Question

我有两个列表，我想以元素方式连接它们。其中一个列表在连接之前经过字符串格式化。

例如：

a = [0, 1, 5, 6, 10, 11] 
b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']

在这种情况下，a会受到字符串格式化。也就是说，新的a或aa应为：

aa = [00, 01, 05, 06, 10, 11]

最终输出应为：

c = ['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']

有人可以告诉我该怎么做？

Answer 1

使用zip：

>>> ["{}{:02}".format(b_, a_) for a_, b_ in zip(a, b)]
['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']

Answer 2

使用zip

[m+str(n) for m,n in zip(b,a)]

输出

['asp10', 'asp11', 'asp15', 'asp16', 'asp210', 'asp211']

Answer 3

其他解决方案（更喜欢 printf formating 样式而不是.format()用法），它也更小：

>>> ["%s%02d" % t for t in zip(b, a)]
['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']

Answer 4

可以用地图和邮编优雅地完成：

map(lambda (x,y): x+y, zip(list1, list2))

示例：

In [1]: map(lambda (x,y): x+y, zip([1,2,3,4],[4,5,6,7]))
Out[1]: [5, 7, 9, 11]

Answer 5

不使用zip。我不知道，我认为这是明显的方法。也许我刚开始学习C：）

c=[]
for i in xrange(len(a)):
    c.append("%s%02d" % (b[i],a[i]))

Answer 6

b = ['asp1', 'asp1', 'asp1', 'asp1', 'asp2', 'asp2']
aa = [0, 1, 5, 6, 10, 11]
new_list =[]
if len(aa) != len(b):
     print 'list length mismatch'
else:
    for each in range(0,len(aa)):
        new_list.append(b[each] + str(aa[each]))
print new_list

Answer 7

输入：

[Serializable]
public partial class SongItem : UserControl,Form
{
    private String songName = "Song Name";
    private String artistName = "Artist Name";
    private Image thumbNail;
    private String length;
    private int maxLengthSongName = 25;
    private int maxLengthArtistName = 25;
    private Color colorHoverOn = Color.FromArgb(53,53,53);
    private Color colorNormal = Color.FromArgb(53,53,53);
    private SongData songData;
    public SongItem()
    {
        InitializeComponent();
        this.MouseClick += Control_MouseClick;
        MouseEvents(this);
    }
    private void SongItem_Load(object sender, EventArgs e)
    {
        try
        {
            LoadData();
        }
        catch { }
        LoadDataToUI();

    }

    #region GettersAndSetters

    public int MaxLengthSongName
    {
        get { return maxLengthSongName; }
        set { maxLengthSongName = value; }
    }
    public int MaxLengthArtistName
    {
        get { return maxLengthArtistName; }
        set { maxLengthArtistName = value; }
    }

    public Color ColorHoverOn
    {
        get { return colorHoverOn; }
        set { colorHoverOn = value; }
    }
    public Color ColorNormal
    {
        get { return colorNormal; }
        set { colorNormal = value; }
    }

    public SongData SongData {
        get{return songData; }
        set { songData = value; } 
    }
    #endregion

    public void LoadData()
    {
        songName = songData.SongName;
        artistName = songData.ArtistName;
        thumbNail = songData.ThumbNail;
        length = songData.Length;
    }

    void MouseEvents(Control container)
    {
        foreach (Control c in container.Controls)
        {
            c.MouseEnter += (s, e) => SongItem_MouseEnter(e);
            c.MouseLeave += (s, e) => SongItem_MouseLeave(e);
            c.MouseClick += Control_MouseClick;
            MouseEvents(c);
        };
    }


    private void SongNameLbl_MouseHover(object sender, EventArgs e)
    {
        if (songName.Length > maxLengthSongName) {
            toolTip1.SetToolTip(songNameLbl, songName);
        }
    }

    private void ArtistNameLbl_MouseHover(object sender, EventArgs e)
    {
        if (artistName.Length > maxLengthArtistName) {
            toolTip1.SetToolTip(artistNameLbl, artistName);
        }
    }

    private void SongNameLbl_Click(object sender, EventArgs e)
    {
        Clipboard.SetText(artistName + " " +songName);
    }

    #endregion

    #region CurrentlySelected
    public event EventHandler<EventArgs> WasClicked;

    private void Control_MouseClick(object sender, MouseEventArgs e)
    {
        var wasClicked = WasClicked;
        if (wasClicked != null)
        {
            WasClicked(this, EventArgs.Empty);
        }
        IsSelected = true;
    }

    private bool _isSelected;
    public bool IsSelected
    {
        get { return _isSelected; }
        set
        {
            _isSelected = value;
            this.BorderStyle = IsSelected ? BorderStyle.FixedSingle : BorderStyle.None;
        }
    }
    #endregion


    private void Fovourite_Click(object sender, EventArgs e)
    {
        Main newMain = new Main();
       // newMain.AddSongToFavorite();
        newMain.listBox1.Items.Add("Test");
    }
}

输出：

public void AddSongToFavorite() {
        listBox1.Items.Add("Test");
        //songList2.AddSong("Dire Straits - Sultans Of Swing");
        MessageBox.Show("Hello", "Test");
       // flowLayoutPanel1.Controls.Add(song);
    }

Answer 8

如果要串联任意数量的列表，可以执行以下操作：

In [1]: lists = [["a", "b", "c"], ["m", "n", "o"], ["p", "q", "r"]] # Or more

In [2]: lists
Out[2]: [['a', 'b', 'c'], ['m', 'n', 'o'], ['p', 'q', 'r']]    

In [4]: list(map("".join, zip(*lists)))
Out[4]: ['amp', 'bnq', 'cor']

Answer 9

我最终使用了一个临时的 DataFrame，它具有可读性和快速性：

a = ["a", "b", "c"]
b = ["1", "2", "3"]

df = pd.DataFrame({"a": a, "b": b})
df["c"] = df.a + df.b
result = df.c.values

输出：

$ result 
["a1", "b2", "c3"]

在幕后，DataFrames 使用 numpy，因此结果是高效的。

和函数一样：

import pandas as pd
from typing import List
def _elementwise_concat(self, a: List[str], b: List[str]) -> List[str]:
    """
    Elementwise concatenate.
    :param a: List of strings.
    :param b: List of strings.
    :return: List, same length, strings concatenated.
    """
    df = pd.DataFrame({"a": a, "b": b})
    df["c"] = df.a + df.b
    return df.c.values

Answer 10

列表理解/zip()/使用zfill()来格式化。

print ([y+str(x).zfill(2) for x,y in zip(a,b)])

输出：

['asp100', 'asp101', 'asp105', 'asp106', 'asp210', 'asp211']

Answer 11

使用 Result Price ID 1 left 100 2 right -10 3 left 60 和 #Full Table with Original Values df_pivot[result.columns] = result print(df_pivot)：

Side  left  right Result  Price
ID                             
1      100      0   left    100
2       80     90  right    -10
3      110     50   left     60

出：

lambda

如何在Python中连接元素两个列表？

11 个答案: