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從 Python 中的大文件中刪除重復(fù)的行

Remove duplicate rows from a large file in Python(從 Python 中的大文件中刪除重復(fù)的行)
本文介紹了從 Python 中的大文件中刪除重復(fù)的行的處理方法,對(duì)大家解決問題具有一定的參考價(jià)值,需要的朋友們下面隨著小編來一起學(xué)習(xí)吧!

問題描述

我有一個(gè) csv 文件,我想從中刪除重復(fù)的行,但它太大而無法放入內(nèi)存.我找到了一種方法來完成它,但我的猜測(cè)是這不是最好的方法.

I've a csv file that I want to remove duplicate rows from, but it's too large to fit into memory. I found a way to get it done, but my guess is that it's not the best way.

每行包含 15 個(gè)字段和數(shù)百個(gè)字符,并且需要所有字段來確定唯一性.我不是比較整行來查找重復(fù)項(xiàng),而是比較 hash(row-as-a-string) 以嘗試節(jié)省內(nèi)存.我設(shè)置了一個(gè)過濾器,將數(shù)據(jù)劃分為大致相等的行數(shù)(例如一周中的幾天),并且每個(gè)分區(qū)足夠小,以至于該分區(qū)的哈希值查找表將適合內(nèi)存.我為每個(gè)分區(qū)傳遞一次文件,檢查唯一行并將它們寫入第二個(gè)文件(偽代碼):

Each row contains 15 fields and several hundred characters, and all fields are needed to determine uniqueness. Instead of comparing the entire row to find a duplicate, I'm comparing hash(row-as-a-string) in an attempt to save memory. I set a filter that partitions the data into a roughly equal number of rows (e.g. days of the week), and each partition is small enough that a lookup table of hash values for that partition will fit in memory. I pass through the file once for each partition, checking for unique rows and writing them out to a second file (pseudo code):

import csv

headers={'DayOfWeek':None, 'a':None, 'b':None}
outs=csv.DictWriter(open('c:dedupedFile.csv','wb')
days=['Mon','Tue','Wed','Thu','Fri','Sat','Sun']

outs.writerows(headers)

for day in days:
    htable={}
    ins=csv.DictReader(open('c:igfile.csv','rb'),headers)
    for line in ins:
        hvalue=hash(reduce(lambda x,y:x+y,line.itervalues()))
        if line['DayOfWeek']==day:
            if hvalue in htable:
                pass
            else:
                htable[hvalue]=None
                outs.writerow(line)

我想加快速度的一種方法是找到更好的過濾器來減少必要的通過次數(shù).假設(shè)行的長度是均勻分布的,也許不是

One way I was thinking to speed this up is by finding a better filter to reduce the number of passes necessary. Assuming the length of the rows is uniformly distributed, maybe instead of

for day in days: 

if line['DayOfWeek']==day:

我們有

for i in range(n):

if len(reduce(lambda x,y:x+y,line.itervalues())%n)==i:

在內(nèi)存允許的范圍內(nèi),'n' 盡可能小.但這仍然使用相同的方法.

where 'n' as small as memory will allow. But this is still using the same method.

Wayne Werner在下面提供了一個(gè)很好的實(shí)用解決方案;從算法的角度來看,我很好奇是否有更好/更快/更簡單的方法來做到這一點(diǎn).

Wayne Werner provided a good practical solution below; I was curious if there was better/faster/simpler way to do this from an algorithm perspective.

附:我僅限于 Python 2.5.

P.S. I'm limited to Python 2.5.

推薦答案

如果你想要一個(gè)非常簡單的方法來做到這一點(diǎn),只需創(chuàng)建一個(gè) sqlite 數(shù)據(jù)庫:

If you want a really simple way to do this, just create a sqlite database:

import sqlite3
conn = sqlite3.connect('single.db')
cur = conn.cursor()
cur.execute("""create table test(
f1 text,
f2 text,
f3 text,
f4 text,
f5 text,
f6 text,
f7 text,
f8 text,
f9 text,
f10 text,
f11 text,
f12 text,
f13 text,
f14 text,
f15 text,
primary key(f1,  f2,  f3,  f4,  f5,  f6,  f7,  
            f8,  f9,  f10,  f11,  f12,  f13,  f14,  f15))
"""
conn.commit()

#simplified/pseudo code
for row in reader:
    #assuming row returns a list-type object
    try:
        cur.execute('''insert into test values(?, ?, ?, ?, ?, ?, ?, 
                       ?, ?, ?, ?, ?, ?, ?, ?)''', row)
        conn.commit()
    except IntegrityError:
        pass

conn.commit()
cur.execute('select * from test')

for row in cur:
    #write row to csv file

那么您自己就不必?fù)?dān)心任何比較邏輯 - 只需讓 sqlite 為您處理.它可能不會(huì)比散列字符串快得多,但它可能要容易得多.當(dāng)然,如果需要,您可以修改存儲(chǔ)在數(shù)據(jù)庫中的類型,或者視情況而定.當(dāng)然,由于您已經(jīng)將數(shù)據(jù)轉(zhuǎn)換為字符串,因此您可以只使用一個(gè)字段.這里有很多選擇.

Then you wouldn't have to worry about any of the comparison logic yourself - just let sqlite take care of it for you. It probably won't be much faster than hashing the strings, but it's probably a lot easier. Of course you'd modify the type stored in the database if you wanted, or not as the case may be. Of course since you're already converting the data to a string you could just have one field instead. Plenty of options here.

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