Python 搜索按整数时间戳排序的列表的简单方法
我有一个表格的日志条目列表:Python 搜索按整数时间戳排序的列表的简单方法,python,search,Python,Search,我有一个表格的日志条目列表: [{'time': 199920331000, 'message': 'message1'}, {'time': 199920331001, 'message': 'message2'}...] 其中时间值始终在列表中增加。如果我想获取晚于给定时间戳的日志,我可以遍历元素,直到看到大于给定时间戳的时间戳: def getLog(timestamp): global logs for x in range(len(logs)): if
[{'time': 199920331000, 'message': 'message1'}, {'time': 199920331001, 'message': 'message2'}...]
其中时间值始终在列表中增加。如果我想获取晚于给定时间戳的日志,我可以遍历元素,直到看到大于给定时间戳的时间戳:
def getLog(timestamp):
global logs
for x in range(len(logs)):
if logs[x]['time'] > timestamp:
return logs[x:]
return []
我想Python3中已经有了一种快速搜索机制,但不知道去哪里查找。如果我理解正确,您正在查找,它实现了一种高效算法,用于查找排序列表中的值大于或小于给定值的点 您的日志条目需要是一个实现某种形式的排序的类。大概是这样的:
from functools import total_ordering
@total_ordering
class LogEntry(object):
def __init__(self, time, message):
self.time = time
self.message = message
def __eq__(self, other):
if not isinstance(other, self.__class__):
return NotImplemented
return self.time == other.time and self.message == other.message
def __lt__(self, other):
if not isinstance(other, self.__class__):
return NotImplemented
if self.time == other.time:
return self.message < other.message
return self.time < other.time
def binary_search(log_list, timestamp, lo=0, hi=None):
if hi is None:
hi = len(log_list)
while lo < hi:
mid = (lo+hi)//2
midval = log_list[mid]['time']
if midval < timestamp:
lo = mid+1
elif midval > timestamp:
hi = mid
else:
return mid
return -1
请注意,我们不需要声明
日志
全局,因为您没有分配给它。如果您知道时间总是在增加,您可以保证您的列表已排序。
然后我会使用来自的答案并尝试调整它,如下所示:
from functools import total_ordering
@total_ordering
class LogEntry(object):
def __init__(self, time, message):
self.time = time
self.message = message
def __eq__(self, other):
if not isinstance(other, self.__class__):
return NotImplemented
return self.time == other.time and self.message == other.message
def __lt__(self, other):
if not isinstance(other, self.__class__):
return NotImplemented
if self.time == other.time:
return self.message < other.message
return self.time < other.time
def binary_search(log_list, timestamp, lo=0, hi=None):
if hi is None:
hi = len(log_list)
while lo < hi:
mid = (lo+hi)//2
midval = log_list[mid]['time']
if midval < timestamp:
lo = mid+1
elif midval > timestamp:
hi = mid
else:
return mid
return -1
def二进制搜索(日志列表,时间戳,lo=0,hi=None):
如果hi为无:
hi=len(日志列表)
当lo时间戳:
高=中
其他:
中途返回
返回-1
(尚未测试)鉴于Python尝试
b.\uu gt\uuu(a)
当a.\uu lt\uu(b)
未实现时,不需要更改日志条目的类,提供足够智能的密钥就足够了:
import bisect
from functools import total_ordering
from operator import itemgetter
log = [
{'time': 199920331000, 'message': 'message1'},
{'time': 199920331001, 'message': 'message2'},
# ...
]
@total_ordering
class Key(object):
def __init__(self, keyfunc, keyval):
self.keyfunc = keyfunc
self.keyval = keyval
def __eq__(self, other):
return self.keyval == self.keyfunc(other)
def __lt__(self, other):
return self.keyval < self.keyfunc(other)
start = bisect.bisect(log, Key(itemgetter("time"), 199920331000))
print log[start:]
(这是从中剥离出来的)