-1

我有 5,000,000 个以这种方式格式化的无序字符串（Name.Name.Day-Month-Year 24hrTime）：

"John.Howard.12-11-2020 13:14"
"Diane.Barry.29-07-2020 20:50"
"Joseph.Ferns.08-05-2020 08:02"
"Joseph.Ferns.02-03-2020 05:09"
"Josephine.Fernie.01-01-2020 07:20"
"Alex.Alexander.06-06-2020 10:10"
"Howard.Jennings.07-07-2020 13:17"
"Hannah.Johnson.08-08-2020 00:49"
...

找到时间 t 在某个 n 和 m 之间的所有字符串的最快方法是什么？（即删除所有时间 < n || m < time 的字符串的最快方法）

这种过滤将在不同的范围内进行多次。时间范围必须始终在同一天，并且开始时间始终早于结束时间。

在 java 中，这是我目前的方法，给出了一些时间字符串 M 和 N 以及 500 万个字符串列表：

ArrayList<String> finalSolution = new ArrayList<>();

String[] startingMtimeArr = m.split(":");
String[] startingNtimeArr = n.split(":");
Integer startingMhour = Integer.parseInt(startingMtimeArr[0]);
Integer startingMminute = Integer.parseInt(startingMtimeArr[1]);
Integer endingNhour = Integer.parseInt(startingNtimeArr[0]);
Integer endingNminute = Integer.parseInt(startingNtimeArr[1]);

for combinedString in ArraySizeOf5Million{
  String[] arr = combinedString.split(".");
  String[] subArr = arr[2].split(" ");
  String[] timeArr = subArr[1].split(":");
  String hour = timeArr[0];
  String minute = timeArr[1];

   If hour >= startingMhour 
        && minute >= startingMminute 
        && hour <= endingNhour 
        && minute <= endingNminute {
    finalSolution.add(hour)
   } 
}

Java 是我的母语，但任何其他语言也可以。更好/更快的逻辑是我所追求的

4

3 回答 3

0

正如@Paddy3118 已经指出的那样，二进制搜索可能是要走的路。

  1. （如果您的数据在磁盘上）：加载输入数据并按日期/时间排序。
  2. i0 是结果集的开始索引，i1 是结果集的结束索引（都是从二进制搜索获得的）：枚举结果条目。

我使用的代码（在 Lisp 中）显示在此答案的末尾。它没有丝毫优化（我想通过一些优化工作可以使加载和初始排序更快）。

这就是我的交互式会话的样子（包括时间信息，我的 foo.txt 输入文件包含 500 万个条目）。

rlwrap sbcl --dynamic-space-size 2048
这是 SBCL 2.1.1.debian，ANSI Common Lisp 的实现。有关 SBCL 的更多信息，请访问http://www.sbcl.org/。SBCL 是免费软件，按原样提供，绝对不提供任何担保。它主要在公共领域；有些部分是在 BSD 风格的许可证下提供的。有关详细信息，请参阅发行版中的 CREDITS 和 COPYING 文件。
(ql:quickload :cl-ppcre)
加载“cl-ppcre”：
加载 1 个 ASDF 系统：
cl-ppcre
；加载“cl-ppcre”
..
(:CL-PPCRE)
(load "fivemillion.lisp")
T
(time (defparameter data (load-input-for-queries "foo.txt")))
"sorting..."
评估花费：
实时
32.091 秒总运行时间 32.090620 秒（31.386722 用户，0.703898 系统）
[运行时间包括 2.641 秒 GC 时间和 29.450 秒非 GC 时间。]
100.00% CPU
15 lambdas 转换
115,308,171,684 个处理器周期
6,088,198,752 字节 consed
DATA
(time (defparameter output (query-interval data '(2018 1 1) '(2018 1 2))))
评估花费：
0.000 秒的实时
0.000111 秒总运行时间（0.000109 个用户，0.000002 个系统）
100.00% CPU
395,172 个处理器周期
65,536 个字节 consed
OUTPUT
（时间（defparameteroutput (query-interval data '(2018 1 1) '(2018 1 2 8))))
评估时间：
0.000 秒的实时时间
0.000113 秒的总运行时间（0.000110 用户，0.000003 系统）
100.00% CPU
399,420 处理器周期
65,536字节 consed
OUTPUT
(time (defparameter output (query-interval data '(2018 1 1) '(2019 1 1))))
评估花费：
0.020 秒的实时
0.022469 秒的总运行时间（0.022469 用户，0.000000 系统）
110.00 % CPU
80,800,092 个处理器周期
15,958,016 字节 consed
OUTPUT

因此，虽然加载和排序时间（一次完成）没什么好写的（但可以优化），但(query-interval ...)调用速度非常快。查询的结果集越大，函数返回的列表越长（conses 越多，运行时间越长）。我本可以更聪明，只返回结果集的开始和结束索引，并将条目的收集留给调用者。

这里是源代码，其中还包括生成我使用的测试数据集的代码：

(defun random-uppercase-character ()
  (code-char (+ (char-code #\A) (random 26))))
(defun random-lowercase-character ()
  (code-char (+ (char-code #\a) (random 26))))
(defun random-name-part (nchars)
  (with-output-to-string (stream)
    (write-char (random-uppercase-character) stream)
    (loop repeat (- nchars 1) do
      (write-char (random-lowercase-character) stream))))
(defun random-day-of-month ()
  "Assumes every month has 31 days, because it does not matter
for this exercise."
  (+ 1 (random 31)))
(defun random-month-of-year ()
  (+ 1 (random 12)))
(defun random-year ()
  "Some year between 2017 and 2022"
  (+ 2017 (random 5)))
(defun random-hour-of-day ()
  (random 24))
(defun random-minute-of-hour ()
  (random 60))
(defun random-entry (stream)
  (format stream "\"~a.~a.~d-~d-~d ~d:~d\"~%"
      (random-name-part 10)
      (random-name-part 10)
      (random-day-of-month)
      (random-month-of-year)
      (random-year)
      (random-hour-of-day)
      (random-minute-of-hour)))
(defun generate-input (entry-count file-name)
  (with-open-file (stream
           file-name
           :direction :output
           :if-exists :supersede)
    (loop repeat entry-count do
      (random-entry stream))))

(defparameter *line-scanner*
  (ppcre:create-scanner
   "\"(\\w+).(\\w+).(\\d+)-(\\d+)-(\\d+)\\s(\\d+):(\\d+)\""))
;;      0       1      2      3      4        5      6
;;      fname   lname  day    month  year     hour   minute

(defun decompose-line (line)
  (let ((parts (nth-value
        1
        (ppcre:scan-to-strings
         *line-scanner*
         line))))
    (make-array 7 :initial-contents
        (list (aref parts 0)
              (aref parts 1)
              (parse-integer (aref parts 2))
              (parse-integer (aref parts 3))
              (parse-integer (aref parts 4))
              (parse-integer (aref parts 5))
              (parse-integer (aref parts 6))))))
(defconstant +fname-index+ 0)
(defconstant +lname-index+ 1)
(defconstant +day-index+ 2)
(defconstant +month-index+ 3)
(defconstant +year-index+ 4)
(defconstant +hour-index+ 5)
(defconstant +minute-index+ 6)
(defvar *compare-<-criteria*
  (make-array 5 :initial-contents
          (list +year-index+
            +month-index+
            +day-index+
            +hour-index+
            +minute-index+)))

(defun compare-< (dl1 dl2)
  (labels ((comp (i)
         (if (= i 5)
         nil
         (let ((index (aref *compare-<-criteria* i)))
           (let ((v1 (aref dl1 index))
             (v2 (aref dl2 index)))
             (cond
               ((< v1 v2) t)
               ((= v1 v2) (comp (+ i 1)))
               (t nil)))))))
    (comp 0)))
           
(defun time-stamp-to-index (hours minutes)
  (+ minutes (* 60 hours)))

(defun load-input-for-queries (file-name)
  (let* ((decomposed-line-list
       (with-open-file (stream file-name :direction :input)
         (loop for line = (read-line stream nil nil)
           while line
           collect (decompose-line line))))
     (number-of-lines (length decomposed-line-list))
     (decomposed-line-array (make-array number-of-lines
                        :initial-contents
                        decomposed-line-list)))
    (print "sorting...") (terpri)
    (sort decomposed-line-array #'compare-<)))

(defun unify-date-list (date)
  (let ((date-length (length date)))
    (loop
      for i below 5
      collecting (if (> date-length i) (nth i date) 0))))

(defun decomposed-line-date<date-list (decomposed-line date-list)
  (labels ((comp (i)
         (if (= i 5)
         nil
         (let ((index (aref *compare-<-criteria* i)))
           (let ((v1 (aref decomposed-line index))
             (v2 (nth i date-list)))
             (cond
               ((< v1 v2) t)
               ((= v1 v2) (comp (+ i 1)))
               (t nil)))))))
    (comp 0)))

(defun index-before (data key predicate
             &key (left 0) (right (length data)))
  (if (and (< left right) (> (- right left) 1))
      (if (funcall predicate (aref data left) key)
      (let ((mid (+ left (floor (- right left) 2))))
        (if (funcall predicate (aref data mid) key)
        (index-before data key predicate
                  :left mid
                  :right right)
        (index-before data key predicate
                  :left left
                  :right mid)))
      left)
      right))

(defun query-interval (data start-date end-date)
  "start-date and end-date are given as lists of the form:
'(year month day hour minute) or shorter versions e.g.
'(year month day hour), omitting trailing values which will be
appropriately defaulted."
  (let ((d0 (unify-date-list start-date))
    (d1 (unify-date-list end-date)))
    (let* ((start-index (index-before
             data
             d0
             #'decomposed-line-date<date-list))
       (end-index (index-before
               data
               d1
               #'decomposed-line-date<date-list
               :left (cond
                   ((< start-index 0) 0)
                   ((>= start-index (length data))
                (length data))
                   (t start-index)))))
      (loop for i from start-index below end-index
        collecting (aref data i)))))
于 2021-05-24T17:32:07.017 回答
0

由于数据将被多次搜索，我首先解析字符串以便多次搜索 = 参见by_date.

我使用二进制搜索来查找特定日期的第一个字符串，然后迭代增加次数，在filtered函数变量中收集适当的字符串strings_between。

# -*- coding: utf-8 -*-
"""
https://stackoverflow.com/questions/67562250/fastest-string-filtering-algorithm

Created on Tue May 18 09:20:11 2021

@author: Paddy3118
"""

strings = """\
John.Howard.12-11-2020 13:14
Diane.Barry.29-07-2020 20:50
Joseph.Ferns.08-05-2020 08:02
Joseph.Ferns.02-03-2020 05:09
Josephine.Fernie.01-01-2020 07:20
Alex.Alexander.06-06-2020 10:10
Howard.Jennings.07-07-2020 13:17
Hannah.Johnson.08-08-2020 00:49
Josephine.Fernie.08-08-2020 07:20
Alex.Alexander.08-08-2020 10:10
Howard.Jennings.08-08-2020 13:17
Hannah.Johnson.08-08-2020 09:49\
"""

## First parse the date information once for all future range calcs

def to_mins(hr_mn='00:00'):
    hr, mn = hr_mn.split(':')
    return int(hr) * 60 + int(mn)


by_date = dict()    # Keys are individual days, values are time-sorted
for s in strings.split('\n'):
    name_day, time = s.strip().split()
    name, day = name_day.rsplit('.', 1)
    minutes = to_mins(time)
    if day not in by_date:
        by_date[day] = [(minutes, s)]
    else:
        by_date[day].append((minutes, s))
for day_info in by_date.values():
    day_info.sort()


## Now rely on dict search for day then binary +linear search within day.

def _bisect_left(a, x):
    """Return the index where to insert item x in list a, assuming a is sorted.
    The return value i is such that all e in a[:i] have e < x, and all e in
    a[i:] have e >= x.  So if x already appears in the list, a.insert(x) will
    insert just before the leftmost x already there.

    'a' is a list of tuples whose first item is assumed sorted and searched apon.
    """

    lo, hi = 0, len(a)
    while lo < hi:
        mid = (lo+hi)//2
        # Use __lt__ to match the logic in list.sort() and in heapq
        if a[mid][0] < x: lo = mid+1
        else: hi = mid
    return lo


def strings_between(day="01-01-2020", start="00:00", finish="23:59"):
    global by_date

    if day not in by_date:
        return []
    day_data = by_date[day]
    start, finish = to_mins(start), to_mins(finish)
    from_index = _bisect_left(day_data, start)

    filtered = []
    for time, s in day_data[from_index:]:
        if time <= finish:
            filtered.append(s)
        else:
            break
    return filtered


## Example data

assert by_date == {
 '12-11-2020': [(794, 'John.Howard.12-11-2020 13:14')],
 '29-07-2020': [(1250, 'Diane.Barry.29-07-2020 20:50')],
 '08-05-2020': [(482, 'Joseph.Ferns.08-05-2020 08:02')],
 '02-03-2020': [(309, 'Joseph.Ferns.02-03-2020 05:09')],
 '01-01-2020': [(440, 'Josephine.Fernie.01-01-2020 07:20')],
 '06-06-2020': [(610, 'Alex.Alexander.06-06-2020 10:10')],
 '07-07-2020': [(797, 'Howard.Jennings.07-07-2020 13:17')],
 '08-08-2020': [(49, 'Hannah.Johnson.08-08-2020 00:49'),
                (440, 'Josephine.Fernie.08-08-2020 07:20'),
                (589, 'Hannah.Johnson.08-08-2020 09:49'),
                (610, 'Alex.Alexander.08-08-2020 10:10'),
                (797, 'Howard.Jennings.08-08-2020 13:17')]}

## Example queries from command line
"""
In [7]: strings_between('08-08-2020')
Out[7]:
['Hannah.Johnson.08-08-2020 00:49',
 'Josephine.Fernie.08-08-2020 07:20',
 'Hannah.Johnson.08-08-2020 09:49',
 'Alex.Alexander.08-08-2020 10:10',
 'Howard.Jennings.08-08-2020 13:17']

In [8]: strings_between('08-08-2020', '09:30', '24:00')
Out[8]:
['Hannah.Johnson.08-08-2020 09:49',
 'Alex.Alexander.08-08-2020 10:10',
 'Howard.Jennings.08-08-2020 13:17']

In [9]: strings_between('08-08-2020', '09:49', '10:10')
Out[9]: ['Hannah.Johnson.08-08-2020 09:49', 'Alex.Alexander.08-08-2020 10:10']

In [10]:
"""
于 2021-05-18T09:33:57.173 回答
0

Python中每分钟使用索引的一些示例：

from pprint import pprint
from itertools import groupby

big_list = [
    "John.Howard.12:14",
    "Diane.Barry.13:50",
    "xxxDiane.Barryxxx.13:50",  # <-- added a name in the same HH:MM
    "Joseph.Ferns.08:02",
    "Joseph.Ferns.05:09",
    "Josephine.Fernie.07:20",
    "Alex.Alexander.10:10",
    "Howard.Jennings.12:17",
    "Hannah.Johnson.00:49",
]

# 1. sort the list by time HH:MM
big_list = sorted(big_list, key=lambda k: k[-5:])

# the list is now:

# ['Hannah.Johnson.00:49',
#  'Joseph.Ferns.05:09',
#  'Josephine.Fernie.07:20',
#  'Joseph.Ferns.08:02',
#  'Alex.Alexander.10:10',
#  'John.Howard.12:14',
#  'Howard.Jennings.12:17',
#  'Diane.Barry.13:50',
#  'xxxDiane.Barryxxx.13:50']

# 2. create an index (for every minute in a day)
index = {}

times = []
for i, item in enumerate(big_list):
    times.append(int(item[-5:-3]) * 60 + int(item[-2:]))

last = 0
cnt = 0
for v, g in groupby(times):
    for i in range(last, v):
        index[i] = [cnt, cnt]
    s = sum(1 for _ in g)
    index[v] = [cnt, cnt + s]
    cnt += s
    last = v + 1

for i in range(last, 60 * 24):
    index[i] = [cnt, cnt]


# 3. you can now do a fast query using the index
def find_all_strings(n, m):
    n = int(n[-5:-3]) * 60 + int(n[-2:])
    m = int(m[-5:-3]) * 60 + int(m[-2:])

    return big_list[index[n][0] : index[m][1]]


print(find_all_strings("00:10", "00:30"))  # []
print(find_all_strings("00:30", "00:50"))  # ['Hannah.Johnson.00:49']
print(find_all_strings("12:00", "13:55"))  # ['John.Howard.12:14', 'Howard.Jennings.12:17', 'Diane.Barry.13:50', 'xxxDiane.Barryxxx.13:50']
print(find_all_strings("13:00", "13:55"))  # ['Diane.Barry.13:50', 'xxxDiane.Barryxxx.13:50']
print(find_all_strings("15:00", "23:00"))  # []
于 2021-05-17T00:49:23.603 回答