有没有更好的方法可以将字符串转换为Python中的数据集？

Question

我刚刚在Python上完成了一个课程作业，运行得很好，我对此很满意，但代码看起来实在太丑了！我已经提交了这段代码，因为我们并不需要根据外观来评分，只要它能正常运行就行。不过，我希望能得到一些关于如何将字符串转换为数据集的建议，以便在未来的项目中使用。

输入是一个由节点和边组成的网格，举个例子：

"4:(1,2;4),(2,6;3),(3,7;15),(4,8;1),(5,6;1),(6,7;1),(5,9;9),(6,10;2),(7,11;1),(8,12;23),(9,10;5),(9,13;7),(10,14;6),(11,15;3),(12,16;3),(13,14;4),(15,16;7)"

在这个例子中，冒号前面的第一个数字是网格的大小（4x4），而(1,2;4)表示从节点1到节点2的边，费用是4。接下来的代码将这个输入转换成一个数组，其中array[0]是网格的大小，array[1]是一个字典，格式是(node1,node2)=费用。

def partitionData(line):
finalDic = dict()
#partition the data around the formating
line = line.split(":")
line[1] = line[1].split("),(")
#clean up data some more
line[1][0] = line[1][0][1:]
end = len(line[1])-1
line[1][end] = line[1][end][:len(line[1][end])-2]
#simplify data and organize into a list
for i in range(len(line[1])):
    line[1][i] = line[1][i].split(",")
    line[1][i][1] = line[1][i][1].split(";")
    #clean up list
    for j in range(len(line[1][i])):
        line[1][i].append(line[1][i][1][j])
    del line[1][i][1]
#convert everything to integer to simplify algorithm
for i in range(len(line[1])):
    for j in range(len(line[1][i])):
        line[1][i][j] = int(line[1][i][j])
line[0] = int(line[0])
newData = dict()
for i in range(len(line[1])):
    newData[(line[1][i][0],line[1][i][1])] = line[1][i][2]
line[1] = newData
for i in line[1]:
    if not ((min(i),max(i)) in finalDic):
        finalDic[(min(i),max(i))] = line[1][i]
    else:
        print "There is a edge referenced twice!"
        exit()  
line[1] = finalDic
return line

我一开始写的代码更简洁，但没有考虑到数字可能会大于9。我觉得现在的代码非常丑，肯定有更好看的方法来实现这个功能。

Answer 1

已经有人提出了一些解决方案：

• 使用解析器：太复杂了
• 使用正则表达式：我喜欢这个，但需要懂正则
• 使用ast模块：很有趣，但也需要了解它

。

我考虑用最简单的方法来解决这个问题，方便初学者理解。而且，我的解决方案表明，Python自带的功能就足够完成这个任务。

。

首先，我会给你展示一下你代码的修改版本，WhiteDawn，这样你就能看到一些非常基础的点，明白它们可以用Python的特性来简化。

比如说，seq是一个序列，seq[len(seq)-1]是它的最后一个元素，但seq[-1]也是最后一个元素。顺便提一下，你的代码里有个错误：我觉得是

line[1][end] = line[1][end][:len(line[1][end])-1]
# not:
line[1][end] = line[1][end][:len(line[1][end])-2]

否则在执行时会出错

另外要注意一个很棒的函数enumerate()

你还需要学习列表的切片：如果li = [45, 12, 78, 96]，那么li[2:3] = [2, 5, 8]会把li变成li = [45, 12, 2, 5, 8, 96]

y = "4:(1,2;4),(2,6;3),(3,7;15),(4,8;1),(5,6;1),(6,7;1),(5,9;9),(6,10;2),(7,11;1),(8,12;23),(9,10;5),(9,13;7),(10,14;6),(11,15;3),(12,16;3),(13,14;4),(15,16;7)"


def partitionData(line):
    finalDic = dict()

    #partition the data around the formating
    print 'line==',line
    line = line.split(":")
    print '\ninstruction :  line = line.split(":")'
    print 'line==',line
    print 'len of line==',len(line),'  (2 strings)'

    print '---------------------'
    line[1] = line[1].split("),(")
    print '\ninstruction :  line[1] = line[1].split("),(")'
    print 'line[1]==',line[1]

    #clean up data some more
    line[1][0] = line[1][0][1:]
    print 'instruction :  line[1][0] = line[1][0][1:]'
    line[1][-1] = line[1][-1][0:-1]
    print 'instruction :  line[1][-1] = line[1][-1][0:-1]'
    print 'line[1]==',line[1]

    print '---------------------'
    #simplify data and organize into a list
    for i,x in enumerate(line[1]):
        line[1][i] = x.split(",")
        line[1][i][1:] = line[1][i][1].split(";")
    print 'loop to clean the data in line[1]'
    print 'line[1]==',line[1]
    print '---------------------'
    #convert everything to integer to simplify algorithm
    print 'convert everything to integer to simplify algorithm'
    for i,x in enumerate(line[1]):
        line[1][i] = map(int,x)

    line[0] = int(line[0])
    print 'line==',line
    print '---------------------'
    newData = dict()
    for a,b,c in line[1]:
        newData[(a,b)] = c
    line[1] = newData
    print 'line==',line



    print '---------------------'
    for i in line[1]:
        print 'i==',i,'  (min(i),max(i))==',(min(i),max(i))
        if not ((min(i),max(i)) in finalDic):
            finalDic[(min(i),max(i))] = line[1][i]
        else:
            print "There is a edge referenced twice!"
            exit()
    line[1] = finalDic
    print '\nline==',line
    return line


print partitionData(y)

。

其次，我的解决方案：

y = "4:(1,2;4),(2,6;3),(3,7;15),(4,8;1),(5,6;1),(6,7;1),(5,9;9),(6,10;2),(7,11;1),(8,12;23),(9,10;5),(9,13;7),(10,14;6),(11,15;3),(12,16;3),(13,14;4),(15,16;7)"


# line[1]== {(1, 2): 4, (5, 9): 9, (2, 6): 3, (6, 7): 1, (4, 8): 1, (5, 6): 1, (6, 10): 2, (9, 10): 5, (13, 14): 4, (11, 15): 3, (10, 14): 6, (9, 13): 7, (12, 16): 3, (7, 11): 1, (3, 7): 15, (8, 12): 23, (15, 16): 7}

def partitionData(line):
    finalDic = dict()
    #partition the data around the formating
    print '\nline==',line

    line = line.split(":")
    print '\ninstruction:\n   line = line.split(":")'
    print 'result:\n   line==',line
    print '\n----------------------------------------------------'

    print '\nline[1]==',line[1]

    line[1] = line[1][1:-1].replace(";",",")
    print '\ninstruction:\n   line[1] = line[1][1:-1].replace(";",",")'
    print 'result:\n   line[1]==',line[1]

    line[1] = [ x.split(",") for x in line[1].split("),(") ]
    print '\ninstruction:\n   line[1] = [ x.split(",") for x in line[1].split("),(") ]'
    print 'result:\n   line[1]==',line[1]

    line = [int(line[0]),dict( ((int(a),int(b)),int(c)) for (a,b,c) in line[1] ) ]
    print '\ninstruction:\n   line = [int(line[0],dict( ((int(a),int(b)),int(c)) for (a,b,c) in line[1] ) ]'
    print 'result:\n   line[1]==',line[1]         


    for i in line[1]:
        if not ((min(i),max(i)) in finalDic):
            finalDic[(min(i),max(i))] = line[1][i]
        else:
            print "There is a edge referenced twice!"
            exit()
    line[1] = finalDic
    print '\nline[1]==',line[1]


    return line


print partitionData(y)

我让最后的FinalDict保持不变，因为我不明白这段代码的作用。如果i是一对整数，那么(min(i),max(i))其实就是这对整数本身。

Answer 2

import re
data = "4:(1,2;4),(2,6;3),(3,7;15),(4,8;1),(5,6;1),(6,7;1),(5,9;9),(6,10;2),(7,11;1),(8,12;23),(9,10;5),(9,13;7),(10,14;6),(11,15;3),(12,16;3),(13,14;4),(15,16;7)"
temp = data.split(":")    # split into grid size and rest
array = [int(temp[0]),{}] # first item: grid size
# split the rest of the string (from the second to the second-to-last characters)
# along the delimiters ");("
for item in temp[1][1:-1].split("),("):
    numbers = re.split("[,;]", item)          # split item along delimiters , or ;
    k1, k2, v = (int(num) for num in numbers) # and convert to int
    array[1][(k1,k2)] = v                     # populate the array
print array    

[4, {(1, 2): 4, (5, 9): 9, (2, 6): 3, (6, 7): 1, (4, 8): 1, (5, 6): 1, (6, 10):2, (9, 10): 5, (13, 14): 4, (11, 15): 3, (10, 14): 6, (9, 13): 7, (12, 16): 3, (7, 11): 1, (3, 7): 15, (8, 12): 23, (15, 16): 7}]

结果是

Answer 3

import re

# regular expression for matching a (node1,node2;cost)
EDGE = re.compile(r'\((\d+),(\d+);(\d+)\)')

def parse(s):
    # Separate size from the list of edges
    size, edges = s.split(':')

    # Build a dictionary
    edges = dict(
        # ...where key is (node1,node2) and value is (cost)
        # (all converted to integers)
        ((int(node1),int(node2)),int(cost))

        # ...by iterating the edges using the regular expression
        for node1,node2,cost in EDGE.findall(edges))

    return int(size),edges

例子：

>>> test = "4:(1,2;4),(2,6;3),(3,7;15),(4,8;1),(5,6;1),(6,7;1),(5,9;9),(6,10;2),(7,11;1),(8,12;23),(9,10;5),(9,13;7),(10,14;6),(11,15;3),(12,16;3),(13,14;4),(15,16;7)"
>>> parse(test)
(4, {(1, 2): 4, (5, 9): 9, (2, 6): 3, (6, 7): 1, (4, 8): 1, (5, 6): 1, (6, 10): 2, (9, 10): 5, (13, 14): 4, (11, 15): 3, (10, 14): 6, (9, 13): 7, (12, 16): 3, (7, 11): 1, (3, 7): 15, (8, 12): 23, (15, 16): 7})