Python 基于最大和最小范围内的项目值的字典列表

因此，我有一个字典列表，它以字典列表的形式构成框架信息。列表按帧编号键排序

[
    { "frame_num": "1","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "2","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "3","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "4","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "6","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "8","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "10","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "15","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "16","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "18","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "20","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "22","width": "1920","height":"1080","other_info": "some_info" }
]

因此，我想获取一个帧列表，其frame_num值在4:14的范围内，这是具有此范围的列表的一部分 我的答案是

[
    { "frame_num": "4","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "6","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "8","width": "1920","height":"1080","other_info": "some_info" },
    { "frame_num": "10","width": "1920","height":"1080","other_info": "some_info" },
]

一种可能的方法是借助列表理解。但是我想要一个有效的方法，因为这个列表可能有数百万帧信息

---

我希望得到一些类似python的方法来实现这一点。

我假设最后一帧12是一个打字错误，因为您没有在输出中使用它

这是相当容易的列表理解。首先，定义范围

r = range(4, 14)

然后，使用列表的过滤器来选择帧：

selected_frames = [f for f in frames if int(f["frame_num"]) in r]
# [{'frame_num': '4', 'width': '1920', 'height': '1080', 'other_info': 'some_info'}, {'frame_num': '6', 'width': '1920', 'height': '1080', 'other_info': 'some_info'}, {'frame_num': '8', 'width': '1920', 'height': '1080', 'other_info': 'some_info'}, {'frame_num': '10', 'width': '1920', 'height': '1080', 'other_info': 'some_info'}]

别担心，在r是。但是转换为int需要一些时间

如果您的帧已排序，且范围连续无步长，则可以在退出范围时使用常规循环和中断：

selected_frames = []
for f in frames:
    if int(f["frame_num"]) in r:
        selected_frames.append(f)
    elif selected_frames: # not in range anymore
        break

如果有步骤，仍然可以尝试测试范围边界，然后应用该步骤

我想您可以从同一帧列表中选取多个范围，也许为该列表编制索引是一个好主意：

nums = [int(f["frame_num"]) for f in frames]
# [1, 2, 3, 4, 6, 8, 10, 15, 16, 18, 20]

现在，您可以找到具有以下内容的元素：

---

谢谢，@jferard。这个答案确实帮助我更多地了解python。“继续开导吧！”罗珊乔汉很乐意帮忙。如果你认为这回答了你的问题，请接受我的回答，这是对StackOverflow表示感谢的另一种方式。

import bisect
f = bisect.bisect_left(nums, 4)
t = bisect.bisect_left(nums, 14)
selected_frames = frames[f:t]