Error> The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()

huangapple
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英文:

Error> The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()

问题

  1. 我想要在 Pandas 数据框中*水平*对每行中值为1的列进行求和。我不是在垂直方向上求和列，而是在水平方向上。看起来互联网上的每个例子都是垂直求和列。
  3. 这是我的简单函数：
  5.     def isOne(x):
  6.         if (x == 1):
  7.             return 1 
  8.         else:
  9.             return 0
  11. 这是我的应用语句：
  13.     df.apply(lambda column: sum(isOne(column)), axis=1)
  15. 这是我收到的错误：
  17.     ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
  18.     
  19. 我尝试在布尔表达式上附加了 .any()、.all() 和 .item()，但没有效果。
英文:

I want to sum the columns in a Pandas dataframe(array) horizontally along each row that only have a value of 1. I am not summing the columns vertically. I am summing them horizontally. It seems like every example on the Internet sums column vertically.

Here is my simple function:

  1. def isOne(x):
  2.     if (x == 1):
  3.         return 1 
  4.     else:
  5.         return 0

Here is my apply statement

  1. df.apply(lambda column: sum(isOne(column)), axis=1)

This is the error that I receive:

  1. ValueError                                Traceback (most recent call last)
  2. Cell In[30], line 2
  3. 	  1 tally = 0
  4. ----> 2 df.apply(lambda column: sum(isOne(column)), axis=1
  5. 	  3         )
  7. File ~\AppData\Local\Programs\Python\Python311\Lib\site-packages\pandas\core\frame.py:9568, in DataFrame.apply(self, func, axis, raw, result_type, args, **kwargs)
  8.    9557 from pandas.core.apply import frame_apply
  9.    9559 op = frame_apply(
  10.    9560     self,
  11.    9561     func=func,
  12.    (...)
  13.    9566     kwargs=kwargs,
  14.    9567 )
  15. -> 9568 return op.apply().__finalize__(self, method="apply")
  17. File ~\AppData\Local\Programs\Python\Python311\Lib\site-packages\pandas\core\apply.py:764, in FrameApply.apply(self)
  18. 	761 elif self.raw:
  19. 	762     return self.apply_raw()
  20. --> 764 return self.apply_standard()
  22. File ~\AppData\Local\Programs\Python\Python311\Lib\site-packages\pandas\core\apply.py:891, in FrameApply.apply_standard(self)
  23. 	890 def apply_standard(self):
  24. --> 891     results, res_index = self.apply_series_generator()
  25. 	893     # wrap results
  26. 	894     return self.wrap_results(results, res_index)
  28. File ~\AppData\Local\Programs\Python\Python311\Lib\site-packages\pandas\core\apply.py:907, in FrameApply.apply_series_generator(self)
  29. 	904 with option_context("mode.chained_assignment", None):
  30. 	905     for i, v in enumerate(series_gen):
  31. 	906         # ignore SettingWithCopy here in case the user mutates
  32. --> 907         results[i] = self.f(v)
  33. 	908         if isinstance(results[i], ABCSeries):
  34. 	909             # If we have a view on v, we need to make a copy because
  35. 	910             #  series_generator will swap out the underlying data
  36. 	911             results[i] = results[i].copy(deep=False)
  38. Cell In[30], line 2, in <lambda>(column)
  39. 	  1 tally = 0
  40. ----> 2 df.apply(lambda column: sum(isOne(column)), axis=1
  41. 	  3         )
  43. Cell In[29], line 2, in isOne(x)
  44. 	  1 def isOne(x):
  45. ----> 2     if (x == 1):
  46. 	  3         return 1 
  47. 	  4     else:
  49. File ~\AppData\Local\Programs\Python\Python311\Lib\site-packages\pandas\core\generic.py:1527, in NDFrame.__nonzero__(self)
  50.    1525 @final
  51.    1526 def __nonzero__(self) -> NoReturn:
  52. -> 1527     raise ValueError(
  53.    1528         f"The truth value of a {type(self).__name__} is ambiguous. "
  54.    1529         "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
  55.    1530     )
  57. ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

I tried appending .any(), .all() and .item to boolean expression, buthtat had no effect.

答案1

得分: 0

请尝试这个：

  1. def isOne(arr):
  2.     j = 0
  3.     for i in arr:
  4.         if (i == 1):
  5.             j += 1
  6.     return j
  8. df.apply(isOne, axis=1)
英文:

Try this:

  1. def isOne(arr):
  2.     j = 0
  3.     for i in arr:
  4.         if (i == 1):
  5.             j += 1
  6.     return j
  9. df.apply(isOne, axis=1)

答案2

得分: 0

如果你想要计算每行中1的数量，请使用向量化的代码。速度会更快：

  1. df.eq(1).sum(axis=1)

如果你想要使用lambda函数练习：在axis=1情况下，x是包含行值的Series。像“series等于1？”这样的陈述是含糊的。你是指Series是否包含任何1，还是全部为1？可以尝试这样做：

  1. def count_ones(x: pd.Series) -> int:
  2.     return (x == 1).sum()
  4. df.apply(count_ones, axis=1)
英文:

If you want to count the number of 1's in each row, use vectorized code. It's a lot faster:

  1. df.eq(1).sum(axis=1)

If you want to use a lambda as practice: with axis=1, x is a Series containing the values for the row. Statements like "series equal to 1?" are ambiguous. Do you mean the series contains any 1, or all 1? Try this:

  1. def count_ones(x: pd.Series) -> int:
  2.     return (x == 1).sum()
  4. df.apply(count_ones, axis=1)

huangapple
  • 本文由 发表于 2023年2月27日 10:32:13
  • 转载请务必保留本文链接：https://go.coder-hub.com/75576333.html
  • apply
  • dataframe
  • pandas
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