如何从现有列表startTid生成仅保留每个条目前2个元素的列表dayList

Hey there! Let's tweak your code so it creates and returns the dayList you need, instead of just printing each entry. I'll show you two approaches—one that's easy to follow for clarity, and a more Pythonic version that's cleaner.

Basic Approach (Explicit Loop) #

This version uses an explicit loop to build your list step by step, which is great for understanding exactly what's happening:

def getDays(filnavn):
    # 获取原始数据
    startTid, stopTid, forbruk = getAMSdata(filnavn)
    dayList = []  # 初始化空列表来存储结果
    
    for entry in startTid:  # 直接遍历startTid的元素，比range(len())更直观
        trimmed_entry = entry[:2]  # 取每个条目的前2个元素
        dayList.append(trimmed_entry)  # 将处理后的条目添加到列表
        # print(trimmed_entry)  # 可选：保留打印验证，不需要可删除
    
    return dayList  # 返回最终生成的列表

# 调用函数并接收返回值
result_day_list = getDays(FIL)
# 可以打印结果确认
print(result_day_list)

Pythonic Approach (List Comprehension) #

If you want a more concise and efficient way, list comprehensions are the way to go—they let you build the list in a single line:

def getDays(filnavn):
    startTid, stopTid, forbruk = getAMSdata(filnavn)
    # 一行生成目标列表：遍历每个entry，取前2个元素
    dayList = [entry[:2] for entry in startTid]
    
    # 可选打印验证
    # for item in dayList:
    #     print(item)
    
    return dayList

result_day_list = getDays(FIL)

Key Notes: #

• I replaced range(len(startTid)) with directly iterating over startTid—this is more readable and avoids unnecessary index management.
• The slice entry[:2] works exactly like entry[0:2] (omitting the start index defaults to 0).
• We initialize dayList as an empty list, then populate it with each trimmed entry before returning it.

内容的提问来源于stack exchange，提问作者angelsen