Question

我需要进行一系列文本输入，并将它们与数字和某些格式组合在一起。问题在于，我不知道输入的系列将预先包含多少个元素。我知道对于一定数量的输入而言，这样做很容易，但是我需要一个可以迭代尽可能多的函数。

所以基本上我需要服用“苹果香蕉樱桃...”和“ 5” 并输出：

str('{"apple": 5, "banana": 5, "cherry": 5, "...": 5}')

这是我到目前为止所拥有的：

print("Enter fruits:")
fruits = [x for x in input().split()]
print("Enter Maximum Fruits per Day")
maxfruit = input()
def maxfruitfuncstep1(x,y):
    return str('"' + x + '"' + ": " + y)
for i in fruits: 
    print("{" + maxfruitfuncstep1(i,maxfruit) + "}")

但这只是给我输出：

{"apple": 5}
{"banana": 5}
{"cherry": 5}

如何使功能在打印输出中水平运行？我尝试使用“，”。join，但这给了我

",a,p,p,l,e,",:, ,5

Answer 1

这就是您想要的：

int load(std::string filename, GLfloat vertexArray[][3], GLuint faces[][3]) {
//open file
std::cout << " -- Read file started -- " << std::endl;

std::ifstream file(filename);

if (file.is_open())
{
    std::cout << " --File Opened --" << std::endl;
    std::string line;
    int ln = 0;
    int vertNum = 1; //starts at one as faces starts looking at index 1
    int faceNum = 0;
    while (getline(file, line))
    {
        //std::cout << "Reading Line: " << ln << " : " << line << std::endl;
        ln++;
        if (!line.empty())
        {
            if (line.at(0) == 'v')
            {
                float temp;
                float temparr[3];
                std::stringstream ss;
                ss << line;

                int i = 0;
                std::string t;
                while (!ss.eof()) {
                    ss >> t;
                    //std::cout << "Token: " << std::endl;
                    if (std::stringstream(t) >> temp && i <= 3) {
                        //std::cout << "Store: " << temp << std::endl;
                        temparr[i] = temp;
                        i++;
                    }
                    t = "";
                }

                if (true) //put out of bounds checking
                {
                    for (int i = 0; i < 3; i++)
                    {
                        vertexArray[vertNum][i] = temparr[i];
                    }
                }
                else
                {
                    std::cout << "ERROR: Could not add vertex to vertexArray" << std::endl;
                    return 0;
                }
                vertNum++;
            }


            if (line.at(0) == 'f')
            {
                int temp;
                int temparr[3];
                std::stringstream ss;
                ss << line;

                int i = 0;
                std::string t;
                while (!ss.eof()) {
                    ss >> t;
                    //std::cout << "Token: " << t << std::endl;
                    if (std::stringstream(t) >> temp && i <= 3) {
                        temparr[i] = temp;
                        i++;
                    }
                    t = "";
                }

                if (true) //put out of bounds checking here
                {
                    for (int i = 0; i < 3; i++)
                    {
                        faces[faceNum][i] = temparr[i];
                    }
                }
                else
                {
                    std::cout << "ERROR: Could not add face to faces" << std::endl;
                    return 0;
                }
                faceNum++;
            }
        }
    }
    file.close();
    std::cout << "Done!" << std::endl;
    return 1;
}
else
{
    std::cout << " ERROR: Cannot open file " << filename << std::endl;
    return -1;
}

另一个解决方案很简单：

fruits = ["apple", "banana", "cherry"]
maxfruit ="5"
print("{" + ", ".join(fruit + ": " + maxfruit for fruit in fruits) + "}")

Answer 2

一种方法是：

fruits = ['banana', 'strawberry', 'cherry']

max_num = 5

from collections import defaultdict

result = defaultdict(dict)

for x in fruits:
    result[x] = max_num

str(dict(result))

"{'banana': 5, 'strawberry': 5, 'cherry': 5}"

获取循环迭代以在同一行上输出

2 个答案: