Python 列表列表：替换和添加子列表的项

我有一个列表，让我们这样说：

tripInfo_csv = [['1','2',6,2], ['a','h',4,2], ['1','4',6,1], ['1','8',18,3], ['a','8',2,1]]

LineT;Line;Route;Day;Start_point;End_point;Adults;Children;First_visit
SM55;5055;3;Weekend;15;87;21;4;0 
SM02;5002;8;Weekend;AF3;89;5;0;1 
...

将子列表视为trips:[起点、终点、成人数量、儿童数量]

我的目标是得到一个列表，在这个列表中，起点和终点重合的旅行可以得到它们的第三个和第四个值。起始值和结束值应始终为1到8之间的数字。如果它们是字母，则应替换为相应的数字（a=1，b=2，依此类推）

这是我的密码。这是可行的，但我相信可以改进。对我来说，主要问题是性能。我有很多这样的列表，还有更多的子列表

dicPoints = {'a':'1','b':'2','c':'3', 'd':'4', 'e':'5', 'f':'6', 'g':'7', 'h':'8'}
def getTrips (trips):
    okTrips = []
    for trip in trips:
        if not trip[0].isdigit():
            trip[0] = dicPoints[trip[0]]
        if not trip[1].isdigit():
            trip[1] = dicPoints[trip[1]]

        if len(okTrips) == 0:
            okTrips.append(trip)
        else:
            for i, stop in enumerate(okTrips):
                if stop[0] == trip[0] and stop[1] == trip[1]:
                    stop[2] += trip[2]
                    stop[3] += trip[3]
                    break
                else:
                    if i == len(okTrips)-1:
                        okTrips.append(trip)

正如eguaio提到的，上面的代码有一个bug。应该是这样的：

def getTrips (trips):
    okTrips = []
    print datetime.datetime.now()
    for trip in trips:
        if not trip[0].isdigit():
            trip[0] = dicPoints[trip[0]]
        if not trip[1].isdigit():
            trip[1] = dicPoints[trip[1]]

        if len(okTrips) == 0:
            okTrips.append(trip)
        else:
            flag = 0
            for i, stop in enumerate(okTrips):
                if stop[0] == trip[0] and stop[1] == trip[1]:
                    stop[2] += trip[2]
                    stop[3] += trip[3]
                    flag = 1
                    break

            if flag == 0:
                okTrips.append(trip)

---

由于eguaio的回答，我得到了一个改进的版本，我想与大家分享。这是我根据他的答案写的剧本。 现在我的数据和要求比第一次告诉我的更复杂，所以我做了一些更改。

CSV文件如下所示：

tripInfo_csv = [['1','2',6,2], ['a','h',4,2], ['1','4',6,1], ['1','8',18,3], ['a','8',2,1]]

LineT;Line;Route;Day;Start_point;End_point;Adults;Children;First_visit
SM55;5055;3;Weekend;15;87;21;4;0 
SM02;5002;8;Weekend;AF3;89;5;0;1 
...

脚本：

import os, csv, psycopg2

folder = "F:/route_project/routes"

# Day type
dicDay = {'Weekday':1,'Weekend':2,'Holiday':3}

# Dictionary with the start and end points of each route
#  built from a Postgresql table (with coumns: line_route, start, end)
conn = psycopg2.connect (database="test", user="test", password="test", host="###.###.#.##")
cur = conn.cursor()
cur.execute('select id_linroute, start_p, end_p from route_ends')
recs = cur.fetchall()
dicPoints = {rec[0]: rec[1:] for rec in recs}

# When point labels are text, replace them with a number label in dicPoints
# Text is not important: they are special text labels for start and end
#  of routes (for athletes), so we replace them with labels for start or
#  the end of each route
def convert_point(line, route, point, i):
    if point.isdigit():
        return point
    else:
        return dicPoints["%s_%s" % (line,route)][i]

# Points with text labels mean athletes made the whole or part of this route,
#  we keep them as adults but also keep this number as an extra value
#  for further purposes
def num_athletes(start_p, end_p, adults):
    if not start_p.isdigit() or not end_p.isdigit():
        return adults
    else:
        return 0

# Data is taken for CSV files in subfolders
for root, dirs, files in os.walk(folder):
    for file in files:
        if file.endswith(".csv"):
            file_path = (os.path.join(root, file))
            with open(file_path, 'rb') as csvfile:
                rows = csv.reader(csvfile, delimiter=';', quotechar='"')
                # Skips the CSV header row
                rows.next()
                # linT is not used, yet it's found in every CSV file
                # There's an unused last column in every file, I take advantage out of it
                #  to store the number of athletes in the generator
                gen =((lin, route, dicDay[tday], convert_point(lin,route,s_point,0), convert_point(lin,route,e_point,1), adults, children, num_athletes(s_point,e_point,adults)) for linT, lin, route, tday, s_point, e_point, adults, children, athletes in rows)
                dicCSV = {}
                for lin, route, tday, s_point, e_point, adults, children, athletes in gen:
                    visitors = dicCSV.get(("%s_%s_%s" % (lin,route,s_point), "%s_%s_%s" % (lin,route,e_point), tday), (0, 0, 0))
                    dicCSV[("%s_%s_%s" % (lin,route,s_point), "%s_%s_%s" % (lin,route,e_point), tday)] = (visitors[0] + int(adults), visitors[1] + int(children), visitors[2] + int(athletes))

for k,v in dicCSV.iteritems():
    print k, v

看看这是否有帮助

trips = [['1','2',6,2], ['a','h',4,2], ['1','2',6,1], ['1','8',18,3], ['a','h',2,1]]

# To get the equivalent value
def x(n):
    if '1' <= n <= '8':
        return int(n)
    return ord(n) - ord('a')

# To group lists with similar start and end points
from collections import defaultdict


groups = defaultdict(list)

for trip in trips:
    # Grouping based on start and end point.
    groups[(x(trip[0]), x(trip[1]))].append(trip)

grouped_trips = groups.values()

result = []
for group in grouped_trips:
    start = group[0][0]
    end = group[0][1]
    adults = group[0][2]
    children = group[0][3]
    for trip in group[1:]:
        adults += trip [2]
        children += trip [3]
    result += [[start, end, adults, children]]

print result

trips=['1'，'2'，6,2'，'a'，'h'，4,2'，'1'，'2'，6,1'，'1'，'8'，18,3'，'a'，'h'，2,1]]
#得到等值
def x（n）：
如果“1”假设起点和终点在0和n值之间

然后，结果“OkTrip”最多有n^2个元素。然后，函数中的第二个循环的复杂性为O（n^2）。如果空间复杂度没有问题，可以将复杂度降低到O（n）

首先，创建包含n个列表的dict，使k'（th）子列表包含以“k”开头的trips

当您搜索是否有具有相同起点和终点的不同行程时，您只需要搜索相应的子列表，而不需要搜索所有元素

这个想法来自稀疏矩阵存储技术。
我无法检查以下代码的有效性

代码如下：

dicPoints = {'a':'1','b':'2','c':'3', 'd':'4', 'e':'5', 'f':'6', 'g':'7', 'h':'8'}
Temp = {'1':[],'2':[],'3':[],'4':[],'5':[],'6':[],'7':[],'8':[]};
def getTrips (trips):
   okTrips = []
   for trip in trips:
        if not trip[0].isdigit():
            trip[0] = dicPoints[trip[0]]
        if not trip[1].isdigit():
            trip[1] = dicPoints[trip[1]]

        if len(Temp[trip[0]]) == 0:
            Temp[trip[0]].append(trip)
        else:
            for i, stop in enumerate(Temp[trip[0]]):
                if stop[1] == trip[1]:
                   stop[2] += trip[2]
                   stop[3] += trip[3]
                   break
                else:
                   if i == len(Temp[trip[0]])-1:
                       Temp[trip[0]].append(trip)
        print Temp

    for key in Temp:
        okTrips = okTrips + Temp[key];

为了更有效地处理此问题，最好按起点和终点对输入列表进行排序，以便将具有匹配起点和终点的行分组在一起。然后，我们可以轻松地使用
groupby
函数高效地处理这些组

from operator import itemgetter
from itertools import groupby

tripInfo_csv = [
    ['1', '2', 6, 2], 
    ['a', 'h', 4, 2], 
    ['1', '4', 6, 1], 
    ['1', '8', 18, 3], 
    ['a', '8', 2, 1],
]

# Used to convert alphabetic point labels to numeric form
dicPoints = {v:str(i) for i, v in enumerate('abcdefgh', 1)}

def fix_points(seq):
    return [dicPoints.get(p, p) for p in seq]

# Ensure that all point labels are numeric
for row in tripInfo_csv:
    row[:2] = fix_points(row[:2])

# Sort on point labels
keyfunc = itemgetter(0, 1)
tripInfo_csv.sort(key=keyfunc)

# Group on point labels and sum corresponding adult & child numbers
newlist = []
for k, g in groupby(tripInfo_csv, key=keyfunc):
    g = list(g)
    row = list(k) + [sum(row[2] for row in g), sum(row[3] for row in g)]
    newlist.append(row)

# Print the condensed list
for row in newlist:
    print(row)

输出

['1', '2', 6, 2]
['1', '4', 6, 1]
['1', '8', 24, 6]

下面给出了比您更适合大列表合并的时间：对于
tripInfo\u csv*500000
，时间分别为2秒和1分钟。我们使用dict获得几乎线性的复杂度来获得具有恒定查找时间的键。它也更优雅。请注意，
tg
是一个生成器，因此在创建时不会使用有效的时间或内存

def newGetTrips(trips):

    def convert(l):
        return l if l.isdigit() else dicPoints[l]

    tg = ((convert(a), convert(b), c, d) for a, b, c, d in trips)
    okt = {}
    for a, b, c, d in tg:
        # a trick to get (0,0) as default if (a,b) is not a key of the dictionary yet
        t = okt.get((a,b), (0,0)) 
        okt[(a,b)] = (t[0] + c, t[1] + d)
    return [[a,b,c,d] for (a,b), (c,d) in okt.iteritems()]

此外，作为一个副作用，您正在更改行程列表，而此功能使其保持不变。
还有，你有一个bug。每个（开始、结束）对（但不是第一种情况）考虑的第一项求和是两倍。我找不到原因，但在运行示例时，使用您的
getTrips
我得到：

[['1', '2', 6, 2], ['1', '8', 28, 8], ['1', '4', 12, 2]]

[['1', '8', 24, 6], ['1', '2', 6, 2], ['1', '4', 6, 1]]

通过
newGetTrips
我得到：

[['1', '2', 6, 2], ['1', '8', 28, 8], ['1', '4', 12, 2]]

[['1', '8', 24, 6], ['1', '2', 6, 2], ['1', '4', 6, 1]]

具体点。简化你的问题，很好。两条评论。不鼓励超过78个字符的行。你应该读政治公众人物8（）。此外，如果您使用filter而不是“if”语句，则可以降低圈复杂度，代码可读性更高，并且可以使用78个字符中的更多：）而不是
if file.endswith（“.csv”）
您可以编写
files=filter（lambda f:f.endswith（.csv”），文件）
实际上不需要排序，由于使用了groupby，所以添加了它。排序引入了O（n*log（n））时间复杂度。此解决方案比问题中发布的要快得多，但仍然可以改进。谢谢！我从您的代码中学到了很多（主要是operator.itemgetter和itertools.groupby）。我以后会用的。我喜欢它，但我发现eguaio更快，代码行更少。不幸的是，我不能将这个答案标记为有用（没有足够的声誉）。太好了！功能强大，速度快，行数很少。不知道生成器表达式，甚至不知道生成器。它真的很有用。我的数据和需求变得更加复杂，但无论如何，我还是设法将其全部融入其中。