看看numba解压range和xrange的方式，很明显有很多事情不仅仅等同于for循环。例如：
import unittest
class FastOpsTest(unittest.TestCase):
def test_numba_sum(self):
arr = np.array([1,2,3], dtype=float)
self.assertEquals(FastOpsTest.sum(arr), 6)
FastOpsTest.sum.inspect_types()
@staticmethod
@jit(nopython=True, nogil=True)
def sum(arr):
N = len(arr)
result = 0
for i in xrange(N): # range() behaves similarly
result += arr[i]
return result
...给出下面的VM代码。
for循环实际上是调用python范围和xrange C函数。更糟糕的是，看起来它在堆上分配内存（我认为这就是dels所做的）。这看起来非常不理想，特别是对于嵌套循环。
除了手工重构之外：
i =0
while i!=N:
...
i+=1
在numba中优化循环是否有更好的模式？
# File: /home/userx/py/fast_ops.py
# --- LINE 176 ---
# label 0
# del $0.1
# del $0.3
# del $const0.4
@staticmethod
# --- LINE 177 ---
@jit(nopython=True, nogil=True)
# --- LINE 178 ---
def sum(arr):
# --- LINE 179 ---
# arr = arg(0, name=arr) :: array(float64, 1d, C)
# $0.1 = global(len: <built-in function len>) :: Function(<built-in function len>)
# $0.3 = call $0.1(arr, kws=[], args=[Var(arr, /home/userx/py/fast_ops.py (179))], func=$0.1, vararg=None) :: (array(float64, 1d, C),) -> int64
# N = $0.3 :: int64
N = len(arr)
# --- LINE 180 ---
# $const0.4 = const(int, 0) :: int64
# result = $const0.4 :: float64
# jump 18
# label 18
result = 0
# --- LINE 181 ---
# jump 21
# label 21
# $21.1 = global(xrange: <type 'xrange'>) :: Function(<built-in function range>)
# $21.3 = call $21.1(N, kws=[], args=[Var(N, /home/userx/py/fast_ops.py (179))], func=$21.1, vararg=None) :: (int64,) -> range_state_int64
# del N
# del $21.1
# $21.4 = getiter(value=$21.3) :: range_iter_int64
# del $21.3
# $phi31.1 = $21.4 :: range_iter_int64
# del $21.4
# jump 31
# label 31
# $31.2 = iternext(value=$phi31.1) :: pair<int64, bool>
# $31.3 = pair_first(value=$31.2) :: int64
# $31.4 = pair_second(value=$31.2) :: bool
# del $31.2
# $phi54.1 = $31.3 :: int64
# del $phi54.1
# $phi54.2 = $phi31.1 :: range_iter_int64
# del $phi54.2
# $phi34.1 = $31.3 :: int64
# del $31.3
# branch $31.4, 34, 54
# label 34
# del $31.4
# i = $phi34.1 :: int64
# del $phi34.1
# del i
# del $34.5
# del $34.6
for i in xrange(N):
# --- LINE 182 ---
# $34.5 = getitem(index=i, value=arr) :: float64
# $34.6 = inplace_binop(static_rhs=<object object at 0x7fb921f7cbc0>, rhs=$34.5, immutable_fn=+, lhs=result, static_lhs=<object object at 0x7fb921f7cbc0>, fn=+=) :: float64
# result = $34.6 :: float64
# jump 31
# label 54
# del arr
# del $phi34.1
# del $phi31.1
# del $31.4
# jump 55
# label 55
# del result
result += arr[i]
# --- LINE 183 ---
# $55.2 = cast(value=result) :: float64
# return $55.2
return result
inspect_types()返回Numba IR - 我不熟悉它，但我认为没有任何理由期望它贴近实际执行的地图。
在抽象中逐渐减少，您还可以使用inspect_llvm()方法查看LLVM IR，并使用inspect_asm()查看实际执行的内容。在这种情况下，查看LLVM IR可以很清楚地编译成一个非常简单的for循环 - 我相信标签B24:是内循环。
print(next(iter(FastOpsTest.sum.inspect_llvm().values())))
# some parts ommitted
define i32 @"_ZN8__main__11FastOpsTest7sum$242E5ArrayIdLi1E1C7mutable7alignedE"(double* noalias nocapture %retptr, { i8*, i32 }** noalias nocapture readnone %excinfo, i8* noalias nocapture readnone %env, i8* nocapture readnone %arg.arr.0, i8* nocapture readnone %arg.arr.1, i64 %arg.arr.2, i64 %arg.arr.3, double* nocapture readonly %arg.arr.4, i64 %arg.arr.5.0, i64 %arg.arr.6.0) local_unnamed_addr #0 {
entry:
%.98 = icmp sgt i64 %arg.arr.5.0, 0
br i1 %.98, label %B24.preheader, label %B40
B24.preheader: ; preds = %entry
%0 = add i64 %arg.arr.5.0, 1
br label %B24
B24: ; preds = %B24.preheader, %B24
%lsr.iv8 = phi double* [ %arg.arr.4, %B24.preheader ], [ %scevgep, %B24 ]
%lsr.iv = phi i64 [ %0, %B24.preheader ], [ %lsr.iv.next, %B24 ]
%result.07 = phi double [ %.250, %B24 ], [ 0.000000e+00, %B24.preheader ]
%.242 = load double, double* %lsr.iv8, align 8
%.250 = fadd double %result.07, %.242
%lsr.iv.next = add i64 %lsr.iv, -1
%scevgep = getelementptr double, double* %lsr.iv8, i64 1
%.143 = icmp sgt i64 %lsr.iv.next, 1
br i1 %.143, label %B24, label %B40
B40: ; preds = %B24, %entry
%result.0.lcssa = phi double [ 0.000000e+00, %entry ], [ %.250, %B24 ]
store double %result.0.lcssa, double* %retptr, align 8
ret i32 0
}