使用功能创建列表

发布于 2025-02-04 18:38:26 字数 437 浏览 1 评论 0原文

import csv

def readLevel(filename,listname):
    listname=[]
    with open(filename, "r") as element:
        csv_reader = csv.reader(element)
        listname = list(csv_reader)

readLevel('level001.csv','LEVEL_001')
readLevel('level002.csv','LEVEL_002')

print(LEVEL_001)
print(LEVEL_002)

错误：未定义的名称“ Level_001”

我试图创建一个接收文件名 + listName的函数，从而从文件（文件名）中读取数据并将其列入列表（listName）。创建列表，但没有正确的名称。有正确的方法吗？

原文

import csv

def readLevel(filename,listname):
    listname=[]
    with open(filename, "r") as element:
        csv_reader = csv.reader(element)
        listname = list(csv_reader)

readLevel('level001.csv','LEVEL_001')
readLevel('level002.csv','LEVEL_002')

print(LEVEL_001)
print(LEVEL_002)

Error: name 'LEVEL_001' is not defined

I'm trying to create a function that receives a filename + listname and thus reads the data from a file (filename) and put it in a list (listname).
A list is created but not with the correct name.
Is there a correct way to do this ?

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宛菡 2025-02-11 18:38:26

字符串值'Level_001'与称为level_001的变量不同。

您应该返回在函数中创建的值，然后在调用代码中分配一个变量名称：

import csv

def readLevel(filename):
    with open(filename, "r") as element:
        return list(csv.reader(element))


LEVEL_001 = readLevel('level001.csv')
LEVEL_002 = readLevel('level002.csv')

print(LEVEL_001)
print(LEVEL_002)

The string value 'LEVEL_001' is not the same thing as a variable called LEVEL_001.

Instead of trying to pass a variable name as an argument to the function, you should return the value that you create in your function and then assign it a variable name in the calling code:

import csv

def readLevel(filename):
    with open(filename, "r") as element:
        return list(csv.reader(element))


LEVEL_001 = readLevel('level001.csv')
LEVEL_002 = readLevel('level002.csv')

print(LEVEL_001)
print(LEVEL_002)

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