如何在foreach嵌套循环C#中使用列表将字符串的行转换为列 [英] How to transpose rows of string into columns using lists in a foreach nested loop C#
问题描述
我有一个结构如下的字符串:
I have a string structured as following:
RLLR
LRRL
RVVL
RRRR
// string was made like:
string s = "RLLR" + "\n" + "LRRL" + "\n" + "RVVL" + "\n" + "RRRR";
我想对这个表做的是将行转为列，所以它看起来像:
What I want to do to this table is transpose the rows to columns, so it will look like:
RLRR
LRVR
LRVR
RLLR
到目前为止我所做的是将字符串转换为数组，这样我就可以像这样循环遍历它:
What I've done so far is that I converted the string to an array so i can loop through it like so:
List<string> line_value = new List<string>();//list for one line of array
List<string> single_value = new List<string>();//list for individual characters for those lines
string s = "RLLR" + "\n" + "LRRL" + "\n" + "RVVL" + "\n" + "RRRR";
string[] strarray = new string[]{""};
strarray = s.Split("\n");
int z = 0;
foreach(string line in strarray)
{
line_value.Add(line);//adds single line to list
foreach(char letter in line_value[z])
{
Console.WriteLine(letter.ToString());
single_value.Add(letter.ToString());
}
z++;
}
通过这样做，我可以像这样打印出字符串，其中一切都是水平的:
By doing it this way, I can print out the string like so, where everything is horizontal:
R
L
L
R
L
R
R
.
.
.
R
但是，我仍然对如何建立字符串感到困惑，因此它会像这样转置:
However, I am still kind of confused on how to establish the string so it will be transposed like so:
RLRR
LRVR
LRVR
RLLR
我将如何转置字符串以便将行变成列?
How would I transpose the string so it would turn the rows into columns?
推荐答案
这样做的简单方法是不使用 foreach 循环，并滥用 for 循环这一事实，即您可以简单地交换列和行索引.
The easy way of doing this is to do it without a foreach loop, and use a for loop abusing the fact that you can simply swap the column and row indexes.
using System.Text;
static string TransposeRowsToColumns(string rowString)
{
string[] rows = rowString.Split("\n");
StringBuilder columnBuilder = new StringBuilder();
for (int columnIndex = 0; columnIndex < rows[0].Length; columnIndex++)
{
for (int rowIndex = 0; rowIndex < rows.Length; rowIndex++)
{
columnBuilder.Append(rows[rowIndex][columnIndex]);
}
columnBuilder.Append("\n");
}
return columnBuilder.ToString();
}
注意上面的代码依赖于列数是统一的.
如果您想使用带有列表的 foreach 循环来执行此操作，您可以这样做:
If you're wanting to do this with a foreach loop with lists, you can do it like:
static string TransposeRowsToColumnsList(string rowString)
{
string[] rows = rowString.Split("\n");
List<List<string>> grid = new List<List<string>>();
int columnIndex = 0;
foreach (string row in rows)
{
grid.Add(new List<string>());
foreach (string column in rows.Select(r => string.Concat(r.Skip(columnIndex).Take(1))))
{
grid[columnIndex].Add(column);
}
columnIndex++;
}
return string.Join("\n", grid.Select(r => string.Concat(r.Select(c => c))));
}
用法:
string s = "RLLR" + "\n" + "LRRL" + "\n" + "RVVL" + "\n" + "RRRR";
Console.WriteLine(TransposeRowsToColumns(s));
Console.WriteLine(TransposeRowsToColumnsList(s));
编辑
为了将输入更改为基本上按空格拆分列而不是假设它们是单个字符，我们可以将第二种方法更改为:
For changing the input to essentially split the columns by space instead of the assumption that they're a single character, we can alter the second method to be like:
static string TransposeRowsToColumnsList(string inputString, string columnSplitBy = "", string rowSplitBy = "\n")
{
IEnumerable<IEnumerable<string>> inputGrid = inputString.Split(rowSplitBy).Select(r =>
{
return columnSplitBy == "" ? r.Select(c => new string(c, 1)).ToArray() : r.Split(columnSplitBy);
});
List<List<string>> outputGrid = new List<List<string>>();
int columnIndex = 0;
foreach (IEnumerable<string> row in inputGrid)
{
outputGrid.Add(new List<string>());
foreach (string column in inputGrid.Select(r => string.Concat(r.Skip(columnIndex).Take(1))))
{
outputGrid[columnIndex].Add(column);
}
columnIndex++;
}
return string.Join(rowSplitBy, outputGrid.Select(r => string.Concat(string.Join(columnSplitBy, r.Select(c => c)))));
}
虽然这很快就会变得混乱.对于更具扩展性的解决方案，我们可以创建扩展方法来分离算法的每个阶段并吐出所需的结果.
Although this gets messy really quick. For a more scalable solution, we can create extension methods to separate each stage of the algorithm and spit out the desired result.
我们首先定义一个可以将字符串转换为所需类型的接口，并带有转换小数的实现:
We first define an interface that can convert a string to a desired type with an implementation of converting decimals:
public interface IStringConverter<T>
{
T ConvertFromString(string input);
}
public class DecimalConverter : IStringConverter<decimal>
{
public decimal ConvertFromString(string input)
{
return decimal.Parse(input);
}
}
接下来，我们可以定义将网格转置为我们想要的方式所需的所有扩展方法:
Next we can define all the extension methods we'll need to transpose the grid to the way we want:
public static class CustomExtensions
{
public static IEnumerable<string> ForceSplit(this string input, string pattern)
{
return pattern != string.Empty ? input.Split(pattern) : input.Select(x => x.ToString());
}
public static IEnumerable<IEnumerable<string>> ConvertToGrid(this string input, string columnSplit = "", string rowSplit = "\n")
{
return input.Split(rowSplit).Select(r => r.ForceSplit(columnSplit));
}
public static IEnumerable<IEnumerable<T>> ConvertToGrid<T>(this string input, IStringConverter<T> converter, string columnSplit = "", string rowSplit = "\n")
{
return input.Split(rowSplit).Select(r => r.ForceSplit(columnSplit).Select(converter.ConvertFromString));
}
public static IEnumerable<IEnumerable<T>> PivotGrid<T>(this IEnumerable<IEnumerable<T>> input)
{
return input
.SelectMany(r => r.Select((c, index) => new {column = c, index}))
.GroupBy(i => i.index, i => i.column)
.Select(g => g.ToList());
}
public static string ConvertToString<T>(this IEnumerable<IEnumerable<T>> input, string columnSplit = "", string rowSplit = "\n")
{
return string.Join(rowSplit, input.Select(r => string.Join(columnSplit, r)));
}
}
注意事项:
- 我们现在正在通过
ConvertToGrid 将每个元素转换为所需类型的单元格 - 我们能够将网格从行旋转到列(感谢 这个答案)
- 如果需要，我们可以将网格转换回字符串格式
使用
string letters = "RLLR" + "\n" + "LRRL" + "\n" + "RVVL" + "\n" + "RRRR";
string numbers = "25.0 45.7 23" + "\n" + "12.4 67.4 0.0" + "\n" + "0.00 0.00 0.00" + "\n" + "67.8 98.4 0.00";
string transposedLetters = TransposeRowsToColumnsList(letters);
string transposedNumbers = TransposeRowsToColumnsList(numbers, " ");
string pivotedLetters = letters
.ConvertToGrid()
.PivotGrid()
.ConvertToString();
string pivotedNumbers = numbers
.ConvertToGrid(new DecimalConverter(), " ")
.PivotGrid()
.ConvertToString(" ");
我个人认为扩展方法的方法更易于维护和扩展，但原始方法更易于调用.
I personally find the extension method approach more maintainable and more extensible, but the original method is easier to call.
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