Question

如果有一个值，那么它应该是具有空值的键作为列表；如果有多个值，则第一个值将是键，其余的将是字典中的值列表。

>

Ex: Column
    ram
    sneha, vijay, harish
    deva
    babu, dominic

Expected o/p:
{
'ram':[],
'sneha': ['vijay', 'harish'],
'deva' : [],
'babu' : ['dominic']

}

Answer 1

假设您的列数据采用我描述的格式，以下解决方案将提供预期的输出结果

int i=1;
printf("%d, %d and %d\n", i++, i++, i--);

Answer 2

您只需运行一次循环即可一次选择一行，并将该行的第0个索引设置为键，并将其后的值设置为列表中的值。

可以这样做

    74|                 <form name="mobget">
    75|                   <select id="monster">
 >> 76|                   <% for (var i = 0; i < mobs.length; ++i) { %>
    77|                       <option name="mobs" value="<%= mobs[i].id %>"><%= mobs[i].name %></option>
    78|                   <% } %>
    79|                   </select>

mobs is not defined
    at eval (eval at compile (/home/kogadmin/www/node_modules/ejs/lib/ejs.js:618:12), <anonymous>:62:28)
    at returnedFn (/home/kogadmin/www/node_modules/ejs/lib/ejs.js:653:17)
    at tryHandleCache (/home/kogadmin/www/node_modules/ejs/lib/ejs.js:251:36)
    at View.exports.renderFile [as engine] (/home/kogadmin/www/node_modules/ejs/lib/ejs.js:482:10)
    at View.render (/home/kogadmin/www/node_modules/express/lib/view.js:135:8)
    at tryRender (/home/kogadmin/www/node_modules/express/lib/application.js:640:10)
    at Function.render (/home/kogadmin/www/node_modules/express/lib/application.js:592:3)
    at ServerResponse.render (/home/kogadmin/www/node_modules/express/lib/response.js:1008:7)
    at /home/kogadmin/www/server.js:112:9
    at Layer.handle [as handle_request] (/home/kogadmin/www/node_modules/express/lib/router/layer.js:95:5)

Answer 3

您可以这样做：

columns_list = [['ram'], ['sneha', 'vijay', 'harish'],
                ['deva'], ['babu', 'dominic']]

result = {item[0]: item[1:] for item in columns_list}

print(result)
# {'ram': [], 'sneha': ['vijay', 'harish'], 'deva': [], 'babu': ['dominic']}

如果输入为string，则可以这样：

rows = '''ram
sneha, vijay, harish
deva
babu, dominic'''
columns_list = [row.split(',') for row in rows.split("\n")]
# if columns have space at begin and end then `strip` them
columns_list = tuple(map(lambda cols: [c.strip() for c in cols], columns_list))
print(columns_list)
# (['ram'], ['sneha', 'vijay', 'harish'], ['deva'], ['babu', 'dominic'])
result = {item[0]: item[1:] for item in columns_list}
print(result)
# {'ram': [], 'sneha': ['vijay', 'harish'], 'deva': [], 'babu': ['dominic']}

Answer 4

如果您的源数据是字符串对象，那么

data = ["ram", "sneha, vijay, harish", "deva", "babu, dominic"]
res = {}
for i in data:
    val = i.split(",")
    res[val[0]]= list(map(str.strip, val[1:]))
print(res)

输出：

{'babu': ['dominic'], 'deva': [], 'ram': [], 'sneha': ['vijay', 'harish']}

如何在python中将列的多个值转换为字典？

4 个答案: