从嵌套列表中打印出重复项答案

【问题标题】：Printing out duplicate items from nested list从嵌套列表中打印出重复项
【发布时间】：2020-12-13 00:56:32
【问题描述】：

系统要求用户输入糖果的名称。如果列表中不存在该甜点，则应通知用户。如果该甜点在列表中，则每次在列表中提到该特定甜点时，该列表都会打印出列表的所有详细信息。例如，如果我输入吉百利：

# List of transactions [[customer, sweet, amount, money, day, month]
lst_sweets = [
    ["Tom", "Cadbury", 2, 9, 1, 10],
    ["Rachel", "Galaxy", 5, 10, 1, 9],
    ["Rachel", "Smarties", 4, 10, 1, 8],
    ["Lisa", "Gum drops", 10, 9, 1, 7],
    ["Donald", "Cadbury", 1, 5, 1, 6],
    ["Marie", "Smarties", 5, 10, 1, 5]
]
def sweet_purchases():
    x = input("Enter the sweet name")

    p = -1
    sum = 0

    for i in range(len(lst_sweets)):

        item = lst_sweets[i]
        name = item[1]

        if name == x:
            p = i

    if p == -1:
        print("customer not in list")
    else:

        sweetx = lst_sweets[p]
        print("Name:", sweetx[0], "Sweet:", sweetx[1], "amount:", sweetx[2], "money:", sweetx[3], "day:", sweetx[4], "month:", sweetx[5])

期望的输出：姓名：，汤姆，斯威特：，吉百利，金额：，2，钱：，9，日：，1，月：，10 姓名：Donald Sweet：吉百利金额：1、钱：5、日：、1、月：、6

实际输出：姓名:, Tom, Sweet:, 吉百利, 金额:, 2, 钱:, 9, 日:, 1, 月:, 10

有谁知道为什么两者都没有打印出来？谢谢

【问题讨论】：

标签： python list function nested-lists python-3.9

【解决方案1】：

如果您找到多个匹配项，则会覆盖 p，从而只打印一个匹配项。您应该在循环中打印：

lst_sweets = [
    ["Tom", "Cadbury", 2, 9, 1, 10],
    ["Rachel", "Galaxy", 5, 10, 1, 9],
    ["Rachel", "Smarties", 4, 10, 1, 8],
    ["Lisa", "Gum drops", 10, 9, 1, 7],
    ["Donald", "Cadbury", 1, 5, 1, 6],
    ["Marie", "Smarties", 5, 10, 1, 5]
]

fields = ["name", "sweet", "amount", "money", "day", "month"]

x = input("Enter the sweet name: ")
found = False
for sweet in lst_sweets:
    if sweet[1] == x:
        found = True
        print(", ".join(f"{field}: {val}" for field, val in zip(fields, sweet)))
if not found:
    print("sweet not found")

【讨论】：

【解决方案2】：

# List of transactions [[customer, sweet, amount, money, day, month]
lst_sweets = [
    ["Tom", "Cadbury", 2, 9, 1, 10],
    ["Rachel", "Galaxy", 5, 10, 1, 9],
    ["Rachel", "Smarties", 4, 10, 1, 8],
    ["Lisa", "Gum drops", 10, 9, 1, 7],
    ["Donald", "Cadbury", 1, 5, 1, 6],
    ["Marie", "Smarties", 5, 10, 1, 5]
]
def sweet_purchases():
    x = input("Enter the sweet name")

    p = [] # make a list that stores the matching index
    sum = 0

    for i in range(len(lst_sweets)):

        item = lst_sweets[i]
        name = item[1]

        if name == x:
            p.append(i)

    if p == None: # if the list is empty
        print("customer not in list")
    else:
        # Now iterate over the correct list of index and print the values
        for i in p:
            sweetx = lst_sweets[i]
            print("Name:", sweetx[0], "Sweet:", sweetx[1], "amount:", sweetx[2], "money:", sweetx[3], "day:", sweetx[4], "month:", sweetx[5])

【讨论】：

【解决方案3】：

您在每个循环上设置p 的方式是清除上一个匹配项。

def sweet_purchases():
  x = input("Enter the sweet name")
  cnt=0
  for entry in lst_sweets:
    if entry[1] == x:
      print(entry) # or whatever
      cmt+=1
  if not cmt:
    print('No joy')

如果您只想打印一些输出，这是最简单的方法。如果您想在找到所有行后做某事，您需要累积结果。像这样的：

def sweet_purchases():
  x = input("Enter the sweet name")
  hits=[]
  for entry in lst_sweets:
    if entry[1] == x:
      hits.append(entry)
   return hits

但是“pythonic”在不处理列表的情况下查找列表的方法是使用理解：

def sweet_purchases():
  x = input("Enter the sweet name")
  return [e for e in lst_sweets
            if e[1] == x]

【讨论】：

【解决方案4】：

您的循环只会产生一个匹配项。一种更准确（也更简单！）的方法是使用列表推导生成原始列表的过滤版本，然后遍历过滤后的列表：

def sweet_purchases():
    sweet = input("Enter the sweet name")
    filtered_entries = [entry for entry in lst_sweets if entry[1] == sweet]
    if not filtered_entries:
        print("customer not in list")
        return
    for entry in filtered_entries:
        [name, sweet, amount, money, day, month] = entry
        print(f"Name: {name} Sweet: {sweet} amount: {amount} money: {money} day: {day} month: {month}")

【讨论】：