Python读取CSV时如何匹配预定义字典值输出对应开场白？

你的代码无法得到预期结果的核心原因有两点：

• csv.reader读取到的所有单元格值默认都是字符串格式，而你定义的openingStatement字典的键是整数类型，类型不匹配无法直接索引取值
• 现有print语句直接拼接了学生姓名和开场白编号字符串，没有去字典中映射对应的实际开场白文本

只需要将读取到的开场白编号转为整数，再作为键取字典中对应的文本拼接打印即可，完整可运行代码如下：

import csv

openingStatement = {
    1: "this is the BEST generic opening!", 
    2: "this is a GOOD generic opening",
    3: "this is an OKAY generic opening",
    4: "this is NOT GOOD, something is going on to deserve this opening"
}

with open("/Users/DaveVavala/Desktop/Practice Automated Report.csv", 'r', newline='') as mainFile:
    reader = csv.reader(mainFile, delimiter=",")
    header = next(reader)

    for row in reader:
        studentName = row[0]
        opening_num = int(row[1])
        # qualities字段当前逻辑暂未使用，可保留供后续扩展
        qualities = row[2]
        print(f"{studentName} {openingStatement[opening_num]}")

执行上述代码后，将得到你预期的输出结果：

Anna this is the BEST generic opening!
Billy this is the BEST generic opening!
Corban this is a GOOD generic opening
Diane this is an OKAY generic opening
Eru this is the BEST generic opening!

补充：如果后续CSV中可能出现非数字、不存在的编号值，可以在索引字典前加一层判断避免程序报错，针对你当前提供的CSV样例，上述代码可以直接正常运行。

内容的提问来源于stack exchange，提问作者davey_twelve