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pandas numpy:在进行数学运算时使用序列设置数组元素

numpy-ndarray numpy dataframe pandas python

William · 技术社区 · 3 年前

我有一个名为df4的df,你可以购买以下代码:

df4s = """
contract    RB  BeginDate   ValIssueDate    EndDate Valindex0   48  46  47  49  50
2   A00118  46  19850100    19880901    99999999    50  1   2   3   7   7
3   A00118  47  19000100    19880901    19831231    47  1   2   3   7   7
5   A00118  47  19850100    19880901    99999999    50  1   2   3   7   7
6   A00253  48  19000100    19820101    19811231    47  1   2   3   7   7
7   A00253  48  19820100    19820101    19841299    47  1   2   3   7   7
8   A00253  48  19850100    19820101    99999999    50  1   2   3   7   7
9   A00253  50  19000100    19820101    19781231    47  1   2   3   7   7
10  A00253  50  19790100    19820101    19841299    47  1   2   3   7   7
11  A00253  50  19850100    19820101    99999999    50  1   2   3   7   7

"""

df4 = pd.read_csv(StringIO(df4s.strip()), sep='\s+', 
                  dtype={"RB": int, "BeginDate": int, "EndDate": int,'ValIssueDate':int,'Valindex0':int})

结果将是:

contract    RB  BeginDate   ValIssueDate    EndDate Valindex0   48  46  47  49  50
2   A00118  46  19850100    19880901    99999999    50  1   2   3   7   7
3   A00118  47  19000100    19880901    19831231    47  1   2   3   7   7
5   A00118  47  19850100    19880901    99999999    50  1   2   3   7   7
6   A00253  48  19000100    19820101    19811231    47  1   2   3   7   7
7   A00253  48  19820100    19820101    19841299    47  1   2   3   7   7
8   A00253  48  19850100    19820101    99999999    50  1   2   3   7   7
9   A00253  50  19000100    19820101    19781231    47  1   2   3   7   7
10  A00253  50  19790100    19820101    19841299    47  1   2   3   7   7
11  A00253  50  19850100    19820101    99999999    50  1   2   3   7   7

我试图通过以下逻辑构建一个新列,新列的值将基于两个现有列的值:

def test(RB):
    n=1
    for i in np.arange(RB,50):
        n = n * df4[str(i)].values
    return  n


vfunc=np.vectorize(test)
df4['n']=vfunc(df4['RB'].values)

然后收到错误:

    res = array(outputs, copy=False, subok=True, dtype=otypes[0])

ValueError: setting an array element with a sequence.

0 回复 | 直到 3 年前

hpaulj 3 年前

重建数据帧(感谢使用 StringIO (方法)

In [82]: df4['RB'].values
Out[82]: array([46, 47, 47, 48, 48, 48, 50, 50, 50])
In [83]: test(46)
Out[83]: array([42, 42, 42, 42, 42, 42, 42, 42, 42])
In [84]: test(50)
Out[84]: 1
In [85]: [test(i) for i in df4['RB'].values]
Out[85]: 
[array([42, 42, 42, 42, 42, 42, 42, 42, 42]),
 array([21, 21, 21, 21, 21, 21, 21, 21, 21]),
 array([21, 21, 21, 21, 21, 21, 21, 21, 21]),
 array([7, 7, 7, 7, 7, 7, 7, 7, 7]),
 array([7, 7, 7, 7, 7, 7, 7, 7, 7]),
 array([7, 7, 7, 7, 7, 7, 7, 7, 7]),
 1,
 1,
 1]
In [86]: vfunc=np.vectorize(test)
In [87]: vfunc(df4['RB'].values)
TypeError: only size-1 arrays can be converted to Python scalars

The above exception was the direct cause of the following exception:
Traceback (most recent call last):
  File "<ipython-input-87-8db8cd5dc5ab>", line 1, in <module>
    vfunc(df4['RB'].values)
  File "/usr/local/lib/python3.8/dist-packages/numpy/lib/function_base.py", line 2163, in __call__
    return self._vectorize_call(func=func, args=vargs)
  File "/usr/local/lib/python3.8/dist-packages/numpy/lib/function_base.py", line 2249, in _vectorize_call
    res = asanyarray(outputs, dtype=otypes[0])
ValueError: setting an array element with a sequence.

请注意完整的回溯。 vectorize 从这组大小混合的数组创建返回数组时遇到问题。这是猜测 , based on a trial calculation that it should return an int`dtype。

如果我们告诉它返回一个对象数据类型数组,我们会得到:

In [88]: vfunc=np.vectorize(test, otypes=['object'])
In [89]: vfunc(df4['RB'].values)
Out[89]: 
array([array([42, 42, 42, 42, 42, 42, 42, 42, 42]),
       array([21, 21, 21, 21, 21, 21, 21, 21, 21]),
       array([21, 21, 21, 21, 21, 21, 21, 21, 21]),
       array([7, 7, 7, 7, 7, 7, 7, 7, 7]),
       array([7, 7, 7, 7, 7, 7, 7, 7, 7]),
       array([7, 7, 7, 7, 7, 7, 7, 7, 7]), 1, 1, 1], dtype=object)

我们可以将其分配给df列:

In [90]: df4['n']=_
In [91]: df4
Out[91]: 
   contract  RB  BeginDate  ...  49  50                                     n
2    A00118  46   19850100  ...   7   7  [42, 42, 42, 42, 42, 42, 42, 42, 42]
3    A00118  47   19000100  ...   7   7  [21, 21, 21, 21, 21, 21, 21, 21, 21]
5    A00118  47   19850100  ...   7   7  [21, 21, 21, 21, 21, 21, 21, 21, 21]
6    A00253  48   19000100  ...   7   7           [7, 7, 7, 7, 7, 7, 7, 7, 7]
7    A00253  48   19820100  ...   7   7           [7, 7, 7, 7, 7, 7, 7, 7, 7]
8    A00253  48   19850100  ...   7   7           [7, 7, 7, 7, 7, 7, 7, 7, 7]
9    A00253  50   19000100  ...   7   7                                     1
10   A00253  50   19790100  ...   7   7                                     1
11   A00253  50   19850100  ...   7   7                                     1

我们也可以指定 Out[85] 列表

df4['n']=Out[85]

时间差不多:

In [94]: timeit vfunc(df4['RB'].values)
211 Âµs Â± 5.4 Âµs per loop (mean Â± std. dev. of 7 runs, 1000 loops each)
In [95]: timeit [test(i) for i in df4['RB'].values]
217 Âµs Â± 6.06 Âµs per loop (mean Â± std. dev. of 7 runs, 1000 loops each)

通常 矢量化 比较慢,但是 test 它本身可能足够慢,而迭代方法并没有多大区别。记住(必要时重新阅读文档), 矢量化 是不性能工具。它不会“编译”你的函数,也不会让它运行得更快。

返回对象数据类型数组的另一种方法是:

In [96]: vfunc=np.frompyfunc(test,1,1)
In [97]: vfunc(df4['RB'].values)
Out[97]: 
array([array([42, 42, 42, 42, 42, 42, 42, 42, 42]),
       array([21, 21, 21, 21, 21, 21, 21, 21, 21]),
       array([21, 21, 21, 21, 21, 21, 21, 21, 21]),
       array([7, 7, 7, 7, 7, 7, 7, 7, 7]),
       array([7, 7, 7, 7, 7, 7, 7, 7, 7]),
       array([7, 7, 7, 7, 7, 7, 7, 7, 7]), 1, 1, 1], dtype=object)
In [98]: timeit vfunc(df4['RB'].values)
202 Âµs Â± 6.2 Âµs per loop (mean Â± std. dev. of 7 runs, 1000 loops each)