numba - TypingError: cannot determine Numba type of <class 'builtin_function_or_method'>

Question

我有一个简单的函数来对扑克手牌进行排名（手牌是字符串）。

我用 rA,rB = rank(a),rank(b) 调用它，这是我的实现。没有 @jit(nopython=True) 也能很好地工作，但是有了它，它就失败了：

   File "...poker.py", line 190, in <module>
        rA,rB = rank(a),rank(b)
    
      File "C:\Continuum\anaconda3\lib\site-packages\numba\dispatcher.py", line 344, in _compile_for_args
        reraise(type(e), e, None)
    
      File "C:\Continuum\anaconda3\lib\site-packages\numba\six.py", line 658, in reraise
        raise value.with_traceback(tb)

TypingError: cannot determine Numba type of <class 'builtin_function_or_method'>

---

from numba import jit
from numba.types import string

@jit(nopython=True)
def rank(hand):
#    assert(len(hand) == 5)
    rank = "N/A"
    
    p = pd.Series([h[0] for h in hand]).value_counts()
    v = sorted(set(pd.Series([h[0] for h in hand]).values), reverse=True)
    s = sorted(hand, key=lambda k:k[0])    
    z = zip(s,s[1:])
    
    if all(x[0]==y[0]-1 for x,y in z):
        rank = "Straight "
    
    if len(set([h[1] for h in hand])) == 1:
        rank += "Flush "
    
    if "Straight Flush" in rank and sum([h[0] for h in hand]) == sum([10,11,12,13,14]):
        rank = "Royal Flush"
    
    elif p[p.idxmax()] == 4:
        rank = "4 Of A Kind : %d" % p.idxmax()
        
    elif p[p.idxmax()] == 3 and p[p.idxmin()] == 1:
        rank = "3 Of A Kind : %d" % p.idxmax()
        
    elif p[p.idxmax()] == 3 and p[p.idxmin()] == 2:
        rank = "Full House : %d,%d" % (p.idxmax(), p.idxmin())
        
    elif p[p.idxmax()] == 2:
        max2 = p.nlargest(2)
        
        if list(max2) == [2,2]:
            max2 = sorted(list(max2.keys()), reverse=True)
            rank = "2 Pairs : %d,%d" % (max2[0],max2[1])
        else:
            rank = "Pair : %d" % p.idxmax()
    
    else:
        rank = "High Card : %d" % v[0]
    

    return rank

Answer 1

Pandas 和代码中的其他几个函数调用将不适用于 nopython=True。在 nopython 中可与 numba jit 一起使用的可用库相当有限（几乎仅适用于 numpy 数组和某些 python 内置库）。您可以找到更多信息here

Answer 2

根据 deprecation recommendations，不使用 @jit(nopython=True) 编译的代码在没有装饰器的情况下更快是非常合理的.

有趣的是，我发现我在这里研究的这个小例子在简单地将 Pandas 列直接传递到我的函数到 时明显更快，而不是使用 numba 并支付方法开销numpy 数组的列。

但是，它会继续使用预期的参数来清除此警告

If there is benefit to having the @jit decorator present, then to be future proof supply the keyword argument forceobj=True to ensure the function is always compiled in object mode.