Question

我试图在Python 2.7中以这种方式加入两个列表：

a = ['x','y','z']
b = [1,2,3]

，最终结果应为：

c=['x1','y2','z3']

我该怎么做？

Answer 1

c = [p + str(q) for p, q in zip(a, b)]

Answer 2

使用string.format函数更好，而不是连接字符串。您还可以将itertools.starmap与列表的zip版本一起使用：

if(isSearching == true){
    let contactDict :NSDictionary = self.filteredArray?.object(at: indexPath.row) as! NSDictionary;
    let strArray :NSArray = contactDict.object(forKey: kName) as! NSArray
    nameString = strArray.componentsJoined(by: "") as! NSMutableString

    //nameString = (contactDict.object(forKey: kName) as? String as! NSMutableString)
    companyNameString = (contactDict.object(forKey: kCompanyName) as AnyObject).object(at: 0) as? NSString;
    designationString = (contactDict.object(forKey: kDesignation) as AnyObject).object(at: 0) as? NSString;
    profileImage = contactDict.object(forKey: kProfilePic) as? UIImage;
    connectStatus = contactDict.value(forKey: kLinkStatus) as? NSString;

    if(profileImage?.accessibilityIdentifier == "Img_placeholder"){
        profileImage = nil;
    }

或者您可以将它与传说中的列表推导一起用作：

>>> from itertools import starmap
>>> a = ['x','y','z']
>>> b = [1,2,3]

>>> list(starmap("{}{}".format, zip(a, b)))
['x1', 'y2', 'z3']

# Note: `starmap` returns an iterator. If you want to iterate this value
# only once, then there is no need to type-case it to `list`

使用>>> ['{}{}'.format(x, y) for x, y in zip(a, b)] ['x1', 'y2', 'z3']，您无需明确键入format到int。此外，更改列表中所需字符串的格式也更简单。例如：

str

以下是格式>>> ['{} -- {}'.format(x, y) for x, y in zip(a, b)] ['x -- 1', 'y -- 2', 'z -- 3']列表的通用解决方案：

n

Answer 3

您也可以使用map函数根据@ SilverSlash的解决方案尝试这个：

a = ['x','y','z']
b = [1,2,3]

c = list(map(''.join, zip(a, map(str, b))))
print(c)

输出：

['x1', 'y2', 'z3']

基于两个列表的元素的格式化字符串列表

3 个答案: