“NaN” 匹配语句

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英文:

Match statement "NaN"

问题

我正在编写一个用于应用于Pandas数据帧的Python匹配/案例语句。但是数据中包含大量NaN值，我无法弄清楚如何处理它们，以便它们不会在默认情况下处理为_，因为我希望默认情况执行其他操作。

如何匹配NaN值？

这是我尝试过但不起作用的示例代码：

  1. import pandas as pd
  2. import numpy as np
  4. data = {'col': ["foo", "a", "b", np.nan]}
  5. df = pd.DataFrame(data)
  7. def handle_col(n):
  8.     match n:
  9.         case np.nan:
  10.             return "不是数字"
  11.         case "a":
  12.             return "这是字母a"
  13.         case "b":
  14.             return "这是字母b"
  15.         case _:
  16.             return "这是另一个字符串"
  17.     
  18. df["col"].apply(handle_col)

以下是输入数据帧：

  1.     col
  2. 0   foo
  3. 1   a
  4. 2   b
  5. 3   NaN

以及我得到的（错误）答案：

  1. 0    这是另一个字符串
  2. 1         这是字母a
  3. 2         这是字母b
  4. 3    这是另一个字符串
  5. Name: col, dtype: object
英文:

I am writing a python match/case statement that I want to apply to a pandas dataframe. But the data has a bunch of NaN values in it, and I can't figure out how to process them without them being handled in the default case _, because I want the default case to do something else

How can I match on NaN values ?

Here is an example code that I tried and doesn't work:

  1. import pandas as pd
  2. import numpy as np
  4. data = {'col': ["foo", "a", "b", np.nan]}
  5. df = pd.DataFrame(data)
  7. def handle_col(n):
  8.     match n:
  9.         case np.nan:
  10.             return "not a number"
  11.         case "a":
  12.             return "this is letter a"
  13.         case "b":
  14.             return "this is letter b"
  15.         case _:
  16.             return "this is another string"
  17.     
  18. df["col"].apply(handle_col)

Here's the input dataframe

  1. 	col
  2. 0	foo
  3. 1	a
  4. 2	b
  5. 3	NaN

and the (wrong) answer I get:

  1. 0    this is another string
  2. 1          this is letter a
  3. 2          this is letter b
  4. 3    this is another string
  5. Name: col, dtype: object

答案1

得分: 1

问题在于空值不遵循相等性检查，即 np.nan == np.nan 为 False，解决方法是在 case 中使用保护条件：

  1. def handle_col(n):
  2.     match n:
  3.         case _ if pd.isna(n):
  4.             return "不是数字"
  5.         case "a":
  6.             return "这是字母 a"
  7.         case "b":
  8.             return "这是字母 b"
  9.         case _:
  10.             return "这是另一个字符串"
  12. df["col"].apply(handle_col)
  1. 0    这是另一个字符串
  2. 1       这是字母 a
  3. 2       这是字母 b
  4. 3         不是数字
  5. Name: col, dtype: object
英文:

The problem is that null values don't respect equality check i.e, np.nan == np.nan is False, the solution is to use guards in case:

  1. def handle_col(n):
  2.     match n:
  3.         case _ if pd.isna(n):
  4.             return "not a number"
  5.         case "a":
  6.             return "this is letter a"
  7.         case "b":
  8.             return "this is letter b"
  9.         case _:
  10.             return "this is another string"
  12. df["col"].apply(handle_col)
  1. 0    this is another string
  2. 1          this is letter a
  3. 2          this is letter b
  4. 3              not a number
  5. Name: col, dtype: object

答案2

得分: 0

另一种解决方案，比较 number != number：

  1. data = {'col': ["foo", "a", "b", np.nan, 1.4]}
  2. df = pd.DataFrame(data)
  4. def handle_col(n):
  5.     match n:
  6.         case float(n) if n != n:
  7.             return "不是数字"
  8.         case float(n):
  9.             return "是数字"
  10.         case "a":
  11.             return "这是字母a"
  12.         case "b":
  13.             return "这是字母b"
  14.         case _:
  15.             return "这是另一个字符串"
  17. print(df["col"].apply(handle_col))

打印结果：

  1. 0    这是另一个字符串
  2. 1          这是字母a
  3. 2          这是字母b
  4. 3           不是数字
  5. 4            是数字
  6. Name: col, dtype: object
英文:

Another solution, compare number != number:

  1. data = {'col': ["foo", "a", "b", np.nan, 1.4]}
  2. df = pd.DataFrame(data)
  4. def handle_col(n):
  5.     match n:
  6.         case float(n) if n != n:
  7.             return "not a number"
  8.         case float(n):
  9.             return "a number"
  10.         case "a":
  11.             return "this is letter a"
  12.         case "b":
  13.             return "this is letter b"
  14.         case _:
  15.             return "this is another string"
  17. print(df["col"].apply(handle_col))

Prints:

  1. 0    this is another string
  2. 1          this is letter a
  3. 2          this is letter b
  4. 3              not a number
  5. 4                  a number
  6. Name: col, dtype: object

huangapple
  • 本文由 发表于 2023年7月24日 18:34:00
  • 转载请务必保留本文链接：https://go.coder-hub.com/76753624.html
  • pandas
  • python-3.x
  • switch-statement
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