:

pd.DataFrame({"a": [[1], [1], [1]],
              "b": [2, 2, 2],
              "c": [[7, 8, 9], ["a", "b", "c"], [9, 10, 11]],
              "d": [None, None, None]})

     a  b            c     d
0  [1]  2    [7, 8, 9]  None
1  [1]  2    [a, b, c]  None
2  [1]  2  [9, 10, 11]  None

Basically, the columns should duplicate itself until the length of the longest column.

I know in R if i create a dataframe like with NA to indicate the rows i want to duplicate and use tidyr::fill, is there something similar in python?

df = data.frame(
  a = c("a", NA, NA),
  b = c(1, 2, 3)
)
tidyr::fill(df, a)

  a b
1 a 1
2 a 2
3 a 3
pandas tidyverse dataframe r python
3 Answers

Here is an example of possible solution:

d = {
    "a": [1],
    "b": 2,
    "c": [[7, 8, 9], ["a", "b", "c"], [9, 10, 11]],
    "d": None,
}

max_len = max(len(l) if isinstance(l, list) else 1 for l in d.values())

for key in d.keys():
  if isinstance(d[key], list):
    if len(d[key]) != max_len:
      d[key] = np.repeat(d[key], max_len).tolist()
  else:
    d[key] = np.repeat(np.array(d[key]), max_len).tolist()

Result:

{
 'a': [1, 1, 1],
 'b': [2, 2, 2],
 'c': [[7, 8, 9], ['a', 'b', 'c'], [9, 10, 11]],
 'd': [None, None, None]
}

But it will work obviously only for a particular case, when all column but one have only one element. To solve this task generally one should also specify how columns of different length should be handled: should the whole column be repeated and rest trimmed on the last iteration, or should only first / last value be repeated, or some other approach.

user_141686230

It is easy to do with datar

>>> from datar.tibble import tibble
>>> from datar.base import NA, c
>>> from datar.tidyr import fill
>>> 
>>> d = {
...     "a": [[1]], # in order to get [1] as element
...     "b": 2,
...     "c": [[7, 8, 9], ["a", "b", "c"], [9, 10, 11]],
...     "d": [None],
... }
>>> 
>>> df = tibble(d)
>>> df
     a  b            c     d
0  [1]  2    [7, 8, 9]  None
1  [1]  2    [a, b, c]  None
2  [1]  2  [9, 10, 11]  None
>>> df = tibble(
...   a = c("a", NA, NA),
...   b = c(1, 2, 3)
... )
>>> 
>>> fill(df, "a")
   a  b
0  a  1
1  a  2
2  a  3

I am the author of the package. Feel free to submit issues if you have any questions.

user_50881650

Your R code can pretty much be translated into python. Its unclear if you are able to change the dictionary to a similar format as your R example, but if you can:

d = {
    "a": [[1], None, None],
    "b": [2, None, None],
    "c": [[7, 8, 9], ["a", "b", "c"], [9, 10, 11]],
    "d": [None, None, None],
}
pd.DataFrame(d).ffill()

returns

     a    b            c     d                                                                                          
0  [1]  2.0    [7, 8, 9]  None                                                                                          
1  [1]  2.0    [a, b, c]  None                                                                                          
2  [1]  2.0  [9, 10, 11]  None
user_76384010
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