Question

我有两个数组如下。

BLoc1 = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'I', 'K', 'L', 'M']
BLoc2 = ['A', 'D', 'E', 'F', 'G', 'I', 'J', 'K', 'L']

我希望输出如下。

BLoc1 = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'I', 'NA', 'K', 'L', 'M']
BLoc2 = ['A', 'NA', 'NA', 'D', 'E', 'F', 'G', 'I', 'J', 'K', 'L','NA']

Answer 1

您尝试实现的结果仅适用于单元格数组，因为否则'NA'将被视为'N' 'A'，即两个字母表都位于不同的索引处。

无论如何，以下代码将帮助您做出决定： -

BLoc1 = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'I', 'K', 'L', 'M'];
BLoc2  = ['A', 'D', 'E', 'F', 'G', 'I', 'J', 'K', 'L'];

%Converting to cell arrays
BLoc1=num2cell(BLoc1,size(BLoc1,2));
BLoc2=num2cell(BLoc2,size(BLoc2,2));

%Union of the two
complete=union(BLoc1,BLoc2);

NewBLoc1=complete;
[~,idx]=ismember(complete,BLoc1);  
idx = find(idx==0);          %finding the indexes where BLoc1 has missing elements
[NewBLoc1{idx}]=deal('NA')   %filling the elements at missing indexes with 'Na'

%Similarly for BLoc2
NewBLoc2=complete;
[~,idx]=ismember(complete,BLoc2);
idx = find(idx==0);
[NewBLoc2{idx}]=deal('NA')

%converting back to simple matrix since that's your requirement in the question 
%but I recommend you to also run the code atleast once without the next two lines too.
NewBLoc1= cell2mat(NewBLoc1) 
NewBLoc2= cell2mat(NewBLoc2)

Answer 2

检查一下：

<uses-permission android:name="android.permission.ACCESS_COARSE_LOCATION"/>
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION"/>
<uses-permission android:name="android.permission.ACCESS_MOCK_LOCATION"/>
<uses-permission android:name="android.permission.INTERNET"/>
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE"/>

在matlab中对两个数组进行特殊排序并进行比较

2 个答案: