Working with missing data#

In this section, we will discuss missing (also referred to as NA) values in cudf. cudf supports having missing values in all dtypes. These missing values are represented by <NA>. These values are also referenced as “null values”.

How to Detect missing values#

To detect missing values, you can use isna() and notna() functions.

import numpy as np

import cudf

df = cudf.DataFrame({"a": [1, 2, None, 4], "b": [0.1, None, 2.3, 17.17]})

df
a b
0 1 0.1
1 2 <NA>
2 <NA> 2.3
3 4 17.17 
df.isna()
a b
0 False False
1 False True
2 True False
3 False False 
df["a"].notna()

0     True
1     True
2    False
3     True
Name: a, dtype: bool

One has to be mindful that in Python (and NumPy), the nan’s don’t compare equal, but None’s do. Note that cudf/NumPy uses the fact that np.nan != np.nan, and treats None like np.nan.

None == None

True

np.nan == np.nan

False

So as compared to above, a scalar equality comparison versus a None/np.nan doesn’t provide useful information.

df["b"] == np.nan

0    False
1     <NA>
2    False
3    False
Name: b, dtype: bool

s = cudf.Series([None, 1, 2])

s

0    <NA>
1       1
2       2
dtype: int64

s == None

0    <NA>
1    <NA>
2    <NA>
dtype: bool

s = cudf.Series([1, 2, np.nan], nan_as_null=False)

s

0    1.0
1    2.0
2    NaN
dtype: float64

s == np.nan

0    False
1    False
2    False
dtype: bool

Float dtypes and missing data#

Because NaN is a float, a column of integers with even one missing values is cast to floating-point dtype. However this doesn’t happen by default.

By default if a NaN value is passed to Series constructor, it is treated as <NA> value.

cudf.Series([1, 2, np.nan])

0       1
1       2
2    <NA>
dtype: int64

Hence to consider a NaN as NaN you will have to pass nan_as_null=False parameter into Series constructor.

cudf.Series([1, 2, np.nan], nan_as_null=False)

0    1.0
1    2.0
2    NaN
dtype: float64

Datetimes#

For datetime64 types, cudf doesn’t support having NaT values. Instead these values which are specific to numpy and pandas are considered as null values(<NA>) in cudf. The actual underlying value of NaT is min(int64) and cudf retains the underlying value when converting a cudf object to pandas object.

import pandas as pd

datetime_series = cudf.Series(
    [pd.Timestamp("20120101"), pd.NaT, pd.Timestamp("20120101")]
)
datetime_series

0    2012-01-01 00:00:00.000000000
1                              NaT
2    2012-01-01 00:00:00.000000000
dtype: datetime64[ns]

datetime_series.to_pandas()

0   2012-01-01
1          NaT
2   2012-01-01
dtype: datetime64[ns]

any operations on rows having <NA> values in datetime column will result in <NA> value at the same location in resulting column:

datetime_series - datetime_series

0    0 days 00:00:00
1                NaT
2    0 days 00:00:00
dtype: timedelta64[ns]

Calculations with missing data#

Null values propagate naturally through arithmetic operations between pandas objects.

df1 = cudf.DataFrame(
    {
        "a": [1, None, 2, 3, None],
        "b": cudf.Series([np.nan, 2, 3.2, 0.1, 1], nan_as_null=False),
    }
)

df2 = cudf.DataFrame(
    {"a": [1, 11, 2, 34, 10], "b": cudf.Series([0.23, 22, 3.2, None, 1])}
)

df1
a b
0 1 NaN
1 <NA> 2.0
2 2 3.2
3 3 0.1
4 <NA> 1.0 
df2
a b
0 1 0.23
1 11 22.0
2 2 3.2
3 34 <NA>
4 10 1.0 
df1 + df2
a b
0 2 NaN
1 <NA> 24.0
2 4 6.4
3 37 <NA>
4 <NA> 2.0

While summing the data along a series, NA values will be treated as 0.

df1["a"]

0       1
1    <NA>
2       2
3       3
4    <NA>
Name: a, dtype: int64

df1["a"].sum()

6

Since NA values are treated as 0, the mean would result to 2 in this case (1 + 0 + 2 + 3 + 0)/5 = 2

df1["a"].mean()

2.0

To preserve NA values in the above calculations, sum & mean support skipna parameter. By default it’s value is set to True, we can change it to False to preserve NA values.

df1["a"].sum(skipna=False)

nan

df1["a"].mean(skipna=False)

nan

Cumulative methods like cumsum and cumprod ignore NA values by default.

df1["a"].cumsum()

0       1
1    <NA>
2       3
3       6
4    <NA>
Name: a, dtype: int64

To preserve NA values in cumulative methods, provide skipna=False.

df1["a"].cumsum(skipna=False)

0       1
1    <NA>
2    <NA>
3    <NA>
4    <NA>
Name: a, dtype: int64

Sum/product of Null/nans#

The sum of an empty or all-NA Series of a DataFrame is 0.

cudf.Series([np.nan], nan_as_null=False).sum()

0.0

cudf.Series([np.nan], nan_as_null=False).sum(skipna=False)

nan

cudf.Series([], dtype="float64").sum()

0.0

The product of an empty or all-NA Series of a DataFrame is 1.

cudf.Series([np.nan], nan_as_null=False).prod()

1.0

cudf.Series([np.nan], nan_as_null=False).prod(skipna=False)

nan

cudf.Series([], dtype="float64").prod()

1.0

NA values in GroupBy#

NA groups in GroupBy are automatically excluded. For example:

df1
a b
0 1 NaN
1 <NA> 2.0
2 2 3.2
3 3 0.1
4 <NA> 1.0 
df1.groupby("a").mean()
b
a
3 0.1
1 NaN
2 3.2

It is also possible to include NA in groups by passing dropna=False

df1.groupby("a", dropna=False).mean()
b
a
3 0.1
1 NaN
2 3.2
<NA> 1.5

Inserting missing data#

All dtypes support insertion of missing value by assignment. Any specific location in series can made null by assigning it to None.

series = cudf.Series([1, 2, 3, 4])

series

0    1
1    2
2    3
3    4
dtype: int64

series[2] = None

series

0       1
1       2
2    <NA>
3       4
dtype: int64

Filling missing values: fillna#

fillna() can fill in NA & NaN values with non-NA data.

df1
a b
0 1 NaN
1 <NA> 2.0
2 2 3.2
3 3 0.1
4 <NA> 1.0 
df1["b"].fillna(10)

0    10.0
1     2.0
2     3.2
3     0.1
4     1.0
Name: b, dtype: float64

Filling with cudf Object#

You can also fillna using a dict or Series that is alignable. The labels of the dict or index of the Series must match the columns of the frame you wish to fill. The use case of this is to fill a DataFrame with the mean of that column.

import cupy as cp

dff = cudf.DataFrame(cp.random.randn(10, 3), columns=list("ABC"))

dff.iloc[3:5, 0] = np.nan

dff.iloc[4:6, 1] = np.nan

dff.iloc[5:8, 2] = np.nan

dff
A B C
0 -0.000980 0.551560 -1.443735
1 -1.832599 -0.513116 -0.681301
2 0.359273 0.166505 -0.269154
3 NaN -0.149245 -1.083291
4 NaN NaN 0.588775
5 -0.382155 NaN NaN
6 2.042128 1.788889 NaN
7 0.269661 1.122631 NaN
8 0.624938 -1.074668 -0.098387
9 0.064479 1.647954 0.556011 
dff.fillna(dff.mean())
A B C
0 -0.000980 0.551560 -1.443735
1 -1.832599 -0.513116 -0.681301
2 0.359273 0.166505 -0.269154
3 0.143093 -0.149245 -1.083291
4 0.143093 0.442564 0.588775
5 -0.382155 0.442564 -0.347297
6 2.042128 1.788889 -0.347297
7 0.269661 1.122631 -0.347297
8 0.624938 -1.074668 -0.098387
9 0.064479 1.647954 0.556011 
dff.fillna(dff.mean()[1:3])
A B C
0 -0.000980 0.551560 -1.443735
1 -1.832599 -0.513116 -0.681301
2 0.359273 0.166505 -0.269154
3 NaN -0.149245 -1.083291
4 NaN 0.442564 0.588775
5 -0.382155 0.442564 -0.347297
6 2.042128 1.788889 -0.347297
7 0.269661 1.122631 -0.347297
8 0.624938 -1.074668 -0.098387
9 0.064479 1.647954 0.556011

Dropping axis labels with missing data: dropna#

Missing data can be excluded using dropna():

df1
a b
0 1 NaN
1 <NA> 2.0
2 2 3.2
3 3 0.1
4 <NA> 1.0 
df1.dropna(axis=0)
a b
2 2 3.2
3 3 0.1 
df1.dropna(axis=1)
0
1
2
3
4

An equivalent dropna() is available for Series.

df1["a"].dropna()

0    1
2    2
3    3
Name: a, dtype: int64

Replacing generic values#

Often times we want to replace arbitrary values with other values.

replace() in Series and replace() in DataFrame provides an efficient yet flexible way to perform such replacements.

series = cudf.Series([0.0, 1.0, 2.0, 3.0, 4.0])

series

0    0.0
1    1.0
2    2.0
3    3.0
4    4.0
dtype: float64

series.replace(0, 5)

0    5.0
1    1.0
2    2.0
3    3.0
4    4.0
dtype: float64

We can also replace any value with a <NA> value.

series.replace(0, None)

0    <NA>
1     1.0
2     2.0
3     3.0
4     4.0
dtype: float64

You can replace a list of values by a list of other values:

series.replace([0, 1, 2, 3, 4], [4, 3, 2, 1, 0])

0    4.0
1    3.0
2    2.0
3    1.0
4    0.0
dtype: float64

You can also specify a mapping dict:

series.replace({0: 10, 1: 100})

0     10.0
1    100.0
2      2.0
3      3.0
4      4.0
dtype: float64

For a DataFrame, you can specify individual values by column:

df = cudf.DataFrame({"a": [0, 1, 2, 3, 4], "b": [5, 6, 7, 8, 9]})

df
a b
0 0 5
1 1 6
2 2 7
3 3 8
4 4 9 
df.replace({"a": 0, "b": 5}, 100)
a b
0 100 100
1 1 6
2 2 7
3 3 8
4 4 9

String/regular expression replacement#

cudf supports replacing string values using replace API:

d = {"a": list(range(4)), "b": list("ab.."), "c": ["a", "b", None, "d"]}

df = cudf.DataFrame(d)

df
a b c
0 0 a a
1 1 b b
2 2 . <NA>
3 3 . d 
df.replace(".", "A Dot")
a b c
0 0 a a
1 1 b b
2 2 A Dot <NA>
3 3 A Dot d 
df.replace([".", "b"], ["A Dot", None])
a b c
0 0 a a
1 1 <NA> <NA>
2 2 A Dot <NA>
3 3 A Dot d

Replace a few different values (list -> list):

df.replace(["a", "."], ["b", "--"])
a b c
0 0 b b
1 1 b b
2 2 -- <NA>
3 3 -- d

Only search in column ‘b’ (dict -> dict):

df.replace({"b": "."}, {"b": "replacement value"})
a b c
0 0 a a
1 1 b b
2 2 replacement value <NA>
3 3 replacement value d

Numeric replacement#

replace() can also be used similar to fillna().

df = cudf.DataFrame(cp.random.randn(10, 2))

df[np.random.rand(df.shape[0]) > 0.5] = 1.5

df.replace(1.5, None)
0 1
0 <NA> <NA>
1 <NA> <NA>
2 1.579393975 1.824094745
3 -1.818202592 -1.076327404
4 -1.005274936 -0.993429099
5 1.916061967 0.093062522
6 0.023991821 1.07013605
7 -0.599455766 -0.566379641
8 0.029506187 0.847134773
9 -0.750757995 -0.868897574

Replacing more than one value is possible by passing a list.

df00 = df.iloc[0, 0]

df.replace([1.5, df00], [5, 10])
0 1
0 10.000000 10.000000
1 10.000000 10.000000
2 1.579394 1.824095
3 -1.818203 -1.076327
4 -1.005275 -0.993429
5 1.916062 0.093063
6 0.023992 1.070136
7 -0.599456 -0.566380
8 0.029506 0.847135
9 -0.750758 -0.868898

You can also operate on the DataFrame in place:

df.replace(1.5, None, inplace=True)

df
0 1
0 <NA> <NA>
1 <NA> <NA>
2 1.579393975 1.824094745
3 -1.818202592 -1.076327404
4 -1.005274936 -0.993429099
5 1.916061967 0.093062522
6 0.023991821 1.07013605
7 -0.599455766 -0.566379641
8 0.029506187 0.847134773
9 -0.750757995 -0.868897574
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