Question

我有一个字典d这些键是n整数。密钥不一定是连续的数字，但是，d包含范围num_negative中的(-num_negative,0)个连续密钥和范围num_positive中的(0,num_positive)个连续密钥。我想从它创建另一个字典d2，其键是[0, n]范围内的连续整数，给出以下约束：

（1）num_positive中来自d组的密钥的元素在d2中具有相同的密钥。

（2）i中num_negative组中具有否定键d的元素将在n+i中具有键d2。< / p>

示例：

# num_positive = 2, num_negative = 3, n=8

# num_positive group = {0,1}
# num_negative gorup = {-3,-2,-1}

d = {-6: 'h', -3: 'a', -2: 'b', -1: 'c', 0: 'd', 1: 'e', 68: 'f', 99: 'g'}
d2 = {0: 'd', 1: 'e', 2: 'g', 3: 'h', 4: 'f', 5: 'a', 6: 'b', 7: 'c'}

Answer 1

如果我理解正确，具有连续范围之外的键的项目在结果字典中具有未定义的位置。因此，在我的解决方案中，使用在连续范围[0, len(d))中仍然空闲的任何键，以任意顺序将它们插入到结果字典中。有关更多详细信息，请参阅注释。

from collections import OrderedDict

d = {-6: 'h', -3: 'a', -2: 'b', -1: 'c', 0: 'd', 1: 'e', 68: 'f', 99: 'g'}
remainder = {}

rPos = range(0, 2 + 1) # Positive consecutive range.
rNeg = range(-3, 0) # Negative consecutive range.
n = len(d)

d2 = OrderedDict([(i, None) for i in range(0, n)])

for k, v in d.items():
    if k in rNeg: # Checks if in negative consecutive range.
        d2[n + k] = v
    elif k in rPos: # Checks if in positive negative range.
        d2[k] = v
    else: # Key is outside of either range.
        remainder[k] = v

for k, v in d2.items():
    if v is None: # Finds a key still available for use.
        # Pops an arbitrary element from the remainder and inserts its
        # value into the new dict using an available key.
        d2[k] = remainder.popitem()[1]

Input: {-6: 'h', -3: 'a', -2: 'b', -1: 'c', 0: 'd', 1: 'e', 68: 'f', 99: 'g'}
Output: OrderedDict([(0, 'd'), (1, 'e'), (2, 'g'), (3, 'f'), (4, 'h'), (5, 'a'), (6, 'b'), (7, 'c')])

Answer 2

以下代码应该这样做。

d1={}

for key,val in d.items(): 
   if key in pos and key<=n :               #the elements that had a key from the group num_positive in d, will have the same key in d2
       d1[key] = val 
   elif key in neg and ((key+n) in range(n+1)) :             #the elements that had a negative key i from the group num_negative in d, will have the key n+i in d2
       d1[key+n] = val

#driver values：

IN : n = 8
IN : pos = [0,1]
IN : neg = [-3,-2,-1]
IN : d = {-6: 'h', -3: 'a', -2: 'b', -1: 'c', 0: 'd', 1: 'e', 68: 'f', 99: 'g'} 

OUT : d1 = {5: 'a', 6: 'b', 7: 'c', 0: 'd', 1: 'e'}

注意： OP的输出存在一些差异。其中很少有人表现得与定义的不同。

因此，忽略不遵循给定约束的键。

Python dict：在约束下将键映射到不同的范围

2 个答案: