很抱歉打扰，但我发现了一篇有趣的文章“Mortaz, E. (2020)。多类不平衡分类问题中模型选择的不平衡准确度度量。基于知识的系统, 210, 106490” (@987654321 @) 并且他们在那里计算这个度量（IAM）（公式在论文中，我理解了），但我想问：我怎样才能在 R 上复制它？
对于这个愚蠢的问题，我提前道歉。感谢您的关注！
【问题讨论】：
标签： classification metrics multilabel-classification multiclass-classification
文中提供的IAM公式为：IAM formula
其中 cij 是分类器混淆矩阵 (c) 中的元素 (i,j)。 k是指分类中的类数（k>=2）。结果表明，该度量可以作为多类模型选择的单独度量。
下面提供的python实现IAM（不平衡准确度指标）的代码：
def IAM(c):
'''
c is a nested list presenting the confusion matrix of the classifier (len(c)>=2)
'''
l = len(c)
iam = 0
for i in range(l):
sum_row = 0
sum_col = 0
sum_row_no_i = 0
sum_col_no_i = 0
for j in range(l):
sum_row += c[i][j]
sum_col += c[j][i]
if j is not i:
sum_row_no_i += c[i][j]
sum_col_no_i += c[j][i]
iam += (c[i][i] - max(sum_row_no_i, sum_col_no_i))/max(sum_row, sum_col)
return iam/l
c = [[2129, 52, 0, 1],
[499, 70, 0, 2],
[46, 16, 0, 1],
[85, 18, 0, 7]]
IAM(c) = -0.5210576475801445
在 R 中实现 IAM（不平衡准确度指标）的代码如下：
IAM <- function(c) {
# c is a matrix representing the confusion matrix of the classifier.
l <- nrow(c)
result = 0
for (i in 1:l) {
sum_row = 0
sum_col = 0
sum_row_no_i = 0
sum_col_no_i = 0
for (j in 1:l){
sum_row = sum_row + c[i,j]
sum_col = sum_col + c[j,i]
if(i != j) {
sum_row_no_i = sum_row_no_i + c[i,j]
sum_col_no_i = sum_col_no_i + c[j,i]
}
}
result = result + (c[i,i] - max(sum_row_no_i, sum_col_no_i))/max(sum_row, sum_col)
}
return(result/l)
}
c <- matrix(c(2129,52,0,1,499,70,0,2,46,16,0,1,85,18,0,7), nrow=4, ncol=4)
IAM(c) = -0.5210576475801445
另一个来自鸢尾花数据集（3类问题）和sklearn的例子：
from sklearn.datasets import load_iris
from sklearn.linear_model import LogisticRegression
from sklearn.metrics import confusion_matrix
X, y = load_iris(return_X_y=True)
clf = LogisticRegression(max_iter = 1000).fit(X, y)
pred = clf.predict(X)
c = confusion_matrix(y, pred)
print('confusion matrix:')
print(c)
print(f'accuarcy : {clf.score(X, y)}')
print(f'IAM : {IAM(c)}')
confusion matrix:
[[50 0 0]
[ 0 47 3]
[ 0 1 49]]
accuarcy : 0.97
IAM : 0.92
【讨论】：