如何重新排序重复极数据框架的答案
原标题：How to re order duplicates answers on polars dataframe
我有一个包含多个问题和答案的极地数据框。 问题在于每个答案都包含在它自己的栏目中, 这意味着我有很多多余的信息。 因此, 我希望为问题和答案只有一个列。 下面是数据的例子 : 数据 = {“ ID ” : [1, 1, 1, “ 问题 ” : [“ A”, B”, C, “ 问题 ”, “ 问题 A ” : [“ Answer A, “ Answer A”, “ Answer B ” : [ Answer B, “ Answer B”, “ Answer C”, “ Ans” (d) a. (d) af. fre, (d) d. (d) d. (d) a. (d) a. (d) (d) a. (d) (d) (d) a. (d) y. (i) (d) (i) (y. (d) ) (y. a f) a. (d) a f) a f.
最佳回答
TLDR.
import polars.selectors as cs
(
df
.unpivot(
on=cs.starts_with("Answer"),
index=["ID", "Question"],
variable_name="Source",
value_name="Answer",
)
.filter(
pl.col("Question") == pl.col("Source").str.strip_prefix("Answer ")
)
.drop("Source")
)
shape: (3, 3)
┌─────┬──────────┬───────────────────┐
│ ID ┆ Question ┆ Answer │
│ --- ┆ --- ┆ --- │
│ i64 ┆ str ┆ str │
╞═════╪══════════╪═══════════════════╡
│ 1 ┆ A ┆ Some answer │
│ 1 ┆ B ┆ Some other answer │
│ 1 ┆ C ┆ Another answer │
└─────┴──────────┴───────────────────┘
Explanation.
An approach that is a bit more general is to melt (pl.DataFrame.unpivot) the dataframe on the answer columns. This gives you a long format dataframe, which for each original row contains one row for each answer column.
import polars.selectors as cs
(
df
.unpivot(
on=cs.starts_with("Answer"),
index=["ID", "Question"],
variable_name="Source",
value_name="Answer",
)
)
shape: (9, 4)
┌─────┬──────────┬──────────┬───────────────────┐
│ ID ┆ Question ┆ Source ┆ Answer │
│ --- ┆ --- ┆ --- ┆ --- │
│ i64 ┆ str ┆ str ┆ str │
╞═════╪══════════╪══════════╪═══════════════════╡
│ 1 ┆ A ┆ Answer A ┆ Some answer │
│ 1 ┆ B ┆ Answer A ┆ Some answer │
│ 1 ┆ C ┆ Answer A ┆ Some answer │
│ 1 ┆ A ┆ Answer B ┆ Some other answer │
│ 1 ┆ B ┆ Answer B ┆ Some other answer │
│ 1 ┆ C ┆ Answer B ┆ Some other answer │
│ 1 ┆ A ┆ Answer C ┆ Another answer │
│ 1 ┆ B ┆ Answer C ┆ Another answer │
│ 1 ┆ C ┆ Answer C ┆ Another answer │
└─────┴──────────┴──────────┴───────────────────┘
From here, it is easy to (after some transformation) filter for rows in which the Source column matches the question (see TLDR).
问题回答
If there are a limited number of answer columns, this can be done with a simple pl.when().then() chain:
df2 = df.select(
"ID",
"Question",
pl.when(pl.col("Question") == "A")
.then("Answer A")
.when(pl.col("Question") == "B")
.then("Answer B")
.when(pl.col("Question") == "C")
.then("Answer C")
.alias("Answer"),
)
This will set the value from the column Answer A as Answer if Question == "A" and so on.
Using functools.reduce you can generalize Dogbert s solution to an arbitrary number of columns:
import polars as pl
from functools import reduce
answer_cols = ["A", "B", "C"]
answer = reduce(
lambda expr, q: expr.when(Question=q).then(pl.col(f"Answer {q}")),
answer_cols,
pl.when(False).then(None) # initial expression / dummy condition
)
res = df.select("ID", "Question", answer.alias("Answer"))
Or if you prefer not to use functools.reduce:
answer = pl.when(False).then(None)
for q in answer_cols:
answer = answer.when(Question=q).then(pl.col(f"Answer {q}"))
Note that the initial condition pl.when(False).then(None) is always false, and is only needed to initialize the when-then chain expression
Output:
>>> res
shape: (3, 3)
┌─────┬──────────┬──────────┐
│ ID ┆ Question ┆ Answer │
│ --- ┆ --- ┆ --- │
│ i64 ┆ str ┆ str │
╞═════╪══════════╪══════════╡
│ 1 ┆ A ┆ Answer A │
│ 1 ┆ B ┆ Answer B │
│ 1 ┆ C ┆ Answer C │
└─────┴──────────┴──────────┘
