用Python迭代地写入文件，在每次迭代中创建一个新字符串

我想我忽略了显而易见的事实，但如果有人能指出这一点，我将不胜感激。这是一个小时的语法游戏，我很沮丧

以下是一个MWE。运行不到1秒：

from itertools import product

PARAM_GRID = list(product(
range(64, 68),
range(64, 68),
range(132, 136),
range(132, 136),
))

SCOREFILE = './fake_scores.csv'

def search_distributed():

    for param_ix in range(len(PARAM_GRID)):

        params = PARAM_GRID[param_ix]
        score = 0.9

        with open(SCOREFILE, 'a') as f:
            #NOT MINE, and doesn't do what I want
            #f.write('{",".join(map(str, params + (score,)))}\n')

            #MINE, and still doesn't do what I want.
            paramstr = str(params) 
            scorestr = str(score) 
            f.write("\n".join({paramstr + ' ' + scorestr}))

def main():
    search_distributed()

if __name__ == '__main__':
    main()

而不是这样创建的，即包含以下内容的文件：

(64, 64, 132, 132) 0.9(64, 64, 132, 133) 0.9(64, 64, 132, 134) 0.9...

(64, 64, 132, 132) 0.9
(64, 64, 132, 133) 0.9
(64, 64, 132, 134) 0.9

我想要一个包含以下内容的文件：

(64, 64, 132, 132) 0.9(64, 64, 132, 133) 0.9(64, 64, 132, 134) 0.9...

(64, 64, 132, 132) 0.9
(64, 64, 132, 133) 0.9
(64, 64, 132, 134) 0.9

---

显然，我需要以某种方式修改join命令。我已经在上面展示了我的尝试。有人能解决这个问题吗？

下面是您的代码的一个小改动，可以解决您的问题：

from itertools import product

PARAM_GRID = list(product(
range(64, 68),
range(64, 68),
range(132, 136),
range(132, 136),
))

SCOREFILE = './fake_scores.csv'

def search_distributed():

    for param_ix in range(len(PARAM_GRID)):

        params = PARAM_GRID[param_ix]
        score = 0.9

        with open(SCOREFILE, 'a') as f:
            #NOT MINE, and doesn't do what I want
            #f.write('{",".join(map(str, params + (score,)))}\n')

            #MINE, and still doesn't do what I want.
            paramstr = str(params) 
            scorestr = str(score) 
            f.write(paramstr + ' ' + scorestr + '\n')

def main():
    search_distributed()

if __name__ == '__main__':
    main()

由于您只提供一个元素{paramstr+''+scorestr}只是一个字符串，而不是join方法的元素列表，因此它不会添加任何元素。您只需要手动将其添加到每一行的末尾