Java 如何用不同的子字符串替换多个子字符串?
我从文本文件中获得了这行和弦。比如说,Java 如何用不同的子字符串替换多个子字符串?,java,string,replace,Java,String,Replace,我从文本文件中获得了这行和弦。比如说, String chordLine = "C G Am C"; String transposedChordLine; 接下来,我需要使用下面的类,使用两个参数将弦线转换为一个新的转换弦线,一个字符串和弦和转置的整数增量。例如,转置(“C”,2)将返回D public class Transposer{ private int inc; private static ArrayList<String> keysS
String chordLine = "C G Am C";
String transposedChordLine;
接下来,我需要使用下面的类,使用两个参数将弦线
转换为一个新的转换弦线
,一个字符串
和弦和转置的整数增量。例如,转置(“C”,2)
将返回D
public class Transposer{
private int inc;
private static ArrayList<String> keysSharp;
private static ArrayList<String> keysFlat;
Transposer(){
keysSharp = new ArrayList<String>(Arrays.asList("C", "C#", "D", "D#","E", "F","F#", "G","G#", "A","A#", "B"));
keysFlat = new ArrayList<String>(Arrays.asList("C", "Db", "D", "Eb","E", "F","Gb", "G","Ab", "A","Bb", "B"));
}
public String transpose(String chord,int inc){
this.inc = inc;
String newChord;
if(chord.contains("/")){
String[] split = chord.split("/");
newChord = transposeKey(split[0]) + "/" + transposeKey(split[1]);
}else
newChord = transposeKey(chord);
return newChord;
}
private String transposeKey(String key){ // C#m/D# must pass C#m or D#
String nKey, tempKey;
if(key.length()>1){
nKey = key.substring(0, 2);
}
else{ nKey = key; }
int oldIndex, newIndex;
if(key.contains("b")){
oldIndex = (keysFlat.indexOf(nKey)>-1) ? keysFlat.indexOf(nKey) : keysFlat.indexOf(similarKey(nKey));
newIndex = (oldIndex + inc + keysFlat.size())%keysFlat.size();
tempKey = keysFlat.get(newIndex);
nKey = (key.length() < 3) ? tempKey : key.replace(nKey, tempKey);
//(nKey + key.substring(nKey.length(), key.length()));
}
else if(key.contains("#")){
oldIndex = (keysSharp.indexOf(nKey)>-1) ? keysSharp.indexOf(nKey) : keysSharp.indexOf(similarKey(nKey));
newIndex = (oldIndex + inc + keysSharp.size())%keysSharp.size();
tempKey = keysSharp.get(newIndex);
nKey = (key.length() < 3) ? tempKey : key.replace(nKey, tempKey);
}
else{
nKey = nKey.substring(0, 1);
oldIndex = (keysSharp.indexOf(nKey)>-1) ? keysSharp.indexOf(nKey) : keysSharp.indexOf(similarKey(nKey));
newIndex = (oldIndex + inc + keysSharp.size())%keysSharp.size();
tempKey = keysSharp.get(newIndex);
nKey = (key.length() < 2) ? tempKey : key.replace(nKey, tempKey);
}
return nKey;
}
private String similarKey(String nKey) {
String newKey;
switch(nKey){
case "Cb":
newKey = "B";
break;
case "Fb":
newKey = "E";
break;
case "E#":
newKey = "F";
break;
case "B#":
newKey = "c";
break;
default:
newKey = null;
}
return newKey;
}
}
您的标记化非常不寻常(保留分隔符),因此您可能希望自己进行标记化。基本上,如果您看到与音符匹配的标记,请将其传递给转发器。否则,沿空间传递。使用while循环沿注释导航。下面是实现这一点的代码:
private static final Transposer transposer = new Transposer();
public static void main(String[] args) {
String chordLine = "C G Am C";
String transposed = transposeChordLine(chordLine);
System.out.println(transposed);
}
private static String transposeChordLine(String chordLine) {
char[] chordLineArray = chordLine.toCharArray();
StringBuilder transposed = new StringBuilder();
StringBuilder currentToken = new StringBuilder();
int index = 0;
while(index < chordLine.length()) {
if(chordLineArray[index] == ' ') {
transposed.append(' ');
currentToken = processToken(transposed, currentToken);
} else {
currentToken.append(chordLineArray[index]);
}
index++;
}
processToken(transposed, currentToken);
return transposed.toString();
}
private static StringBuilder processToken(StringBuilder transposed,
StringBuilder currentToken) {
if(currentToken.length() > 0) {
String currentChord = currentToken.toString();
String transposedChord = transposer.transpose(currentChord, 2);
transposed.append(transposedChord);
currentToken = new StringBuilder();
}
return currentToken;
}
以下是较短的解决方案(我将此方法添加到Transposer
类中):
我正在使用正则表达式匹配器创建新字符串。正则表达式匹配和弦名称以及其后的所有空格。为了确保替换具有相同的长度,我使用String.format
并提供类似%-XXs
的格式字符串,其中XX
是带空格的未转置和弦的长度。请注意,如果没有足够的空间,则生成的线将变长
用法:
public static void main(String[] args) {
String chordLine = "C G Am C";
System.out.println(chordLine);
for(int i=0; i<12; i++) {
String result = new Transposer().transposeLine(chordLine, i);
System.out.println(result);
}
}
给定弦线、转置器和转置增量:
String chordLine = "C G Am C";
Transposer tran = new Transposer();
int offset = 2;
要在保留空白的同时获得转置的和弦线,可以使用在空白边界处分割,然后通过转置器有条件地处理结果字符串,如下所示:
String transposed = Arrays.stream(chordLine.split("((?<=\\s)|(?=\\s))")).map( // use regex to split on every whitespace boundary
str -> // for each string in the split
Character.isWhitespace(str.charAt(0)) // if the string is whitespace
? str // then keep the whitespace
: tran.transpose(str, offset) // otherwise, it's a chord, so transpose it
).collect(Collectors.joining()); // re-join all the strings together
很多潜在的解决方案。例如:创建一个对象,
弦线
,它将引用一条弦线。在该对象的构造函数中,可以显式跟踪和弦n_i
和n_i+1
之间的空白量。然后,在换位时,您可以简单地替换输出弦线中的空白
对不起,这与您的问题无关,但您的代码非常糟糕。不要初始化构造函数中的静态成员。将两个static更改为List
,并直接使用Arrays.asList
初始化它们。另外,不要使用实例字段将参数传递给私有方法。如果调用transpose
时inc
可以不同,则将其作为参数传递给私有方法。如果inc
总是相同的,那么将其传递给构造函数,而不是transpose
方法。谢谢你的评论。
public static void main(String[] args) {
String chordLine = "C G Am C";
System.out.println(chordLine);
for(int i=0; i<12; i++) {
String result = new Transposer().transposeLine(chordLine, i);
System.out.println(result);
}
}
C G Am C
C G Am C
C# G# A#m C#
D A Bm D
D# A# Cm D#
E B C#m E
F C Dm F
F# C# D#m F#
G D Em G
G# D# Fm G#
A E F#m A
A# F Gm A#
B F# G#m B
String chordLine = "C G Am C";
Transposer tran = new Transposer();
int offset = 2;
String transposed = Arrays.stream(chordLine.split("((?<=\\s)|(?=\\s))")).map( // use regex to split on every whitespace boundary
str -> // for each string in the split
Character.isWhitespace(str.charAt(0)) // if the string is whitespace
? str // then keep the whitespace
: tran.transpose(str, offset) // otherwise, it's a chord, so transpose it
).collect(Collectors.joining()); // re-join all the strings together
StringBuilder sb = new StringBuilder();
for (String str : chordLine.split("((?<=\\s)|(?=\\s))")) {
sb.append(Character.isWhitespace(str.charAt(0)) ? str : tran.transpose(str, offset));
}
String transposed = sb.toString();