英文:
Sort List of List in Python based on days of the week
问题
我有一个每日温度列表，如下所示：
weekly_temperature = [['Saturday', 100], ['Wednesday', 95], ['Friday', 80],['Monday', 95], ['Sunday', 90], ['Tuesday', 100], ['Thursday', 85]]
我该如何根据星期几对此列表进行排序，从星期一到星期日？
我希望排序后的输出如下所示：
weekly_temperature = [['Monday', 95], ['Tuesday', 100], ['Wednesday', 95],['Thursday', 85], ['Friday', 80], ['Saturday', 100], ['Sunday', 90]]
英文:
I have a list of daily temperature, shown below:
weekly_temperature = [['Saturday', 100], ['Wednesday', 95], ['Friday', 80],['Monday', 95], ['Sunday', 90], ['Tuesday', 100], ['Thursday', 85]]
How do I sort this list based on days of the week, starting from Monday to Sunday?
I would like the sorted output to look like below:
weekly_temperature = [['Monday', 95], ['Tuesday', 100], ['Wednesday', 95],['Thursday', 85], ['Friday', 80], ['Saturday', 100], ['Sunday', 90]]
答案1
得分: 2
你可以使用time.strptime来解析星期几，以便将其用作排序的键：
import timeweekly_temperature = [['Saturday', 100], ['Wednesday', 95], ['Friday', 80],['Monday', 95], ['Sunday', 90], ['Tuesday', 100], ['Thursday', 85]]print(sorted(weekly_temperature, key=lambda t: time.strptime(t[0], '%A')))
这将输出：
[['Monday', 95], ['Tuesday', 100], ['Wednesday', 95], ['Thursday', 85], ['Friday', 80], ['Saturday', 100], ['Sunday', 90]]
请注意，time.strptime取决于您的区域设置，如果要确保安全，可以保存当前的时间区域设置，将其设置为标准的C区域设置，解析您的星期几，然后恢复您的本地设置：
import timeimport localeorig_locale = locale.setlocale(locale.LC_TIME)locale.setlocale(locale.LC_TIME, 'C')weekly_temperature = [['Saturday', 100], ['Wednesday', 95], ['Friday', 80],['Monday', 95], ['Sunday', 90], ['Tuesday', 100], ['Thursday', 85]]print(sorted(weekly_temperature, key=lambda t: time.strptime(t[0], '%A')))locale.setlocale(locale.LC_TIME, orig_locale)
英文:
You can use time.strptime to parse the day of the week, so you can use it as a key for sorting:
import timeweekly_temperature = [['Saturday', 100], ['Wednesday', 95], ['Friday', 80],['Monday', 95], ['Sunday', 90], ['Tuesday', 100], ['Thursday', 85]]print(sorted(weekly_temperature, key=lambda t: time.strptime(t[0], '%A')))
This outputs:
[['Monday', 95], ['Tuesday', 100], ['Wednesday', 95], ['Thursday', 85], ['Friday', 80], ['Saturday', 100], ['Sunday', 90]]
Note that time.strptime depends on your locale settings, and if you want to be on the safe side you can save your current time locale, set it to the standard C locale, parse your day of the week, and then restore your local setting:
import timeimport localeorig_locale = locale.setlocale(locale.LC_TIME)locale.setlocale(locale.LC_TIME, 'C')weekly_temperature = [['Saturday', 100], ['Wednesday', 95], ['Friday', 80],['Monday', 95], ['Sunday', 90], ['Tuesday', 100], ['Thursday', 85]]print(sorted(weekly_temperature, key=lambda t: time.strptime(t[0], '%A')))locale.setlocale(locale.LC_TIME, orig_locale)
答案2
得分: 0
weekly_temperature = [['星期六', 100], ['星期三', 95], ['星期五', 80], ['星期一', 95], ['星期天', 90], ['星期二', 100], ['星期四', 85]] week = sorted(weekly_temperature, key=lambda x: d[x[0]])
英文:
weekly_temperature = [['Saturday', 100], ['Wednesday', 95], ['Friday', 80], ['Monday', 95], ['Sunday', 90], ['Tuesday', 100], ['Thursday', 85]]d = {"Monday": 1,"Tuesday": 2,"Wednesday": 3,"Thursday": 4,"Friday": 5,"Saturday": 6,"Sunday": 7}week = sorted(weekly_temperature, key=lambda x: d[x[0]])
答案3
得分: 0
你所需要的是一个将星期几转换成相应排序键的结构间接引用：
例如：
days = "Monday Tuesday Wednesday Thursday Friday Saturday Sunday"weekly_temperature = [['Saturday', 100], ['Wednesday', 95], ['Friday', 80],['Monday', 95], ['Sunday', 90], ['Tuesday', 100],['Thursday', 85]]weekly_temperature.sort(key=lambda t:days.index(t[0]))[['Monday', 95], ['Tuesday', 100], ['Wednesday', 95], ['Thursday', 85],['Friday', 80], ['Saturday', 100], ['Sunday', 90]]
你可以使用字典或列表来代替字符串，只要你成功将星期几转换为相应的排序键即可。
英文:
What you need is an indirection to a structure that will convert day names into a sort key in the appropriate order:
For example:
days = "Monday Tuesday Wednesday Thursday Friday Saturday Sunday"weekly_temperature = [['Saturday', 100], ['Wednesday', 95], ['Friday', 80],['Monday', 95], ['Sunday', 90], ['Tuesday', 100],['Thursday', 85]]weekly_temperature.sort(key=lambda t:days.index(t[0]))[['Monday', 95], ['Tuesday', 100], ['Wednesday', 95], ['Thursday', 85],['Friday', 80], ['Saturday', 100], ['Sunday', 90]]
You can use a dictionary or a list instead of a string as long as you manage to turn the day name into the appropriate sort key.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论