如何使用timeit.timeit测试多行Python代码且不出现EOL语法错误

报错根因 #

你将包含\n等转义字符的代码直接复制进普通三重引号字符串作为timeit的入参时，字符串解析逻辑会误将代码里的转义符和外层字符串的边界结合，导致解析到字符串末尾时没有找到匹配的引号，触发EOL（行尾扫描字符串字面量）错误。

最优解决方案：封装函数调用 #

完全避开字符串转义问题，后续修改代码也不需要重复复制粘贴，操作步骤如下：

1. 把你的业务逻辑全部封装到一个独立函数中
2. 直接将函数名作为参数传给timeit.timeit即可

示例代码：

# 封装业务逻辑
def run_logic():
    # 此处替换为你的实际输入读取逻辑，以下为示例模拟输入
    input1 = ["2",
              "0 0 1 1 small",
              "3 2 6 5 large",
              "3",
              "2 0 small",
              "0.5 0.6 medium",
              "3 4 large",
              "3",
              "60.0 2.9 111.4 32.9 medium",
              "20.0 82.6 43.4 153.1 small",
              "34.7 9.6 56.6 78.2 large",
              "6",
              "60.0 8.3 small",
              "43.4 153.1 small",
              "13.4 55.9 medium",
              "61.5 68.1 medium",
              "72.1 69.1 large",
              "78.4 13.2 large",
              "0"]

    nm = []
    peanut_descriptions = []
    box_descriptions = []

    for line in input1:
        next_line_index = input1.index(line)+1
        parsed_line = line.split()

        # 读取每组用例的盒子/花生数量
        if len(parsed_line) == 1:
            if int(parsed_line[0]) == 0:
                break
            nm.append(int(parsed_line[0]))

        # 解析花生描述
        if len(parsed_line) == 3:
            peanut_description_dict = {'x': parsed_line[0],
                                       'y': parsed_line[1],
                                       'peanut_size': parsed_line[2]}
            peanut_descriptions.append(peanut_description_dict)

        # 解析盒子描述
        if len(parsed_line) == 5:
            box_description_dict = {'x1': parsed_line[0],
                                    'y1': parsed_line[1],
                                    'x2': parsed_line[2],
                                    'y2': parsed_line[3],
                                    'intended_size': parsed_line[4]}
            box_descriptions.append(box_description_dict)
        previous_line = line

    # 提取花生尺寸列表
    peanut_sizes = []
    for i in peanut_descriptions:
        peanut_size = i['peanut_size']
        peanut_sizes.append(peanut_size)

    # 拆分测试用例
    cases = []
    for i in range(len(nm)):
        if i % 2 != 0:
            continue
        boxes_index = nm[i]
        peanuts_index = nm[i+1]

        boxes_case = box_descriptions[0:boxes_index]
        peanuts_case = peanut_descriptions[0:peanuts_index]
        cases.append([boxes_case, peanuts_case])

        del box_descriptions[0:boxes_index]
        del peanut_descriptions[0:peanuts_index]

    # 匹配逻辑
    all_statuses = []
    for case_index in range(len(cases)):
        case = cases[case_index]
        box = case[0]
        peanut = case[1]
        statuses = []
        for p in range(len(peanut)):
            status = "floor"
            for b in range(len(box)):
                p_x = float(peanut[p]['x'])
                p_y = float(peanut[p]['y'])
                peanut_size = peanut[p]['peanut_size']
                b_x1 = float(box[b]['x1'])
                b_y1 = float(box[b]['y1'])
                b_x2 = float(box[b]['x2'])
                b_y2 = float(box[b]['y2'])
                box_category = box[b]['intended_size']
                if b_x1 <= p_x <= b_x2 and b_y1 <= p_y <= b_y2:
                    status = box_category
                if peanut_size == box_category and status == box_category:
                    status = "correct"
            statuses.append(status)
        all_statuses.append(statuses)

    # 生成输出
    output1 = []
    for i in range(len(all_statuses)):
        case_statuses = all_statuses[i]
        paragraph = ""
        for j in range(len(case_statuses)):
            line = peanut_sizes[j] + " " + case_statuses[j] + "\n"
            paragraph += line
        output1.append(paragraph)
        del peanut_sizes[0:len(case_statuses)]
    output1 = "\n".join(output1)

    # 性能测试时可注释print，避免IO耗时影响测试结果
    # print(output1)

# 性能测试
import timeit
print(timeit.timeit(run_logic, number=100))

备选方案：使用原始字符串 #

如果你坚持要使用字符串传代码的方式，可以在外层三重引号前加r声明为原始字符串，Python会将字符串内所有转义符识别为普通字符，不会触发解析错误：

import timeit
test1 = r"""
# 把你的完整代码直接复制到此处即可
"""
print(timeit.timeit(test1, number=100))

注意事项 #

• 性能测试时建议注释掉print语句，避免IO操作占用过多时间导致测试结果不准
• 实际使用文件输入时，仅需将run_logic函数内的input1替换为你读取文件生成列表的逻辑即可，无需修改测试框架

内容的提问来源于stack exchange，提问作者amab8901