Question

我要做的任务是显示所有其他元素，而不是（Americano）。我尝试使用'Record.remove（“ Americano”），但显示错误

Record.remove("Americano")
ValueError: list.remove(x): x not in list

我该怎么办......

========================这是==================的结果====== Record.append（record [1]）print（Record）->

['Menu_name', 'Cafe_latte', 'Americano', 'Americano', 'Smoothie_queen', 'Americano', 'Americano', 'Cafe_mocha', 'Cafe_latte', 'Americano', 'Amorparty', 'Plain_yogurt', 'Americano', 'Americano', 'Cafe_mocha', 'Berry_smoothie', 'Cafe_mocha', 'Cafe_latte', 'Americano', 'Cafe_latte', 'Berry_smoothie', 'Berry_smoothie', 'Plain_yogurt', 'Berry_smoothie', 'Plane_yougurt', 'Berry_smoothie', 'Cafe_latte', 'Americano', 'Americano', 'Cafe_mocha', 'Plain_yogurt', 'Americano', 'Plain_yogurt', 'Cafe_mocha', 'Plain_yogurt', 'Americano', 'Cofe_latte', 'Plain_yogurt', 'Cafe_mocha', 'Americano', 'Berry_smoothie', 'Cafe_latte', 'Cafe_latte', 'Cafe_mocha', 'Cafe_latte', 'Cafe_latte', 'Cafe_mocha', 'Americano', 'Plain_yogurt', 'Americano', 'Americano', 'Cofe_latte', 'Berry_smoothie', 'Berry_smudie', 'American_air', 'Plain_yogurt', 'Berry_smoothie', 'Berry_smoothie', 'Berry_smoothie', 'Cafe_latte', 'Americano', 'Plain_yogurt', 'Cafe_latte', 'Cafe_mocha', 'Cafe_mocha', 'Plain_yogurt', 'Berry_smoothie', 'Berry_smoothie', 'Cafe_mocha', 'Cafe_mocha', 'Berry_smoothie', 'Cafe_mocha', 'Plain_yogurt']

def load_menu():
    Menu = []
    #========= STEP 1 ==========
    menufile = open("menu.txt", "r")

    for line in menufile:
        menu = line.split()
        Menu.append(menu)

    menufile.close()

    for menu in Menu:
        print(menu[0]," : ",menu[1])
    #========= STEP 1 ==========
    return Menu

def error_check(date):
    #========= STEP 3 ==========
    Record = []
    recordfile = open("ledger_"+date+".txt","r")

    for line in recordfile:
        record = line.split()
        Record.append(record[1])
        Record.remove("Americano")
    print(Record)

    recordfile.close()
    #========= STEP 3 ==========

Answer 1

ValueError: list.remove(x): x not in list

您有错误，因为列表记录在记录中没有“ Americano”。

在将记录添加到“记录”之前，应检查记录是否不是“ Americano”

尝试一下：

def error_check(date):
    #========= STEP 3 ==========
    Record = []
    recordfile = open("ledger_"+date+".txt","r")

    for line in recordfile:
        record = line.split()[1]
        if record != "Americano":
            Record.append(record)
    print(Record)
    recordfile.close()

Answer 2

您应该使用：

def error_check(date):
    #========= STEP 3 ==========
    Record = []
    recordfile = open("ledger_"+date+".txt","r")

    for line in recordfile:
        record = line.rstrip().split()
        Record.append(record[1])
        Record.remove("Americano")
    print(Record)

    recordfile.close()

Answer 3

在删除元素之前，您应检查列表中是否有“ Americano”：

def error_check(date):
    #========= STEP 3 ==========
    Record = []
    recordfile = open("ledger_"+date+".txt","r")

for line in recordfile:
    record = line.split()
    Record.append(record[1])
    if "Americano" in Record:
        Record.remove("Americano")
print(Record)

recordfile.close()
#========= STEP 3 ==========

我无法从列表中删除元素

3 个答案: