Question

这是一个数组。如何根据我的要求交换值。正如您在下面的示例中看到的那样，数组的第一个值和第三个值将是swap。然后第4个值到第6个值等等。

" 198.16666666418314, 199.75",  <- swap this
"203.5, 198.5",
"200, 196.5",                   <- to this
" 199.83333332836628, 217",     <- swap this
"196.33333332836628, 203",
"196.16666665673256, 211.5",    <- to this
" 218.5, 227.5",                <- swap this
"203.5, 222.5",
"211, 225"                      <- to this

我是ios的新手，不知道解决这个问题。

基本上我有这个输出值

2015-08-07 08:45:14.505 ifonter[43240:1784538] VALUE <UIBezierPath: 0x7fd5c8483de0; <MoveTo {232.5, 255.5}>,
<CurveTo {233.58333337306976, 277.58333325386047} {232.5, 263.5} {232.5, 271.5}>,
<CurveTo {239.33333337306976, 290.91666650772095} {234.66666674613953, 283.66666650772095} {236.83333349227905, 287.83333301544189}>,
<CurveTo {247.5, 298} {241.83333325386047, 294} {244.66666650772095, 296}>

我正在努力清理（见下面的代码）

NSError *error = nil;
NSRegularExpression * regex = [NSRegularExpression regularExpressionWithPattern:@"\\<UIBezierPath:(.*?)\\;" options:NSRegularExpressionCaseInsensitive error:&error];
NSString *modifiedString = [regex stringByReplacingMatchesInString: self.fl.strPaths options:0 range: NSMakeRange(0, [self.fl.strPaths length]) withTemplate:@""];

NSString *strMovTo = [modifiedString stringByReplacingOccurrencesOfString:@"<MoveTo {" withString:@"M"];
strMovTo = [strMovTo stringByReplacingOccurrencesOfString:@"}>," withString:@"-"];

strMovTo = [strMovTo stringByReplacingOccurrencesOfString:@"<CurveTo {" withString:@""];
strMovTo = [strMovTo stringByReplacingOccurrencesOfString:@"} {" withString:@"-"];
strMovTo = [strMovTo stringByReplacingOccurrencesOfString:@"}>" withString:@""];
strMovTo = [strMovTo stringByReplacingOccurrencesOfString:@"\n" withString:@""];


NSArray *elements = [strMovTo componentsSeparatedByString:@"-"];

Answer 1

确保它是NSMutableArray。

然后您可以使用exchangeObjectAtIndex:withObjectAtIndex:来交换值。

Answer 2

假设您的数组名称为myArray

NSArray *myArray = @[@" 198.16666666418314, 199.75",@"203.5, 198.5", and so on... ];


NSMutableArray *newArray = [NSMutableArray arrayWithArray:myArray];

for(int i=0;i<[newArray count];i=i+3)
{
  [newArray exchangeObjectAtIndex:i withObjectAtIndex:i+2];
}

交换NSArray目标C的值

2 个答案: