需要比较python中1.5GB左右的超大文件

以上是示例数据。 数据根据电子邮件地址进行排序，文件非常大，大约为1.5Gb

我希望在另一个csv文件中输出类似的内容

"DF","00000000@11111.COM","FLTINT1000130394756","26JUL2010","B2C","6799.2"
"Rail","00000.POO@GMAIL.COM","NR251764697478","24JUN2011","B2C","2025"
"DF","0000650000@YAHOO.COM","NF2513521438550","01JAN2013","B2C","6792"
"Bus","00009.GAURAV@GMAIL.COM","NU27012932319739","26JAN2013","B2C","800"
"Rail","0000.ANU@GMAIL.COM","NR251764697526","24JUN2011","B2C","595"
"Rail","0000MANNU@GMAIL.COM","NR251277005737","29OCT2011","B2C","957"
"Rail","0000PRANNOY0000@GMAIL.COM","NR251297862893","21NOV2011","B2C","212"
"DF","0000PRANNOY0000@YAHOO.CO.IN","NF251327485543","26JUN2011","B2C","17080"
"Rail","0000RAHUL@GMAIL.COM","NR2512012069809","25OCT2012","B2C","5731"
"DF","0000SS0@GMAIL.COM","NF251355775967","10MAY2011","B2C","2000"
"DF","0001HARISH@GMAIL.COM","NF251352240086","22DEC2010","B2C","4006"
"DF","0001HARISH@GMAIL.COM","NF251742087846","12DEC2010","B2C","1000"
"DF","0001HARISH@GMAIL.COM","NF252022031180","09DEC2010","B2C","3439"
"Rail","000AYUSH@GMAIL.COM","NR2151120122283","25JAN2013","B2C","136"
"Rail","000AYUSH@GMAIL.COM","NR2151213260036","28NOV2012","B2C","41"
"Rail","000AYUSH@GMAIL.COM","NR2151313264432","29NOV2012","B2C","96"
"Rail","000AYUSH@GMAIL.COM","NR2151413266728","29NOV2012","B2C","96"
"Rail","000AYUSH@GMAIL.COM","NR2512912359037","08DEC2012","B2C","96"
"Rail","000AYUSH@GMAIL.COM","NR2517612385569","12DEC2012","B2C","96"

---

i、 e如果条目第一次出现，我需要追加1如果第二次出现，我需要追加2，同样，我的意思是我需要计算文件中电子邮件地址的出现次数，如果电子邮件存在两次或两次以上，我希望日期之间存在差异，请记住，日期未排序，因此我们还必须根据特定的日期对其进行排序电子邮件地址，我正在寻找一个python解决方案，使用numpy或pandas库或任何其他库来处理这种类型的海量数据，而不会出现内存不足的情况。我有centos 6.3双核处理器，内存为4GB。使用内置数据库：您可以根据需要插入数据、排序和分组，使用比可用RAM大的文件是没有问题的。

确保您有0.11，阅读以下文档：，以及以下方法：（特别是“合并数百万行”

以下是一个似乎有效的解决方案。以下是工作流程：

1） 按块从csv读取数据并附加到hdfstore 2） 对存储的迭代，创建另一个执行合并器的存储

本质上，我们从表中获取一个块，并与文件其他部分的块组合。combiner函数不会减少，而是计算该块中所有元素之间的函数（以天为单位的差值），在执行过程中消除重复项，并在每次循环后获取最新数据。有点像递归的reduce

这应该是O（块的数量**2）内存和计算时间 在您的情况下，chunksize可以是1m（或更多）

"DF","00000000@11111.COM","FLTINT1000130394756","26JUL2010","B2C","6799.2",1,0 days
"Rail","00000.POO@GMAIL.COM","NR251764697478","24JUN2011","B2C","2025",1,0 days
"DF","0000650000@YAHOO.COM","NF2513521438550","01JAN2013","B2C","6792",1,0 days
"Bus","00009.GAURAV@GMAIL.COM","NU27012932319739","26JAN2013","B2C","800",1,0 days
"Rail","0000.ANU@GMAIL.COM","NR251764697526","24JUN2011","B2C","595",1,0 days
"Rail","0000MANNU@GMAIL.COM","NR251277005737","29OCT2011","B2C","957",1,0 days
"Rail","0000PRANNOY0000@GMAIL.COM","NR251297862893","21NOV2011","B2C","212",1,0 days
"DF","0000PRANNOY0000@YAHOO.CO.IN","NF251327485543","26JUN2011","B2C","17080",1,0 days
"Rail","0000RAHUL@GMAIL.COM","NR2512012069809","25OCT2012","B2C","5731",1,0 days
"DF","0000SS0@GMAIL.COM","NF251355775967","10MAY2011","B2C","2000",1,0 days
"DF","0001HARISH@GMAIL.COM","NF251352240086","09DEC2010","B2C","4006",1,0 days
"DF","0001HARISH@GMAIL.COM","NF251742087846","12DEC2010","B2C","1000",2,3 days
"DF","0001HARISH@GMAIL.COM","NF252022031180","22DEC2010","B2C","3439",3,10 days
"Rail","000AYUSH@GMAIL.COM","NR2151213260036","28NOV2012","B2C","41",1,0 days
"Rail","000AYUSH@GMAIL.COM","NR2151313264432","29NOV2012","B2C","96",2,1 days
"Rail","000AYUSH@GMAIL.COM","NR2151413266728","29NOV2012","B2C","96",3,0 days
"Rail","000AYUSH@GMAIL.COM","NR2512912359037","08DEC2012","B2C","96",4,9 days
"Rail","000AYUSH@GMAIL.COM","NR2512912359037","08DEC2012","B2C","96",5,0 days
"Rail","000AYUSH@GMAIL.COM","NR2517612385569","12DEC2012","B2C","96",6,4 days
"Rail","000AYUSH@GMAIL.COM","NR2517612385569","12DEC2012","B2C","96",7,0 days
"Rail","000AYUSH@GMAIL.COM","NR2151120122283","25JAN2013","B2C","136",8,44 days
"Rail","000AYUSH@GMAIL.COM","NR2151120122283","25JAN2013","B2C","136",9,0 days

另一种可能的（系统管理）方式，避免数据库和SQL查询，以及运行时流程和硬件资源中的大量需求

更新20/04添加了更多代码和简化方法：-

使用电子邮件和这个新字段（即：

sort-k2-k4-n-t，output\u file

）使用UNIX

sort

初始化3个变量，

EMAIL

，

PREV\u TIME

和

COUNT

在每行上交互，如果遇到新电子邮件，请添加“1,0天”。更新

PREV\u TIME=timestamp

，

COUNT=1

，

EMAIL=new\u EMAIL

下一行：3种可能的情况

• a） 如果是同一封电子邮件，不同的时间戳：计算天数，增量计数=1，更新上一次时间，添加“计数，差异天数”
• b） 如果相同的电子邮件，相同的时间戳：递增计数，添加“计数，0天”
• c） 如果是新邮件，从3开始

备选方案1。是添加一个新的字段时间戳，并在打印出该行时将其删除

注意：如果1.5GB太大，无法一次排序，请使用电子邮件作为拆分点，将其拆分为较小的chuck。您可以在不同的机器上并行运行这些块

processing [0] [datastore.h5]
processing [1] [datastore_0.h5]
    count                date  diff                        email
4       1 2011-06-24 00:00:00     0           0000.ANU@GMAIL.COM
1       1 2011-06-24 00:00:00     0          00000.POO@GMAIL.COM
0       1 2010-07-26 00:00:00     0           00000000@11111.COM
2       1 2013-01-01 00:00:00     0         0000650000@YAHOO.COM
3       1 2013-01-26 00:00:00     0       00009.GAURAV@GMAIL.COM
5       1 2011-10-29 00:00:00     0          0000MANNU@GMAIL.COM
6       1 2011-11-21 00:00:00     0    0000PRANNOY0000@GMAIL.COM
7       1 2011-06-26 00:00:00     0  0000PRANNOY0000@YAHOO.CO.IN
8       1 2012-10-25 00:00:00     0          0000RAHUL@GMAIL.COM
9       1 2011-05-10 00:00:00     0            0000SS0@GMAIL.COM
12      1 2010-12-09 00:00:00     0         0001HARISH@GMAIL.COM
11      2 2010-12-12 00:00:00     3         0001HARISH@GMAIL.COM
10      3 2010-12-22 00:00:00    13         0001HARISH@GMAIL.COM
14      1 2012-11-28 00:00:00     0           000AYUSH@GMAIL.COM
15      2 2012-11-29 00:00:00     1           000AYUSH@GMAIL.COM
17      3 2012-12-08 00:00:00    10           000AYUSH@GMAIL.COM
18      4 2012-12-12 00:00:00    14           000AYUSH@GMAIL.COM
13      5 2013-01-25 00:00:00    58           000AYUSH@GMAIL.COM
import pandas as pd
import StringIO
import numpy as np
from time import strptime
from datetime import datetime

# your data
data = """
"DF","00000000@11111.COM","FLTINT1000130394756","26JUL2010","B2C","6799.2"
"Rail","00000.POO@GMAIL.COM","NR251764697478","24JUN2011","B2C","2025"
"DF","0000650000@YAHOO.COM","NF2513521438550","01JAN2013","B2C","6792"
"Bus","00009.GAURAV@GMAIL.COM","NU27012932319739","26JAN2013","B2C","800"
"Rail","0000.ANU@GMAIL.COM","NR251764697526","24JUN2011","B2C","595"
"Rail","0000MANNU@GMAIL.COM","NR251277005737","29OCT2011","B2C","957"
"Rail","0000PRANNOY0000@GMAIL.COM","NR251297862893","21NOV2011","B2C","212"
"DF","0000PRANNOY0000@YAHOO.CO.IN","NF251327485543","26JUN2011","B2C","17080"
"Rail","0000RAHUL@GMAIL.COM","NR2512012069809","25OCT2012","B2C","5731"
"DF","0000SS0@GMAIL.COM","NF251355775967","10MAY2011","B2C","2000"
"DF","0001HARISH@GMAIL.COM","NF251352240086","22DEC2010","B2C","4006"
"DF","0001HARISH@GMAIL.COM","NF251742087846","12DEC2010","B2C","1000"
"DF","0001HARISH@GMAIL.COM","NF252022031180","09DEC2010","B2C","3439"
"Rail","000AYUSH@GMAIL.COM","NR2151120122283","25JAN2013","B2C","136"
"Rail","000AYUSH@GMAIL.COM","NR2151213260036","28NOV2012","B2C","41"
"Rail","000AYUSH@GMAIL.COM","NR2151313264432","29NOV2012","B2C","96"
"Rail","000AYUSH@GMAIL.COM","NR2151413266728","29NOV2012","B2C","96"
"Rail","000AYUSH@GMAIL.COM","NR2512912359037","08DEC2012","B2C","96"
"Rail","000AYUSH@GMAIL.COM","NR2517612385569","12DEC2012","B2C","96"
"""


# read in and create the store
data_store_file = 'datastore.h5'
store = pd.HDFStore(data_store_file,'w')

def dp(x, **kwargs):
    return [ datetime(*strptime(v,'%d%b%Y')[0:3]) for v in x ]

chunksize=5
reader = pd.read_csv(StringIO.StringIO(data),names=['x1','email','x2','date','x3','x4'],
                     header=0,usecols=['email','date'],parse_dates=['date'],
                     date_parser=dp, chunksize=chunksize)

for i, chunk in enumerate(reader):
    chunk['indexer'] = chunk.index + i*chunksize

    # create the global index, and keep it in the frame too
    df = chunk.set_index('indexer')

    # need to set a minimum size for the email column
    store.append('data',df,min_itemsize={'email' : 100})

store.close()

# define the combiner function
def combiner(x):

    # given a group of emails (the same), return a combination
    # with the new data

    # sort by the date
    y = x.sort('date')

    # calc the diff in days (an integer)
    y['diff'] = (y['date']-y['date'].iloc[0]).apply(lambda d: float(d.item().days))
    y['count'] = pd.Series(range(1,len(y)+1),index=y.index,dtype='float64')  

    return y

# reduce the store (and create a new one by chunks)
in_store_file = data_store_file
in_store1 = pd.HDFStore(in_store_file)

# iter on the store 1
for chunki, df1 in enumerate(in_store1.select('data',chunksize=2*chunksize)):
    print "processing [%s] [%s]" % (chunki,in_store_file)

    out_store_file = 'datastore_%s.h5' % chunki
    out_store = pd.HDFStore(out_store_file,'w')

    # iter on store 2
    in_store2 = pd.HDFStore(in_store_file)
    for df2 in in_store2.select('data',chunksize=chunksize):

        # concat & drop dups
        df = pd.concat([df1,df2]).drop_duplicates(['email','date'])

        # group and combine
        result = df.groupby('email').apply(combiner)

        # remove the mi (that we created in the groupby)
        result = result.reset_index('email',drop=True)

        # only store those rows which are in df2!
        result = result.reindex(index=df2.index).dropna()

        # store to the out_store
        out_store.append('data',result,min_itemsize={'email' : 100})
    in_store2.close()
    out_store.close()
    in_store_file = out_store_file

in_store1.close()

# show the reduced store
print pd.read_hdf(out_store_file,'data').sort(['email','diff'])

/usr/bin/gawk-F'''''''''{
拆分（“1-2-3-4-5-6-7-8-9-10-11-12”，月“）；
对于（i=1；i将它们放入数据库。按名称排序，然后按日期排序。这听起来像是按照Map Reduce方法做的事情。我的第一个想法也是将数据放入数据库中，但它不会起作用。我还需要跟踪每个电子邮件地址的首次出现并进行计数，那么如何获得日期的差异？解决方法是什么？？@Jeff我的文件有超过2000万行，唯一值超过500万如果我开始比较这些，我将得到500万*2000万的复杂度，这可能需要几个月才能解决问题我如何从n*n减少复杂度，以便我能够处理如此大的容量第一次出现电子邮件时，它的电子邮件就是参考后续出现的日期？对于第二封电子邮件，很简单，计数为1，天数为天数之差，第三封电子邮件如何。天数是否更新为第三天和第一天之间的差值，或者是否涉及天数（可能第三天是当前日期和第三天-参考日期的最大值？）这将对我已排序的输出执行计数和天数计算。计算完成。谢谢！我仅根据应用了拆分[root@amanka桌面]#gawk'BEGIN{OFS=“，”；COUNT=0；PREV_TIME=0；EMAIL=0；while（（getline第0行）{split（line，a，，”）if（EMAIL！=a[2]）{EMAIL=a[2]；COUNT=1；PREV_TIME=a[7]；打印行，“1,0天”}else{if（PREV_TIME==a[7]）{COUNT=COUNT+1；打印行，COUNT，“0天”；}else{DAYS=（（a[7]-PREV_TIME）/（60*60*24））；PREV_TIME=a[7]；COUNT=COUNT+1；打印行，COUNT，DAYS“天”；}}}}不客气。我很想知道1）gawk+sort使用了多少内存在1.5gb文件上花费了多少时间？我不确定内存，但在10-12分钟左右进行输出排序所花费的时间实际上非常少，大约需要15分钟，而且它比任何其他语言解决方案都要快得令人难以置信，我甚至想单次解析文件O（n）在python中，大约需要35分钟，使用shell脚本将时间减少到一半。对于同一个文件，我有更多的问题。如果您能帮助我，您可以在
/usr/bin/gawk -F'","' ' { 
    split("JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC", month, " "); 
    for (i=1; i<=12; i++) mdigit[month[i]]=i; 
    print $0 "," mktime(substr($4,6,4) " " mdigit[substr($4,3,3)] " " substr($4,1,2) " 00 00 00"
)}' < input.txt |  /usr/bin/sort -k2 -k7 -n -t, > output_file.txt