问MATLAB:使用find函数获取数组中某个值的索引EN版权声明：本文内容由互联网用户自发贡献，该文观点仅代表...
fprintf('value of a: %d \n', a); end 1. 2. 3. 输出结果为: value of a: 10 value of a: 11 value of a: 12 value of a: 13 value of a: 14 value of a: 15 value of a: 16 value of a: 17 value of a: 18 value of a: 19 value of a: 20 1. 2. 3. 4. 5. 6. ...
然后这条find语句做的事是，先把iif的所有输入参数中位于奇数位置的参数取出来形成一个数组[varargin{1}...
Set the initial value to 3000 for the"wheat-onion"and"wheat-lentil"crops that are planted in any dry condition and are plowed traditionally. [idxLand, idxCrop, idxPlough] = findindex(xcrop, ["dry-good","dry-poor"],...["wheat-onion","wheat-lentil"],"tradition"); x0.xcrop(idxLand...
Reward定义的是立即的收益,而value function定义的是长期的收益,它可以看作是累计的reward,常用v来表示。 环境模型(a model of the environment) 预测environment下一步会做出什么样的改变,从而预测agent接收到的状态或者reward是什么。 总之,强化学习作为一个序列决策(Sequential Decision Making)问题,它需要连续选择一些...
FUN accepts % input X and returns a scalar function value F evaluated at X. X0 may be % a scalar, vector, or matrix. % % X = FMINCON(FUN,X0,A,B,Aeq,Beq) minimizes FUN subject to the linear % equalities Aeq*X = Beq as well as A*X <= B. (Set A=[] and B=[] if no...
To find a noninteger value, use a tolerance value based on your data. Otherwise, the result is sometimes an empty matrix due to floating-point roundoff error. Get y = 0:0.1:1 y = 1×11 0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000 Get k = find(y==0.3...
First enter the original matrix in MATLAB: Sort by index type, with benefit index first and cost index second. Find the first column to start as a cost index. Cost_Column plays a very important role in defining the size of the column in the for loop. Automatic operation can be correctly...
One can impute the missing values, i.e., to infer them from the known part of the data. The Imputer function provides basic strategies for imputing missing values, either using the mean, the median or the most frequent value of the column in which the missing values are located, Just ...
k = find(x==13) k = 7 To find a noninteger value, use a tolerance value based on your data. Otherwise, the result is sometimes an empty matrix due to floating-point roundoff error. y = 0:0.1:1 y = 1×11 0 0.1000 0.2000 0.3000 0.4000 0.5000 0.6000 0.7000 0.8000 0.9000 1.0000 ...