complete

In [1]:

%run nb_helpers.py
from datar.all import *

nb_header(complete)

%run nb_helpers.py from datar.all import * nb_header(complete)

Try this notebook on binder.

★ complete

¶

Turns implicit missing values into explicit missing values.¶

Args:¶

  data: A data frame
  *args: columns to expand. Columns can be atomic lists.
    - To find all unique combinations of x, y and z, including       those not present in the data, supply each variable as a
      separate argument: expand(df, x, y, z).

    - To find only the combinations that occur in the data, use       nesting: expand(df, nesting(x, y, z)).

    - You can combine the two forms. For example,       expand(df, nesting(school_id, student_id), date) would
      produce a row for each present school-student combination
      for all possible dates.

  fill: A named list that for each variable supplies a single value
    to use instead of NA for missing combinations.

  explict: Should both implicit (newly created) and explicit
    (pre-existing) missing values be filled by fill? By default,
    this is TRUE, but if set to FALSE this will limit the fill to only
    implicit missing values.

Returns:¶

  Data frame with missing values completed

In [2]:

df = tibble(
  group = c(c[1:2:1], 1),
  item_id = c(c[1:2:1], 2),
  item_name = c("a", "b", "b"),
  value1 = c[1:3:1],
  value2 = c[4:6:1]
)
df >> complete(f.group, nesting(f.item_id, f.item_name))

df = tibble( group = c(c[1:2:1], 1), item_id = c(c[1:2:1], 2), item_name = c("a", "b", "b"), value1 = c[1:3:1], value2 = c[4:6:1] ) df >> complete(f.group, nesting(f.item_id, f.item_name))

Out[2]:

	group	item_id	item_name	value1	value2
	<int64>	<int64>	<object>	<float64>	<float64>
0	1	1	a	1.0	4.0
1	1	2	b	3.0	6.0
2	2	1	a	NaN	NaN
3	2	2	b	2.0	5.0

In [3]:

df >> complete(f.group, nesting(f.item_id, f.item_name), fill=dict(value1=0))

df >> complete(f.group, nesting(f.item_id, f.item_name), fill=dict(value1=0))

Out[3]:

	group	item_id	item_name	value1	value2
	<int64>	<int64>	<object>	<float64>	<float64>
0	1	1	a	1.0	4.0
1	1	2	b	3.0	6.0
2	2	1	a	0.0	NaN
3	2	2	b	2.0	5.0