【转帖】argument after ** must be a mapping, not list

最新推荐文章于 2024-06-26 09:06:00 发布

WebStudy8

最新推荐文章于 2024-06-26 09:06:00 发布

阅读量8.7k

点赞数

分类专栏： Python 文章标签： Python

Python 专栏收录该内容

3 篇文章

订阅专栏

本文解决了一个使用Python格式化字符串的问题，介绍了如何通过遍历字典列表并应用字符串格式化方法来创建新的字符串列表。提供了两种解决方案，一种使用范围迭代，另一种直接遍历列表元素。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

转自：https://teamtreehouse.com/community/format-argument-after-must-be-a-mapping-not-list

问题：

Create a function named string_factory that accepts a list of dictionaries and a string. Return a new list build by using .format() on the string, filled in by each of the dictionaries in the list.

Here's my code: Please help!

dicts = [
    {'name': 'Michelangelo',
     'food': 'PIZZA'},
    {'name': 'Garfield',
     'food': 'lasanga'},
    {'name': 'Walter',
     'food': 'pancakes'},
    {'name': 'Galactus',
     'food': 'worlds'}
]

string = "Hi, I'm {name} and I love to eat {food}!"

def string_factory(dicts, string):
  strings = []
  for items in dicts:
   string.format(**dicts) 
  return strings

print(string_factory(dicts, string))

方案：

Here you can see 2 possible ways of doing it:

First:

def string_factory(dictionary, strings):
  string_list = []
  for i in range(len(dictionary)):
    string_list.append(strings.format(**dictionary[i]))

  return string_list

Second:

def string_factory(dictionary, strings):
  string_list = []
  for element in dictionary:
    string_list.append(strings.format(**element))

  return string_list

采纳：

def string_factory(dicts, string):

  formatted = []
  for data in dicts:
    new_dict = data
    formatted.append(string.format(**new_dict))
  return formatted


print(string_factory(dicts, string))