Question

所以我有一本字典：

LCM_SCS = { 
        (1, "A"): 36, (1, "B"): 60, (1, "C"): 73, (1, "D"): 79, 
        (2, "A"): 36, (2, "B"): 60, (2, "C"): 73, (2, "D"): 79, 
        (3, "A"): 74, (3, "B"): 83, (3, "C"): 88, (3, "D"): 90, 
        (4, "A"): 68, (4, "B"): 79, (4, "C"): 86, (4, "D"): 89, 
        (5, "A"): 30, (5, "B"): 58, (5, "C"): 71, (5, "D"): 78, 
        (6, "A"): 39, (6, "B"): 61, (6, "C"): 74, (6, "D"): 80, 
        (7, "A"): 39, (7, "B"): 61, (7, "C"): 74, (7, "D"): 80, 
        (8, "A"): 39, (8, "B"): 61, (8, "C"): 74, (8, "D"): 80, 
        (10, "A"): 30, (10, "B"): 48, (10, "C"): 65, (10, "D"): 73,

我还有两个组合的数组给出了我上面字典的元组键：

数组1：

array1 = np.array([[1, 1, 1],
          [2, 2, 3],
          [2, 4, 5]])

数组2：

array2 = np.array([["A", "A", "A"],
             ["B", "B", "B"],
             ["C", "C", "C"]])

我的代码是：

Numbers = np.empty_like(array1)


for [x, y], (value1, value2) in np.ndenumerate(izip(array1, array2)):
        CN_numbers[x, y] = LCM_SCS.get((value1, value2))

    return Numbers

此代码不起作用。我想得到的是一个如下所示的数组：

Numbers = array([[36, 36, 36],
          [60, 60, 83],
          [73, 86, 71]])

所以基本上我有两个数组包含要用作查找字典的键的值，我不知道如何在代码中实现它。

任何建议或帮助将不胜感激。

谢谢，

尼克

使用vectorise的解决方案：

a_new = np.empty_like(array1)

def get_CN_numbers(a1, a2):
    return LCM_SCS[(a1, a2)]  # your basic scalar-operation

V_get_CN = np.vectorize(get_CN_numbers)

a_new = V_get_CN(array1, array2)

print a_new

Answer 1

vectorize它。

@numpy.vectorize
def get_CN_numbers(a1, a2):
    return LCM_SCS[(a1,a2)]  # your basic scalar-operation

get_CN_numbers(array1, array2)
=>
array([[36, 36, 36],
       [60, 60, 83],
       [73, 86, 71]])

通常，使用vectorize是扩展标量操作的简单方法（在您的情况下，通过标量键从dict获取值）来处理数组（在您的情况下，两个键的数组））。正如你已经发现的那样，vectorize照顾你的棘手部分是保持形状。

这提供了简单性，但不一定是速度，因为vectorize是使用python-space循环实现的。

Answer 2

试试这个：

>>> new_array = np.rec.fromarrays((array1,array2),names='x,y')     # This will generate all keys that you'll look value for.
>>> print new_array
[[(1, 'A') (1, 'A') (1, 'A')]
 [(2, 'B') (2, 'B') (3, 'B')]
 [(2, 'C') (4, 'C') (5, 'C')]]
>>> result = np.zeros([3,3],dtype=int)      #Having issue to modify directly on new_array so I initialized a new numpy array to store result
>>> for (x,y), value in np.ndenumerate(new_array):
    result[x][y] = LCM_SCS[tuple(value)]

>>> print result
[[36 36 36]
 [60 60 83]
 [73 86 71]]

Answer 3

LCM_SCS = { 
        (1, "A"): 36, (1, "B"): 60, (1, "C"): 73, (1, "D"): 79, 
        (2, "A"): 36, (2, "B"): 60, (2, "C"): 73, (2, "D"): 79, 
        (3, "A"): 74, (3, "B"): 83, (3, "C"): 88, (3, "D"): 90, 
        (4, "A"): 68, (4, "B"): 79, (4, "C"): 86, (4, "D"): 89, 
        (5, "A"): 30, (5, "B"): 58, (5, "C"): 71, (5, "D"): 78, 
        (6, "A"): 39, (6, "B"): 61, (6, "C"): 74, (6, "D"): 80, 
        (7, "A"): 39, (7, "B"): 61, (7, "C"): 74, (7, "D"): 80, 
        (8, "A"): 39, (8, "B"): 61, (8, "C"): 74, (8, "D"): 80, 
        (10, "A"): 30, (10, "B"): 48, (10, "C"): 65, (10, "D"): 73}

array1 = np.array([[1, 1, 1], [2, 2, 3], [2, 4, 5]]).tolist()
array2 = np.array([["A", "A", "A"], ["B", "B", "B"], ["C", "C", "C"]]).tolist()

array1 = [y for sub in array1 for y in sub]
array2 = [y for sub in array2 for y in sub]
results = [LCM_SCS[(array1[k], array2[k])] for k in range(len(array1))]

<强>输出：

[36, 36, 36, 60, 60, 83, 73, 86, 71]

您可以根据需要将其转换为列表列表，甚至可以将其设为np.array()。

迭代两个数组在dict中查找值

3 个答案: