Question

我有一个具有这样布局的多维列表

> set.seed(1)

> res <- testOutMat(ncols=5,nrows=5,col_prob = runif(20, 0.1, 0.2))
> res
     [,1] [,2]
[1,]    0    0
[2,]    3    1

我需要删除每个项目中第一个索引中名称前后的空格。

我已经尝试过：

[['Company Name      ', ['companyemail@email.com']]
['Company Name   ', ['companyemail@email.com','companyemail@email.com']]
['Company Name', ['companyemail@email.com']]
['Company Name          ', ['companyemail@email.com']]]

但是，当我打印列表时，除了一个字母之外，它什么都不显示

我的预期结果是：

def name_filter(vendors):
    unfiltered=vendors
    filtered=[]
    count=0
    for i in unfiltered:
        filtered = [x.strip(' ') for x in unfiltered[count][0]]
        count+=1
    return filtered

Answer 1

我将递归执行此操作-如果每个元素是字符串，则剥离每个元素；如果它是列表，则重复：

def recursive_strip(lst):
    return [(elem.strip() if type(elem) == str else recursive_strip(elem)) for elem in lst]

您可能需要考虑列表可以包含哪些数据类型（以上仅假设字符串或列表）。

Answer 2

问题是您在每个列表的第一项上使用了for循环，然后去除了空格。结果，您将获得单个字符。您不需要列表理解内的第二个for循环。只需剥离第一个字符串元素并同时附加第二个。附注：我也将嵌套列表vendors定义为输入，因为我认为这是基于您的代码逻辑的正确输入结构

vendors = [['Company Name      ', ['companyemail@email.com']],
           ['Company Name   ', 
           ['companyemail@email.com','companyemail@email.com']],
           ['Company Name', ['companyemail@email.com']],
           ['Company Name          ', ['companyemail@email.com']]]

def name_filter(vendors):
    unfiltered=vendors
    filtered=[]
    for i in unfiltered:
        print (i)
        filtered.append([i[0].strip(' '), i[1]]) # <--- This is the key line

    return filtered

name_filter(vendors)

# [['Company Name', ['companyemail@email.com']],
#  ['Company Name', ['companyemail@email.com', 'companyemail@email.com']],
#  ['Company Name', ['companyemail@email.com']],
#  ['Company Name', ['companyemail@email.com']]]

Answer 3

您可以使用str.rstrip()

vendors = \
    [['Company Name      ', ['companyemail@email.com']],
    ['Company Name   ', ['companyemail@email.com','companyemail@email.com']],
    ['Company Name', ['companyemail@email.com']],
    ['Company Name          ', ['companyemail@email.com']]]


filtered_vendors = [[vendor[0].rstrip(),vendor[1]] for vendor in vendors]

Answer 4

您可以在列表理解内完成

vendors = [['Company Name      ', ['companyemail@email.com']],
           ['Company Name   ', ['companyemail@email.com','companyemail@email.com']],
           ['Company Name', ['companyemail@email.com']],
           ['Company Name          ', ['companyemail@email.com']]]

vendors = [ [company[0].strip(),*company[1:]] for company in vendors ]

供应商中的每个项目本身就是一个列表，您需要输出一个列表，其中第一项内容被理解。

Answer 5

我认为您忘记了逗号

unfiltered = [['Company Name      ', ['companyemail@email.com']], ['Company Name   ', ['companyemail@email.com','companyemail@email.com']], ['Company Name', ['companyemail@email.com']] ,['Company Name          ', ['companyemail@email.com']]]
filtered = []
count=0
text = ''
for i in unfiltered:
    print(i[0])
    test = str((i[0])).strip()
    filtered.append( (str(i)).replace(i[0],test))


print(filtered)

Answer 6

for (int i = 0; i < response.length(); i++) {
    JSONObject cake = response.getJSONObject(i);

    String cakeId = cake.optString("id");
    String cakeName = cake.optString("name");

    JSONArray JSONingredients = cake.optJSONArray("ingredients");

    List<Ingredients> ingredients = new List<Ingredients>();

    for (int j = 0; j < JSONingredients.length(); j++) {
        JSONObject item = JSONingredients.getObject(j);

        String measure = item.getString(“measure”);
        String ingredient = item.getString(“ingredient”);
        Double quantity = item.getDouble(“quantity”);

        Ingredients item2 = new Ingredients(measure, ingredient, quantity);
        ingredients.add(item2);
      } 


      mCakeList.add(new CakesItem(cakeId,ingredients, cakeName));
    }

无法在多维列表中使用.strip（）

6 个答案: