In pandas, the following code will split the string from col1 into many columns. is there a way to do this in polars?
d = {'col1': ["a/b/c/d", "a/b/c/d"]}
df= pd.DataFrame(data=d)
df[["a","b","c","d"]]=df["col1"].str.split('/',expand=True)
Here's an algorithm that will automatically adjust for the required number of columns -- and should be quite performant.
Let's start with this data. Notice that I've purposely added the empty string "" and a null value - to show how the algorithm handles these values. Also, the number of split strings varies widely.
import polars as pl
df = pl.DataFrame(
{
"my_str": ["cat", "cat/dog", None, "", "cat/dog/aardvark/mouse/frog"],
}
)
df
shape: (5, 1)
┌─────────────────────────────┐
│ my_str │
│ --- │
│ str │
╞═════════════════════════════╡
│ cat │
├╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ cat/dog │
├╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ null │
├╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ │
├╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ cat/dog/aardvark/mouse/frog │
└─────────────────────────────┘
The Algorithm
The algorithm below may be a bit more than you need, but you can edit/delete/add as you need.
(
df
.with_row_count('id')
.with_column(pl.col("my_str").str.split("/").alias("split_str"))
.explode("split_str")
.with_column(
("string_" + pl.arange(0, pl.count()).cast(pl.Utf8).str.zfill(2))
.over("id")
.alias("col_nm")
)
.pivot(
index=['id', 'my_str'],
values='split_str',
columns='col_nm',
)
.with_column(
pl.col('^string_.*$').fill_null("")
)
)
shape: (5, 7)
┌─────┬─────────────────────────────┬───────────┬───────────┬───────────┬───────────┬───────────┐
│ id ┆ my_str ┆ string_00 ┆ string_01 ┆ string_02 ┆ string_03 ┆ string_04 │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ u32 ┆ str ┆ str ┆ str ┆ str ┆ str ┆ str │
╞═════╪═════════════════════════════╪═══════════╪═══════════╪═══════════╪═══════════╪═══════════╡
│ 0 ┆ cat ┆ cat ┆ ┆ ┆ ┆ │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ cat/dog ┆ cat ┆ dog ┆ ┆ ┆ │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 2 ┆ null ┆ ┆ ┆ ┆ ┆ │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ ┆ ┆ ┆ ┆ ┆ │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ cat ┆ dog ┆ aardvark ┆ mouse ┆ frog │
└─────┴─────────────────────────────┴───────────┴───────────┴───────────┴───────────┴───────────┘
How it works
We first assign a row number id (which we'll need later), and use split to separate the strings. Note that the split strings form a list.
(
df
.with_row_count('id')
.with_column(pl.col("my_str").str.split("/").alias("split_str"))
)
shape: (5, 3)
┌─────┬─────────────────────────────┬────────────────────────────┐
│ id ┆ my_str ┆ split_str │
│ --- ┆ --- ┆ --- │
│ u32 ┆ str ┆ list[str] │
╞═════╪═════════════════════════════╪════════════════════════════╡
│ 0 ┆ cat ┆ ["cat"] │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ cat/dog ┆ ["cat", "dog"] │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 2 ┆ null ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ ┆ [""] │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ ["cat", "dog", ... "frog"] │
└─────┴─────────────────────────────┴────────────────────────────┘
Next, we'll use explode to put each string on its own row. (Notice how the id column tracks the original row that each string came from.)
(
df
.with_row_count('id')
.with_column(pl.col("my_str").str.split("/").alias("split_str"))
.explode("split_str")
)
shape: (10, 3)
┌─────┬─────────────────────────────┬───────────┐
│ id ┆ my_str ┆ split_str │
│ --- ┆ --- ┆ --- │
│ u32 ┆ str ┆ str │
╞═════╪═════════════════════════════╪═══════════╡
│ 0 ┆ cat ┆ cat │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ cat/dog ┆ cat │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ cat/dog ┆ dog │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 2 ┆ null ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ ┆ │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ cat │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ dog │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ aardvark │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ mouse │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ frog │
└─────┴─────────────────────────────┴───────────┘
In the next step, we're going to generate our column names. I chose to call each column string_XX where XX is the offset with regards to the original string.
I've used the handy zfill expression so that 1 becomes 01. (This makes sure that string_02 comes before string_10 if you decide to sort your columns later.)
You can substitute your own naming in this step as you need.
(
df
.with_row_count('id')
.with_column(pl.col("my_str").str.split("/").alias("split_str"))
.explode("split_str")
.with_column(
("string_" + pl.arange(0, pl.count()).cast(pl.Utf8).str.zfill(2))
.over("id")
.alias("col_nm")
)
)
shape: (10, 4)
┌─────┬─────────────────────────────┬───────────┬───────────┐
│ id ┆ my_str ┆ split_str ┆ col_nm │
│ --- ┆ --- ┆ --- ┆ --- │
│ u32 ┆ str ┆ str ┆ str │
╞═════╪═════════════════════════════╪═══════════╪═══════════╡
│ 0 ┆ cat ┆ cat ┆ string_00 │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ cat/dog ┆ cat ┆ string_00 │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ cat/dog ┆ dog ┆ string_01 │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 2 ┆ null ┆ null ┆ string_00 │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ ┆ ┆ string_00 │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ cat ┆ string_00 │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ dog ┆ string_01 │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ aardvark ┆ string_02 │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ mouse ┆ string_03 │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ frog ┆ string_04 │
└─────┴─────────────────────────────┴───────────┴───────────┘
In the next step, we'll use the pivot function to place each string in its own column.
(
df
.with_row_count('id')
.with_column(pl.col("my_str").str.split("/").alias("split_str"))
.explode("split_str")
.with_column(
("string_" + pl.arange(0, pl.count()).cast(pl.Utf8).str.zfill(2))
.over("id")
.alias("col_nm")
)
.pivot(
index=['id', 'my_str'],
values='split_str',
columns='col_nm',
)
)
shape: (5, 7)
┌─────┬─────────────────────────────┬───────────┬───────────┬───────────┬───────────┬───────────┐
│ id ┆ my_str ┆ string_00 ┆ string_01 ┆ string_02 ┆ string_03 ┆ string_04 │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ u32 ┆ str ┆ str ┆ str ┆ str ┆ str ┆ str │
╞═════╪═════════════════════════════╪═══════════╪═══════════╪═══════════╪═══════════╪═══════════╡
│ 0 ┆ cat ┆ cat ┆ null ┆ null ┆ null ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ cat/dog ┆ cat ┆ dog ┆ null ┆ null ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 2 ┆ null ┆ null ┆ null ┆ null ┆ null ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ ┆ ┆ null ┆ null ┆ null ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ cat/dog/aardvark/mouse/frog ┆ cat ┆ dog ┆ aardvark ┆ mouse ┆ frog │
└─────┴─────────────────────────────┴───────────┴───────────┴───────────┴───────────┴───────────┘
All that remains is to use fill_null to replace the null values with an empty string "". Notice that I've used a regex expression in the col expression to target only those columns whose names start with "string_". (Depending on your other data, you may not want to replace null with "" everywhere in your data.)
You can use apply() method
import polars as pl
from polars import col
df = pl.DataFrame({
'col1': ["a/b/c/d", "e/f/j/k"]
})
print(df)
df:
shape: (2, 1)
┌─────────┐
│ col1 │
│ --- │
│ str │
╞═════════╡
│ a/b/c/d │
├╌╌╌╌╌╌╌╌╌┤
│ e/f/j/k │
└─────────┘
With apply()
df = df.with_columns([
col('col1'),
*[col('col1').apply(lambda s, i=i: s.split('/')[i]).alias(col_name)
for i, col_name in enumerate(['a', 'b', 'c', 'd'])]
# or without 'for'
# col('col1').apply(lambda s: s.split('/')[0]).alias('a'),
# col('col1').apply(lambda s: s.split('/')[1]).alias('b'),
# col('col1').apply(lambda s: s.split('/')[2]).alias('c'),
# col('col1').apply(lambda s: s.split('/')[3]).alias('d')
])
print(df)
df:
shape: (2, 5)
┌─────────┬─────┬─────┬─────┬─────┐
│ col1 ┆ a ┆ b ┆ c ┆ d │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ str ┆ str ┆ str ┆ str ┆ str │
╞═════════╪═════╪═════╪═════╪═════╡
│ a/b/c/d ┆ a ┆ b ┆ c ┆ d │
├╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌┤
│ e/f/j/k ┆ e ┆ f ┆ j ┆ k │
└─────────┴─────┴─────┴─────┴─────┘
It works, but probably there is more accurate way)
With this way you do the string split to turn col1 into a list of strings. Then you loop over the lists and use .arr.get to extract each element into a separate column
(df
.with_column(pl.col("col1").str.split("/"))
.with_columns(
[pl.col("col1").arr.get(i).alias(str(i)) for i in range(len(df[0,"col1"].split('/')))
]
)
)
One challenge is whether you will have the same number of elements in the list in each row. In this solution I've assumed you have and have taken the length of the list in the first row to do the loop.
You can use struct datatype, as described in this post: https://stackoverflow.com/a/74219166:
import pandas as pl
df = pl.DataFrame({
"my_str": ["cat", "cat/dog", None, "", "cat/dog/aardvark/mouse/frog"],
})
df.select(pl.col('my_str').str.split('/')
.arr.to_struct(n_field_strategy="max_width")).unnest('my_str')
Notice you must use n_field_strategy="max_width", otherwise, unnest() will create only 1 column.
import polars as pl
#Create new column list(can be created dynamically as well)
new_cols=['new_col1','new_col2','new_col3',.....,new_coln]
#Define expression
expr = [pl.col('col1').str.split('/').arr.get(i).alias(col)
for i,col in enumerate(new_cols)
]
#Apply Expression
df.with_columns(expr)
Related
String columns can be selected with:
df.select(pl.col(pl.Utf8))
and a Dataframe's rows can be filtered with a regex pattern for a single column, with something like:
df.filter(pl.col("feature").str.contains("dangerous"))
How can a DataFrame be filtered with a list of regex patterns that could appear in any string column? I.e., if any string in a row matches any regex pattern, then keep that entire row, discard the rest.
EDIT 1
Here's a generated df and patterns to test functionality and performance.
import random
from faker import Faker
import polars as pl
random.seed(42)
Faker.seed(42)
faker = Faker()
df_len = 10000
df = pl.DataFrame(
[
pl.Series("a", [random.randint(0, 511) for _ in range(df_len)]).cast(pl.Binary),
pl.Series("b", [random.randint(0, 1) for _ in range(df_len)]).cast(pl.Boolean),
pl.Series("c", faker.sentences(df_len), pl.Utf8),
pl.Series("d", [random.randint(0, 255) for _ in range(df_len)], pl.UInt8),
pl.Series("e", faker.words(df_len), pl.Utf8),
pl.Series(
"f",
[random.randint(0, 255) * random.TWOPI for _ in range(df_len)],
pl.Float32,
),
pl.Series("g", faker.words(df_len), pl.Utf8),
]
)
patterns = [r"(?i)dangerous", r"always", r"(?i)prevent"]
print(df) yields:
shape: (10000, 7)
┌───────────────┬───────┬─────────────────────────────────────┬─────┬───────────┬────────────┬──────────┐
│ a ┆ b ┆ c ┆ d ┆ e ┆ f ┆ g │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ binary ┆ bool ┆ str ┆ u8 ┆ str ┆ f32 ┆ str │
╞═══════════════╪═══════╪═════════════════════════════════════╪═════╪═══════════╪════════════╪══════════╡
│ [binary data] ┆ false ┆ Agent every development say. ┆ 164 ┆ let ┆ 980.17688 ┆ yard │
│ [binary data] ┆ true ┆ Beautiful instead ahead despite ... ┆ 210 ┆ reach ┆ 458.672516 ┆ son │
│ [binary data] ┆ false ┆ Information last everything than... ┆ 230 ┆ arm ┆ 50.265484 ┆ standard │
│ [binary data] ┆ false ┆ Choice whatever from behavior be... ┆ 29 ┆ operation ┆ 929.911438 ┆ final │
│ ... ┆ ... ┆ ... ┆ ... ┆ ... ┆ ... ┆ ... │
│ [binary data] ┆ true ┆ Building sign recently avoid upo... ┆ 132 ┆ practice ┆ 282.743347 ┆ big │
│ [binary data] ┆ false ┆ Paper will board. ┆ 72 ┆ similar ┆ 376.991119 ┆ just │
│ [binary data] ┆ true ┆ Technology money worker spring m... ┆ 140 ┆ sign ┆ 94.24778 ┆ audience │
│ [binary data] ┆ false ┆ A third traditional ago. ┆ 40 ┆ available ┆ 615.752136 ┆ always │
└───────────────┴───────┴─────────────────────────────────────┴─────┴───────────┴────────────┴──────────┘
EDIT 2
Using #jqurious's answer (the fastest so far), the correct output of df.filter(pl.any(pl.col(pl.Utf8).str.contains(regex))) is:
shape: (146, 7)
┌───────────────┬───────┬─────────────────────────────────────┬─────┬───────────┬─────────────┬──────────┐
│ a ┆ b ┆ c ┆ d ┆ e ┆ f ┆ g │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ binary ┆ bool ┆ str ┆ u8 ┆ str ┆ f32 ┆ str │
╞═══════════════╪═══════╪═════════════════════════════════════╪═════╪═══════════╪═════════════╪══════════╡
│ [binary data] ┆ true ┆ During prevent accept seem show ... ┆ 137 ┆ various ┆ 471.238892 ┆ customer │
│ [binary data] ┆ true ┆ Ball always it focus economy bef... ┆ 179 ┆ key ┆ 471.238892 ┆ guy │
│ [binary data] ┆ false ┆ Admit attack energy always. ┆ 175 ┆ purpose ┆ 1281.769775 ┆ wonder │
│ [binary data] ┆ false ┆ Beyond prevent entire staff. ┆ 242 ┆ hair ┆ 904.778687 ┆ around │
│ ... ┆ ... ┆ ... ┆ ... ┆ ... ┆ ... ┆ ... │
│ [binary data] ┆ true ┆ Your sure piece simple always so... ┆ 247 ┆ recently ┆ 1055.575073 ┆ laugh │
│ [binary data] ┆ false ┆ Difference all machine let charg... ┆ 178 ┆ former ┆ 1061.858276 ┆ always │
│ [binary data] ┆ true ┆ Morning carry event tell prevent... ┆ 3 ┆ entire ┆ 1432.566284 ┆ hit │
│ [binary data] ┆ false ┆ A third traditional ago. ┆ 40 ┆ available ┆ 615.752136 ┆ always │
└───────────────┴───────┴─────────────────────────────────────┴─────┴───────────┴─────────────┴──────────┘
You can turn the list into a single regex.
regex = "|".join(
f"(?:{pattern})" for pattern in
sorted(patterns, key=len, reverse=True)
)
df.filter(pl.any(pl.col(pl.Utf8).str.contains(regex)))
To check regex patterns for each string column, you can use .fold() method
df = pl.DataFrame({
"a": ["foo", "fo", "foa"],
"b": ["foa", "fo", "foo"]
})
df.filter(
pl.fold(acc=pl.lit(False),
f=lambda acc, col: acc | col.str.contains("foo"),
exprs=pl.col(pl.Utf8))
)
┌─────┬─────┐
│ a ┆ b │
│ --- ┆ --- │
│ str ┆ str │
╞═════╪═════╡
│ foo ┆ foa │
│ foa ┆ foo │
└─────┴─────┘
Another way to do it - concat all string columns into a single one and then apply regex expressions:
df.filter(
pl.concat_str(pl.col(pl.Utf8)).str.contains("foo")
)
I have a small use case and here is a polars dataframe.
df_names = pl.DataFrame({'LN'['Mallesham','Bhavik','Mallesham','Bhavik','Mahesh','Naresh','Sharath','Rakesh','Mallesham'],
'FN':['Yamulla','Yamulla','Yamulla','Yamulla','Dayala','Burre','Velmala','Uppu','Yamulla'],
'SSN':['123','456','123','456','893','111','222','333','123'],
'Address':['A','B','C','D','E','F','G','H','S']})
Here I would like to group on LN,FN,SSN and create a new column in which how many number of observations for this group combination and below is the expected output.
'Mallesham','Yamulla','123' is appeared 3 times, hence LN_FN_SSN_count field is filled up with 3.
You can use an expression using over (which is like grouping, aggregating and self-joining in other libs, but without the need for the join):
df_names.with_column(pl.count().over(['LN', 'FN', 'SSN']).alias('LN_FN_SSN_count'))
┌───────────┬─────────┬─────┬─────────┬─────────────────┐
│ LN ┆ FN ┆ SSN ┆ Address ┆ LN_FN_SSN_count │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ str ┆ str ┆ str ┆ str ┆ u32 │
╞═══════════╪═════════╪═════╪═════════╪═════════════════╡
│ Mallesham ┆ Yamulla ┆ 123 ┆ A ┆ 3 │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Bhavik ┆ Yamulla ┆ 456 ┆ B ┆ 2 │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Mallesham ┆ Yamulla ┆ 123 ┆ C ┆ 3 │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Bhavik ┆ Yamulla ┆ 456 ┆ D ┆ 2 │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ ... ┆ ... ┆ ... ┆ ... ┆ ... │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Naresh ┆ Burre ┆ 111 ┆ F ┆ 1 │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Sharath ┆ Velmala ┆ 222 ┆ G ┆ 1 │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Rakesh ┆ Uppu ┆ 333 ┆ H ┆ 1 │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Mallesham ┆ Yamulla ┆ 123 ┆ S ┆ 3 │
└───────────┴─────────┴─────┴─────────┴─────────────────┘
I am trying to remove null values across a list of selected columns. But it seems that I might have got the wtih_columns operation not right. What's the right approach if you want to operate the removing only on selected columns?
df = pl.DataFrame(
{
"id": ["NY", "TK", "FD"],
"eat2000": [1, None, 3],
"eat2001": [-2, None, 4],
"eat2002": [None, None, None],
"eat2003": [-9, None, 8],
"eat2004": [None, None, 8]
}
); df
┌─────┬─────────┬─────────┬─────────┬─────────┬─────────┐
│ id ┆ eat2000 ┆ eat2001 ┆ eat2002 ┆ eat2003 ┆ eat2004 │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ str ┆ i64 ┆ i64 ┆ f64 ┆ i64 ┆ i64 │
╞═════╪═════════╪═════════╪═════════╪═════════╪═════════╡
│ NY ┆ 1 ┆ -2 ┆ null ┆ -9 ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┤
│ TK ┆ null ┆ null ┆ null ┆ null ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┤
│ FD ┆ 3 ┆ 4 ┆ null ┆ 8 ┆ 8 │
└─────┴─────────┴─────────┴─────────┴─────────┴─────────┘
col_list = [word for word in df.columns if word.startswith(("eat"))]
(
df
.with_columns([
pl.col(col_list).filter(~pl.fold(True, lambda acc, s: acc & s.is_null(), pl.all()))
])
)
Expected output:
┌─────┬─────────┬─────────┬─────────┬─────────┬─────────┐
│ id ┆ eat2000 ┆ eat2001 ┆ eat2002 ┆ eat2003 ┆ eat2004 │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ str ┆ i64 ┆ i64 ┆ f64 ┆ i64 ┆ i64 │
╞═════╪═════════╪═════════╪═════════╪═════════╪═════════╡
│ NY ┆ 1 ┆ -2 ┆ null ┆ -9 ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┤
│ FD ┆ 3 ┆ 4 ┆ null ┆ 8 ┆ 8 │
└─────┴─────────┴─────────┴─────────┴─────────┴─────────┘
The polars.all and polars.any expressions will perform their operations horizontally (i.e., row-wise) if we supply them with a list of Expressions or a polars.col with a regex wildcard expression. Let's use the latter to simplify our work:
(
df
.filter(
~pl.all(pl.col('^eat.*$').is_null())
)
)
shape: (2, 6)
┌─────┬─────────┬─────────┬─────────┬─────────┬─────────┐
│ id ┆ eat2000 ┆ eat2001 ┆ eat2002 ┆ eat2003 ┆ eat2004 │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ str ┆ i64 ┆ i64 ┆ f64 ┆ i64 ┆ i64 │
╞═════╪═════════╪═════════╪═════════╪═════════╪═════════╡
│ NY ┆ 1 ┆ -2 ┆ null ┆ -9 ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┤
│ FD ┆ 3 ┆ 4 ┆ null ┆ 8 ┆ 8 │
└─────┴─────────┴─────────┴─────────┴─────────┴─────────┘
The ~ in front of the pl.all stands for negation. Notice that we didn't need the col_list.
One caution: the regex expression in the pl.col must start with ^ and end with $. These cannot be omitted, even if the resulting regex expression is otherwise valid.
Alternately, if you don't like the ~ operator:
(
df
.filter(
pl.any(pl.col('^eat.*$').is_not_null())
)
)
Other Notes
As an aside, polars.sum, polars.min, and polars.max will also operate row-wise when supplied with a list of Expression or a wildcard expression in col.
Edit - using fold
FYI, here's how to use the fold method, if that is what you'd prefer. Note the use of pl.col with a regex expression.
(
df
.filter(
~pl.fold(True, lambda acc, s: acc & s.is_null(), exprs=pl.col('^eat.*$'))
)
)
shape: (2, 6)
┌─────┬─────────┬─────────┬─────────┬─────────┬─────────┐
│ id ┆ eat2000 ┆ eat2001 ┆ eat2002 ┆ eat2003 ┆ eat2004 │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ str ┆ i64 ┆ i64 ┆ f64 ┆ i64 ┆ i64 │
╞═════╪═════════╪═════════╪═════════╪═════════╪═════════╡
│ NY ┆ 1 ┆ -2 ┆ null ┆ -9 ┆ null │
├╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌┤
│ FD ┆ 3 ┆ 4 ┆ null ┆ 8 ┆ 8 │
└─────┴─────────┴─────────┴─────────┴─────────┴─────────┘
(
df
.filter(pl.col("eat2000").is_not_null())
)
I have a dataframe as-
pl.DataFrame({'last_name':['Unknown','Mallesham',np.nan,'Bhavik','Unknown'],
'first_name_or_initial':['U',np.nan,'TRUE','yamulla',np.nan],
'number':['003123490','012457847','100030303','','0023004648'],
'date_of_birth':[np.nan,'12/09/1900','12/09/1900','12/09/1900',np.nan]})
Here I would like to add a new column which contains the field names that do hold on any information except NULL/EMPTY/NAN.
For example:
first row: it has last,first and number field information, and dob is NULL, hence a new column conso_field is filled in with these field names such as last_name,first_name_or_initial and number. like wise I need to get this done for all the rows.
Here is an expected output:
First, let's expand the example to show a row with all null/empty fields (to show how the algorithm handles this case).
import polars as pl
import numpy as np
df = pl.DataFrame(
{
"last_name": ["Unknown", "Mallesham", np.nan, "Bhavik", "Unknown", None],
"first_name_or_initial": ["U", np.nan, "TRUE", "yamulla", np.nan, None],
"number": ["003123490", "012457847", "100030303", "", "0023004648", None],
"date_of_birth": [np.nan, "12/09/1900", "12/09/1900", "12/09/1900", np.nan, None],
}
)
df
shape: (6, 4)
┌───────────┬───────────────────────┬────────────┬───────────────┐
│ last_name ┆ first_name_or_initial ┆ number ┆ date_of_birth │
│ --- ┆ --- ┆ --- ┆ --- │
│ str ┆ str ┆ str ┆ str │
╞═══════════╪═══════════════════════╪════════════╪═══════════════╡
│ Unknown ┆ U ┆ 003123490 ┆ null │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Mallesham ┆ null ┆ 012457847 ┆ 12/09/1900 │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ null ┆ TRUE ┆ 100030303 ┆ 12/09/1900 │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Bhavik ┆ yamulla ┆ ┆ 12/09/1900 │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ Unknown ┆ null ┆ 0023004648 ┆ null │
├╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ null ┆ null ┆ null ┆ null │
└───────────┴───────────────────────┴────────────┴───────────────┘
The Algorithm
df = df.with_row_count()
(
df
.join(
df
.melt(id_vars="row_nr")
.filter(pl.col('value').is_not_null() & (pl.col('value') != ""))
.groupby('row_nr')
.agg(pl.col('variable').alias('conso_field'))
.with_column(pl.col('conso_field').arr.join(','))
,
on='row_nr',
how='left'
)
)
shape: (6, 6)
┌────────┬───────────┬───────────────────────┬────────────┬───────────────┬─────────────────────────────────────┐
│ row_nr ┆ last_name ┆ first_name_or_initial ┆ number ┆ date_of_birth ┆ conso_field │
│ --- ┆ --- ┆ --- ┆ --- ┆ --- ┆ --- │
│ u32 ┆ str ┆ str ┆ str ┆ str ┆ str │
╞════════╪═══════════╪═══════════════════════╪════════════╪═══════════════╪═════════════════════════════════════╡
│ 0 ┆ Unknown ┆ U ┆ 003123490 ┆ null ┆ last_name,first_name_or_initial,... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ Mallesham ┆ null ┆ 012457847 ┆ 12/09/1900 ┆ last_name,number,date_of_birth │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 2 ┆ null ┆ TRUE ┆ 100030303 ┆ 12/09/1900 ┆ first_name_or_initial,number,dat... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ Bhavik ┆ yamulla ┆ ┆ 12/09/1900 ┆ last_name,first_name_or_initial,... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ Unknown ┆ null ┆ 0023004648 ┆ null ┆ last_name,number │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 5 ┆ null ┆ null ┆ null ┆ null ┆ null │
└────────┴───────────┴───────────────────────┴────────────┴───────────────┴─────────────────────────────────────┘
Note that the algorithm keeps the last row with all null/empty values.
How it works
To see how it works, let's take it in steps.
First, we'll need to attach a row number to each row. (This is needed in case any row has all null/empty values.)
Then we'll use melt to place each value in each column on a separate row, next to it's column name.
df = df.with_row_count()
(
df
.melt(id_vars="row_nr")
)
shape: (24, 3)
┌────────┬───────────────┬────────────┐
│ row_nr ┆ variable ┆ value │
│ --- ┆ --- ┆ --- │
│ u32 ┆ str ┆ str │
╞════════╪═══════════════╪════════════╡
│ 0 ┆ last_name ┆ Unknown │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ last_name ┆ Mallesham │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 2 ┆ last_name ┆ null │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ last_name ┆ Bhavik │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ ... ┆ ... ┆ ... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 2 ┆ date_of_birth ┆ 12/09/1900 │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ date_of_birth ┆ 12/09/1900 │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ date_of_birth ┆ null │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 5 ┆ date_of_birth ┆ null │
└────────┴───────────────┴────────────┘
Note that columns values will be converted to string values in this step.
Next, we'll filter out any rows with null or "" values.
df = df.with_row_count()
(
df
.melt(id_vars="row_nr")
.filter(pl.col('value').is_not_null() & (pl.col('value') != ""))
)
shape: (14, 3)
┌────────┬───────────────┬────────────┐
│ row_nr ┆ variable ┆ value │
│ --- ┆ --- ┆ --- │
│ u32 ┆ str ┆ str │
╞════════╪═══════════════╪════════════╡
│ 0 ┆ last_name ┆ Unknown │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ last_name ┆ Mallesham │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ last_name ┆ Bhavik │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ last_name ┆ Unknown │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ ... ┆ ... ┆ ... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ number ┆ 0023004648 │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ date_of_birth ┆ 12/09/1900 │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 2 ┆ date_of_birth ┆ 12/09/1900 │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ date_of_birth ┆ 12/09/1900 │
└────────┴───────────────┴────────────┘
In the next step, we'll aggregate up all the remaining rows by row number, keeping only the column names. These represent columns with non-null, non-empty values.
df = df.with_row_count()
(
df
.melt(id_vars="row_nr")
.filter(pl.col('value').is_not_null() & (pl.col('value') != ""))
.groupby('row_nr')
.agg(pl.col('variable').alias('conso_field'))
)
shape: (5, 2)
┌────────┬─────────────────────────────────────┐
│ row_nr ┆ conso_field │
│ --- ┆ --- │
│ u32 ┆ list[str] │
╞════════╪═════════════════════════════════════╡
│ 2 ┆ ["first_name_or_initial", "numbe... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 3 ┆ ["last_name", "first_name_or_ini... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ ["last_name", "number"] │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 0 ┆ ["last_name", "first_name_or_ini... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ ["last_name", "number", "date_of... │
└────────┴─────────────────────────────────────┘
Note that we get a list of column names for each row. (Note: we don't need to worry about the order of the rows at this point. We'll use the row number and a left-join in the last step recombine the values to the original DataFrame.)
Then, it's simply a matter of joining the columns names into one string:
df = df.with_row_count()
(
df
.melt(id_vars="row_nr")
.filter(pl.col('value').is_not_null() & (pl.col('value') != ""))
.groupby('row_nr')
.agg(pl.col('variable').alias('conso_field'))
.with_column(pl.col('conso_field').arr.join(','))
)
shape: (5, 2)
┌────────┬─────────────────────────────────────┐
│ row_nr ┆ conso_field │
│ --- ┆ --- │
│ u32 ┆ str │
╞════════╪═════════════════════════════════════╡
│ 3 ┆ last_name,first_name_or_initial,... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 2 ┆ first_name_or_initial,number,dat... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 4 ┆ last_name,number │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 0 ┆ last_name,first_name_or_initial,... │
├╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 1 ┆ last_name,number,date_of_birth │
└────────┴─────────────────────────────────────┘
From here, we simply use a "left join" to merge the data back to the original dataset (as shown at the beginning.)
I have a problem to merge columns into one. Say I have a dataframe (df) like below:
>> print(df)
shape: (3, 4)
┌─────┬───────┬───────┬───────┐
│ a ┆ b_a_1 ┆ b_a_2 ┆ b_a_3 │
│ --- ┆ --- ┆ --- ┆ --- │
│ i64 ┆ str ┆ str ┆ str │
╞═════╪═══════╪═══════╪═══════╡
│ 1 ┆ a-- ┆ ┆ │
├╌╌╌╌╌┼╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┤
│ 1 ┆ ┆ b-- ┆ │
├╌╌╌╌╌┼╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┤
│ 1 ┆ ┆ ┆ c-- │
└─────┴───────┴───────┴───────┘
And I want to be able to merge the last three (3) columsn into one using python-polars. I have tried and successfully got what I want. However,
>> out = df.select(pl.concat_str(['b_a_1', 'b_a_2', 'b_a_3']).alias('b_a'))
>> print(out)
shape: (3, 1)
┌─────┐
│ b_a │
│ --- │
│ str │
╞═════╡
│ a-- │
├╌╌╌╌╌┤
│ b-- │
├╌╌╌╌╌┤
│ c-- │
└─────┘
when I use regex in selecting the columns, I don't get the above result
>> out = df.select(pl.concat_str('^b_a_\d$'))
>> print(out)
shape: (3, 3)
┌───────┬───────┬───────┐
│ b_a_1 ┆ b_a_2 ┆ b_a_3 │
│ --- ┆ --- ┆ --- │
│ str ┆ str ┆ str │
╞═══════╪═══════╪═══════╡
│ a-- ┆ ┆ │
├╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┤
│ ┆ b-- ┆ │
├╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌┤
│ ┆ ┆ c-- │
└───────┴───────┴───────┘
and nothing when run
>> out = df.select(pl.concat_str('^b_a_*$'))
>> print(out)
shape: (0, 0)
┌┐
╞╡
└┘
How am I to select the columns with regex and combine them into one?
Thank you very much for your time and suggestion.
Sincerely,
Thi An