消除不需要的字符从每一个字符串数组中迅速
问题描述:
我叫coordinateArray是保存这些数据的磁盘阵列:消除不需要的字符从每一个字符串数组中迅速
["(19.452187074041884", " -99.1457748413086)", "(19.443769985032485", " -99.14852142333984)", "(19.443446242121073", " -99.13787841796875)", "(19.450244707639662", " -99.13822174072266)"]
["(19.407780723677718", " -99.18417591514299)", "(19.400373302640162", " -99.18473381461808)", "(19.400049473260434", " -99.18039936485002)", "(19.405433052977592", " -99.17838234367082)"]
["(19.4022042123319", " -99.1457748413086)", "(19.401070819438004", " -99.16139602661133)", "(19.39184146912981", " -99.16268348693848)", "(19.389736456288546", " -99.14706230163574)"]
["(19.42114689571205", " -99.17375564575195)", "(19.425598915444077", " -99.15392875671387)", "(19.414913863184566", " -99.1547)", "(19.41264724664359", " -99.16594505310059)", "(19.412890099927345", " -99.16766166687012)"]
我需要删除从每个括号协调
我该怎么办这在Swift 3.0中?
我试图做到这一点
let coordinateArray = coor.components(separatedBy: ",")
var coordinateArrayF = [String]()
for coordinate in coordinateArray {
let coordinatevar = coordinate.replacingOccurrences(of: "()", with: "")
coordinateArrayF.append(coordinatevar)
}
但它不工作我究竟做错了什么?
答
函数式编程是你的朋友!
var data = ["(19.452187074041884", " -99.1457748413086)", "(19.443769985032485", " -99.14852142333984)", "(19.443446242121073", " -99.13787841796875)", "(19.450244707639662", " -99.13822174072266)"]
let cleanData = data.map {
item in item.replacingOccurrences(of: "(", with: "")
}
+0
感谢我做了一件非常相似,但有一个for循环,这是非常好的了解如何使用功能的编程来做到这一点。什么阻止它的工作是,你需要创建在replacingOcurrences定期ocurrence(作者:“”与“”)方法。我会在我如何解决它的问题上发布答案。 –
更好地修复*生成数组的代码,并使其成为元组或CLLocationCoordinate2D或类似的数组。 –
这是我从一开始就想做的事,但我不知道该怎么做!什么会,我需要做的,使CLLocationCoordinate2D数组: 我使用的组件(“” separatedBy)坐标的巨大串产生的阵列?由于 什么我收到 –