Question

我有一个简单的代码，它读取csv文件，根据前2个列查找重复项，然后将重复项写入另一个csv并在第三个csv中保留唯一值...

我正在使用set：

def my_func():
    area = "W09"

    inf = r'f:\JDo\Cleaned\_merged\\'+ area +'.csv'
    out  = r'f:\JDo\Cleaned\_merged\no_duplicates\\'+area+'_no_duplicates.csv'
    out2 = r'f:\JDo\Cleaned\_merged\duplicates\\'+area+"_duplicates.csv"



    #i = 0
    seen = set()

    with open(inf, 'r') as infile, open(out, 'w') as outfile1, open(out2, 'w') as outfile2:
        reader = csv.reader(infile, delimiter=" ")
        writer1 = csv.writer(outfile1, delimiter=" ")
        writer2 = csv.writer(outfile2, delimiter=" ")
        for row in reader:
            x, y = row[0], row[1]

            x = float(x)
            y = float(y)

            if (x, y) in seen:

                writer2.writerow(row)
                continue
            seen.add((x, y))
            writer1.writerow(row)



    seen.clear()

我想，那套是最好的选择，但是套装的大小是输入文件大小的七倍？（输入文件范围从140 MB到50GB csv）和RAM使用率从1GB到近400 GB（我使用的服务器有768 GB的RAM）：

我还在小样本上使用了探查器

Line #    Mem usage    Increment   Line Contents

 8   21.289 MiB   21.289 MiB   @profile
 9                             def my_func():
10   21.293 MiB    0.004 MiB       area = "W10"
11
12   21.293 MiB    0.000 MiB       inf = r'f:\JDo\Cleaned\_merged\\'+ area +'.csv'
13   21.293 MiB    0.000 MiB       out  = r'f:\JDo\Cleaned\_merged\no_duplicates\\'+area+'_no_duplicates.csv'
14   21.297 MiB    0.004 MiB       out2 = r'f:\JDo\Cleaned\_merged\duplicates\\'+area+"_duplicates.csv"
15
16
17
18                                 #i = 0
19   21.297 MiB    0.000 MiB       seen = set()
20
21   21.297 MiB    0.000 MiB       with open(inf, 'r') as infile, open(out,'w') as outfile1, open(out2, 'w') as outfile2:
22   21.297 MiB    0.000 MiB           reader = csv.reader(infile, delimiter=" ")
23   21.297 MiB    0.000 MiB           writer1 = csv.writer(outfile1, delimiter=" ")
24   21.297 MiB    0.000 MiB           writer2 = csv.writer(outfile2, delimiter=" ")
25 1089.914 MiB   -9.008 MiB           for row in reader:
26 1089.914 MiB   -7.977 MiB               x, y = row[0], row[1]
27
28 1089.914 MiB   -6.898 MiB               x = float(x)
29 1089.914 MiB  167.375 MiB               y = float(y)
30
31 1089.914 MiB  166.086 MiB               if (x, y) in seen:
32                                             #z = line.split(" ",3)[-1]
33                                             #if z == "5284":
34                                             #    print X, Y, z
35
36 1089.914 MiB    0.004 MiB                   writer2.writerow(row)
37 1089.914 MiB    0.000 MiB                   continue
38 1089.914 MiB  714.102 MiB               seen.add((x, y))
39 1089.914 MiB   -9.301 MiB               writer1.writerow(row)
40
41
42
43  690.426 MiB -399.488 MiB       seen.clear()

可能是什么问题？是否有更快的方法来过滤掉结果？还是一种使用较少RAM方式的方法？

csv示例：我们正在考虑将GeoTIFF转换为csv文件，因此它是X Y值

    475596 101832 4926
    475626 101832 4926
    475656 101832 4926
    475686 101832 4926
    475716 101832 4926
    475536 101802 4926
    475566 101802 4926
    475596 101802 4926
    475626 101802 4926
    475656 101802 4926
    475686 101802 4926
    475716 101802 4926
    475746 101802 4926
    475776 101802 4926
    475506 101772 4926
    475536 101772 4926
    475566 101772 4926
    475596 101772 4926
    475626 101772 4926
    475656 101772 4926
    475686 101772 4926
    475716 101772 4926
    475746 101772 4926
    475776 101772 4926
    475806 101772 4926
    475836 101772 4926
    475476 101742 4926
    475506 101742 4926

编辑：所以我尝试了Jean提供的解决方案： https://stackoverflow.com/a/49008391/9418396

结果是，在我的140 MB csv的小集合上，集合的大小现在已经减半，这是一个很好的改进。我将尝试在更大的数据上运行它，看看它做了什么。我无法将它真正链接到探查器，因为探查器会延长执行时间。

Line #    Mem usage    Increment   Line Contents

 8   21.273 MiB   21.273 MiB   @profile
 9                             def my_func():
10   21.277 MiB    0.004 MiB       area = "W10"
11
12   21.277 MiB    0.000 MiB       inf = r'f:\JDo\Cleaned\_merged\\'+ area +'.csv'
13   21.277 MiB    0.000 MiB       out  = r'f:\JDo\Cleaned\_merged\no_duplicates\\'+area+'_no_duplicates.csv'
14   21.277 MiB    0.000 MiB       out2 = r'f:\JDo\Cleaned\_merged\duplicates\\'+area+"_duplicates.csv"
15
16
17   21.277 MiB    0.000 MiB       seen = set()
18
19   21.277 MiB    0.000 MiB       with open(inf, 'r') as infile, open(out,'w') as outfile1, open(out2, 'w') as outfile2:
20   21.277 MiB    0.000 MiB           reader = csv.reader(infile, delimiter=" ")
21   21.277 MiB    0.000 MiB           writer1 = csv.writer(outfile1, delimiter=" ")
22   21.277 MiB    0.000 MiB           writer2 = csv.writer(outfile2, delimiter=" ")
23  451.078 MiB -140.355 MiB           for row in reader:
24  451.078 MiB -140.613 MiB               hash = float(row[0])*10**7 + float(row[1])
25                                         #x, y = row[0], row[1]
26
27                                         #x = float(x)
28                                         #y = float(y)
29
30                                         #if (x, y) in seen:
31  451.078 MiB   32.242 MiB               if hash in seen:
32  451.078 MiB    0.000 MiB                   writer2.writerow(row)
33  451.078 MiB    0.000 MiB                   continue
34  451.078 MiB   78.500 MiB               seen.add((hash))
35  451.078 MiB -178.168 MiB               writer1.writerow(row)
36
37  195.074 MiB -256.004 MiB       seen.clear()

Answer 1

您可以创建自己的哈希函数以避免存储tuple浮点数，但浮动值将浮点数以独特的方式组合在一起。

让我们说坐标不能超过1000万（也许你可以减少到100万），你可以这样做：

hash = x*10**7 + y

（这会在您的花车上执行一种逻辑＆＃34; OR＆＃34;由于值有限，x和y之间不会混淆< / p>

然后将hash放入您的集合中，而不是tuple个浮点数。 10**14 >>> 10**14+1.5 100000000000001.5没有浮动吸收的风险，值得一试：

    for row in reader:
        hash = float(row[0])*10**7 + float(row[1])

        if hash in seen:
            writer2.writerow(row)
            continue
        seen.add(hash)
        writer1.writerow(row)

然后循环变为：

tuple

一个浮点数，即使很大（因为浮点数的大小是固定的），内存至少比2个浮点数的>>> sys.getsizeof((0.44,0.2)) 64 >>> sys.getsizeof(14252362*10**7+35454555.0) 24小2或3倍。在我的机器上：

//a[./preceding-sibling::h2[.='TAG1']][./following-sibling::h2[.='TAG2']]

什么记忆如此？

1 个答案: