Question

我正在尝试打开并写入.txt文件，但是，我想在程序中使用变量名（更改后的名称，以便区分文件）来格式化文件名。

我的整个代码是：

def main():
    num_episodes = 50
    steps = 10000
    learning_rate_lower_limit = 0.02
    learning_rate_array = numpy.linspace(1.0, learning_rate_lower_limit, num_episodes)
    gamma = 1.0
    epsilon = .25
    file = csv_to_array(sys.argv[1])
    grid = build_racetrack(file)
    scenario = sys.argv[2]
    track = sys.argv[3]

    if(track == "right"):
        start_state = State(grid=grid, car_pos=[26, 3])
    else:
        start_state = State(grid=grid, car_pos=[8,1])

    move_array = []
    for episode in range(num_episodes):
        state = start_state
        learning_rate = learning_rate_array[episode]

        total_steps = 0
        for step in range(steps):
            total_steps = total_steps + 1
            action = choose_action(state, epsilon)
            next_state, reward, finished, moves = move_car(state, action, scenario)
            move_array.append(moves)
            if (finished == True):
                print("Current Episode: ", episode, "Current Step: ", total_steps)
                file = open("{}_Track_Episode{}.txt", "w").format(track, episode)
                file.write(str(move_array))
                move_array = []
                break
            else:
                query_q_table(state)[action] = query_q_table(state, action) + learning_rate * (
                            reward + gamma * numpy.max(query_q_table(next_state)) - query_q_table(state, action))
                state = next_state
main()

我当前遇到的错误是：
file = open("{}_Track_Episode{}.txt", "w").format(track, episode) AttributeError: '_io.TextIOWrapper' object has no attribute 'format'

一些研究表明，在写入对象时无法格式化对象。但是，在程序中文件名是动态变量的情况下，如何使用.format创建文件？

Answer 1

您需要先打开文件的格式，然后再尝试打开文件。

file = open("{}_Track_Episode{}.txt".format(track, episode), "w")

由于尝试format（一个open()）返回的对象而TextIOWrapper时出现了该错误。而且该对象没有format方法

Python 3-格式化正在写入的文件的问题

1 个答案: