我想从预测列表中创建一个版本的scikit
我现在有一个列表如下:
[[0,1,0,0,0,1,1,0,0,0],[0,1,0,1,0,1,1,1,0,0],[0,0,0,0,0,1,1,0,0,0]]
我想找出每个列表的第一个值为0或1的概率,然后每个连续值的概率相同
即输出如下:
[[0.33,0.66],[0,1],[0.66,0.3]........etc
我已经写了下面的代码,它工作得很好,但它似乎很笨拙,我相信有一个更好的方法来实现我的目标
有什么建议吗
#create np array from list
ar = np.array([[0,1,0,0,0,1,1,0,0,0],[0,1,0,1,0,1,1,1,0,0],[0,0,0,0,0,1,1,0,0,0]])
#calculate unique values and sort in order
uni = np.unique(ar)
uni.sort()
#create new pred list
new_pred = []
#transpose and iterate
for row in ar.transpose():
# create dic with keys as unique values
val_dic = {k: 0 for k in uni}
#create list for row probabilities
row_pred = []
#iterate row and incremnet dic if found
for val in row:
if val in val_dic.keys():
val_dic[val] = val_dic.get(val, 0) + 1
#calc row total
total = sum(val_dic.values())
#append row list with probabilities
for val in val_dic.values():
row_pred.append(val/total)
#append final output list
new_pred.append(row_pred)
print(new_pred)
输出:
[[1.0, 0.0], [0.3333333333333333, 0.6666666666666666], [1.0, 0.0], [0.6666666666666666, 0.3333333333333333], [1.0, 0.0], [0.0, 1.0], [0.0, 1.0], [0.6666666666666666, 0.3333333333333333], [1.0, 0.0], [1.0, 0.0]]
如果您的
ar
仅由问题中的0
、1
组成,您可以这样做来简化代码:印刷品:
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