根据字典值和列表创建新矩阵

Question

下面是一个简化的例子：

val = [10,23,45,31,78,43,1,67,82]

indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]

samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3

假设我有一本字典 (samp)，它有两个组：第 0 组和第 1 组。 字典中有向量 val 中的值的索引。

我想通过创建一个 8 X 2 矩阵来根据字典中的给定组提取 val 中的所有值， 我在两列中按索引排序的组和值，所以它看起来像这样：

val  group
10   2
23   0
45   2
31   0
87   0
43   1
1    1
67   0
82   1

我该怎么做？

Answer 1

一种获得方法

[(j, next(k for k,v in samp.items() if i in v)) for i,j in enumerate(val)]

输出：

[(10, 2),
 (23, 0),
 (45, 2),
 (31, 1),
 (78, 0),
 (43, 0),
 (1, 1),
 (67, 1),
 (82, 0)]

Answer 2

这是一个不使用 pandas 的解决方案，它输出一个 (8,2) numpy 矩阵：

val = [10,23,45,31,78,43,1,67,82]

indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]

indices = [indx, indx2, indx3]

def get_group(x):
    for i,indx_arr in enumerate(indices):
        if x in indx_arr:
            return i

pairs = [(v,get_group(i)) for i,v in enumerate(val)]
np.asarray(pairs)

array([[10,  2],
   [23,  0],
   [45,  2],
   [31,  1],
   [78,  0],
   [43,  0],
   [ 1,  1],
   [67,  1],
   [82,  0]])

Answer 3

使用dictionary comprehension反转字典中的键值对，然后使用map：

df = pd.DataFrame(val,columns=['val'])
d = {value1:key  for key,value in samp.items() for value1 in value}
df['group'] = df.index.map(d)

print(df)
   val  group
0   10      2
1   23      0
2   45      2
3   31      1
4   78      0
5   43      0
6    1      1
7   67      1

---

print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}

如果值是 numpy 数组怎么办：

indx = np.array([1,4,5,8])
indx2 = np.array([3,6,7])
indx3 = np.array([0,2])

samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3

print(samp)
{0: array([1, 4, 5, 8]), 1: array([3, 6, 7]), 2: array([0, 2])}

d = {value1:key  for key,value in samp.items() for value1 in value}

print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}