这是一些单元测试

Question

我知道我可以想出办法，但我想知道是否有更简洁的解决方案。始终有String.Join(", ", lList)和lList.Aggregate((a, b) => a + ", " + b);，但我想为最后一个添加例外，以", and "作为其加入字符串。 Aggregate()在我可以使用的地方有一些索引值吗？感谢。

Answer 1

你可以这样做

string finalString = String.Join(", ", myList.ToArray(), 0, myList.Count - 1) + ", and " + myList.LastOrDefault();

Answer 2

这是一个解决方案，它使用空列表和列表，其中包含一个项目：

C＃

return list.Count() > 1 ? string.Join(", ", list.Take(list.Count() - 1)) + " and " + list.Last() : list.FirstOrDefault();

VB

Return If(list.Count() > 1, String.Join(", ", list.Take(list.Count() - 1)) + " and " + list.Last(), list.FirstOrDefault())

Answer 3

我使用以下扩展方法（也有一些代码保护）：

public static string OxbridgeAnd(this IEnumerable<String> collection)
{
    var output = String.Empty;

    var list = collection.ToList();

    if (list.Count > 1)
    {
        var delimited = String.Join(", ", list.Take(list.Count - 1));

        output = String.Concat(delimited, ", and ", list.LastOrDefault());
    }

    return output;
}

这是一个单元测试：

 [TestClass]
    public class GrammarTest
    {
        [TestMethod]
        public void TestThatResultContainsAnAnd()
        {
            var test = new List<String> { "Manchester", "Chester", "Bolton" };

            var oxbridgeAnd = test.OxbridgeAnd();

            Assert.IsTrue( oxbridgeAnd.Contains(", and"));
        }
    }

<强> 修改

此代码现在处理null和单个元素：

  public static string OxbridgeAnd(this IEnumerable<string> collection)
    {
        var output = string.Empty;

        if (collection == null) return null;

        var list = collection.ToList();

        if (!list.Any()) return output;

        if (list.Count == 1) return list.First();

        var delimited = string.Join(", ", list.Take(list.Count - 1));

        output = string.Concat(delimited, ", and ", list.LastOrDefault());

        return output;
    }

Answer 4

此版本枚举一次值并使用任意数量的值：

import csv
import pandas as pd

with open('2column.csv','rb') as f:
    reader = csv.reader(f)
    a = list(reader)
gene = []
ratio = []
for t in range(len(a)):
    if '///' in a[t][0]:
        s = a[t][0].split('///')
        gene.append(s[0])
        gene.append(s[1])
        ratio.append(a[t][1])
        ratio.append(a[t][1])
    else:
        gene.append(a[t][0])
        ratio.append(a[t][1])
    gene[t] = gene[t].strip()

newgene = []
newratio = []
for i in range(len(gene)):
    g = gene[i]
    r = ratio[i]
    if g not in newgene:
        newgene.append(g)
    for j in range(i+1,len(gene)):
        if g==gene[j]:
            if ratio[j]>r:
                r = ratio[j]
    newratio.append(r)

for i in range(len(newgene)):
    print newgene[i] + '\t' + newratio[i]

if len(newgene) > len(set(newgene)):
    print 'missionfailed'

Answer 5

此版本仅枚举一次值，并且可以使用任意数量的值。

（改进了@Grastveit answer）

我将其转换为扩展方法并添加了一些单元测试。添加了一些空检查。此外，如果values集合中的某个项目包含null，并且它是最后一个项目，我会修改一个错误，它会被跳过。这与String.Join()在.NET Framework中的行为方式不符。

#region Usings
using System;
using System.Collections.Generic;
using System.Text;
#endregion 

namespace MyHelpers
{
    public static class StringJoinExtensions
    {
        public static string JoinAnd<T>(this IEnumerable<T> values, 
             string separator, string lastSeparator = null)
        {
            if (values == null)
                throw new ArgumentNullException(nameof(values));
            if (separator == null)
                throw new ArgumentNullException(nameof(separator));

            var sb = new StringBuilder();
            var enumerator = values.GetEnumerator();

            if (enumerator.MoveNext())
                sb.Append(enumerator.Current);

            bool objectIsSet = false;
            object obj = null;
            if (enumerator.MoveNext())
            {
                obj = enumerator.Current;
                objectIsSet = true;
            }

            while (enumerator.MoveNext())
            {
                sb.Append(separator);
                sb.Append(obj);
                obj = enumerator.Current;
                objectIsSet = true;
            }

            if (objectIsSet)
            {
                sb.Append(lastSeparator ?? separator);
                sb.Append(obj);
            }

            return sb.ToString();
        }
    }
}

这是一些单元测试

#region Usings
using MyHelpers;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System;
using System.Linq;
#endregion

namespace UnitTests
{
    [TestClass]
    public class StringJoinExtensionsFixture
    {
        [DataTestMethod]
        [DataRow("", "", null, null)]
        [DataRow("1", "1", null, null)]
        [DataRow("1 and 2", "1", "2", null)]
        [DataRow("1, 2 and 3", "1", "2", "3")]
        [DataRow(", 2 and 3", "", "2", "3")]
        public void ReturnsCorrectResults(string expectedResult, 
             string string1, string string2, string string3)
        {
            var input = new[] { string1, string2, string3 }
                .Where(r => r != null);
            string actualResult = input.JoinAnd(", ", " and ");
            Assert.AreEqual(expectedResult, actualResult);
        }

        [TestMethod]
        public void ThrowsIfArgumentNulls()
        {
            string[] values = default;
            Assert.ThrowsException<ArgumentNullException>(() =>
                 StringJoinExtensions.JoinAnd(values, ", ", " and "));

            Assert.ThrowsException<ArgumentNullException>(() =>
               StringJoinExtensions.JoinAnd(new[] { "1", "2" }, null, 
                  " and "));
        }

        [TestMethod]
        public void LastSeparatorCanBeNull()
        {
            Assert.AreEqual("1, 2", new[] { "1", "2" }
               .JoinAnd(", ", null), 
                   "lastSeparator is set to null explicitly");
            Assert.AreEqual("1, 2", new[] { "1", "2" }
               .JoinAnd(", "), 
                   "lastSeparator argument is not specified");
        }

        [TestMethod]
        public void SeparatorsCanBeEmpty()
        {
            Assert.AreEqual("1,2", StringJoinExtensions.JoinAnd(
                new[] { "1", "2" }, "", ","), "separator is empty");
            Assert.AreEqual("12", StringJoinExtensions.JoinAnd(
                 new[] { "1", "2" }, ",", ""), "last separator is empty");
            Assert.AreEqual("12", StringJoinExtensions.JoinAnd(
                 new[] { "1", "2" }, "", ""), "both separators are empty");
        }

        [TestMethod]
        public void ValuesCanBeNullOrEmpty()
        {
            Assert.AreEqual("-2", StringJoinExtensions.JoinAnd(
               new[] { "", "2" }, "+", "-"), "1st value is empty");
            Assert.AreEqual("1-", StringJoinExtensions.JoinAnd(
                 new[] { "1", "" }, "+", "-"), "2nd value is empty");
            Assert.AreEqual("1+2-", StringJoinExtensions.JoinAnd(
                new[] { "1", "2", "" }, "+", "-"), "3rd value is empty");

            Assert.AreEqual("-2", StringJoinExtensions.JoinAnd(
             new[] { null, "2" }, "+", "-"), "1st value is null");
            Assert.AreEqual("1-", StringJoinExtensions.JoinAnd(
             new[] { "1", null }, "+", "-"), "2nd value is null");
            Assert.AreEqual("1+2-", StringJoinExtensions.JoinAnd(
             new[] { "1", "2", null }, "+", "-"), "3rd value is null");
        }
    }
}

Answer 6

我能想到的最简单的方法是这样... print（'，'。join（a [0：-1]）+'和'+ a [-1]）

a = [a, b, c, d]
print(', '.join(a[0:-1]) + ', and ' + a[-1])

a，b，c和d

或者，如果您不喜欢加拿大语法，牛津逗号和多余的花体字

print(', '.join(a[0:-1]) + ' and ' + a[-1])

a，b，c和d

保持简单。

加入List <string>与Commas Plus“和”for Last Element </string>

6 个答案:

C＃

VB

这是一些单元测试