A = [3 NaN 5 6 7 NaN NaN 9]; TF = ismissing(A) 则TF=
举一反三
- A = [3 NaN 5 6 7 NaN NaN 9];TF = ismissing(A)则TF=( ) A: 0 1 0 0 0 1 1 0 B: 2 6 7 C: 1 3 4 5 8 D: 以上都不对
- 如果df1=pd.DataFrame([[1,2,3],[NaN,NaN,2],[NaN,NaN,NaN],[8,8,NaN]])，则 df1.fillna(100)=
- dblVar = [NaN;3;5;7;9;11;13]; singleVar = single([1;NaN;5;7;9;11;13]); cellstrVar = {'one';'three';'';'seven';'nine';'eleven';'thirteen'}; charVar = ['A';'C';'E';' ';'I';'J';'L']; categoryVar = categorical({'red';'yellow';'blue';'violet';'';'ultraviolet';'orange'}); dateVar = [datetime(2015,1:2:10,15) NaT datetime(2015,11,15)]'; stringVar = ["a";"b";"c";"d";"e";"f";missing]; A = table(dblVar,singleVar,cellstrVar,charVar,categoryVar,dateVar,stringVar) TF = ismissing(A) 以下说法不正确的是
- 下面代码输出正确的是（ ） A: 3, 3, 3 B: 3, 3, NaN C: 3, NaN, NaN D: othe
- 【填空题】写出代码输出结果。 null == undefined; //结果是_____ NaN == NaN; //结果是_____ 5 == NaN; //结果是_____ NaN != NaN; //结果是_____ false == 0; //结果是_____ true == 1; //结果是_____ true == 2; //结果是_____ undefined == 0; //结果是_____ null == 0; //结果是_____ "5" == 5; //结果是_____