python - 需要在 python 中比较 1.5GB 左右的非常大的文件

Question

"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"

以上是样本数据。数据根据电子邮件地址排序，文件非常大，大约 1.5Gb

我想在另一个 csv 文件中输出类似这样的内容

"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

即，如果条目第一次出现，我需要附加 1 如果它出现第二次，我需要附加 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更多，我想要差异在日期和记住日期之间没有排序，所以我们还必须根据特定的电子邮件地址对它们进行排序，我正在使用 numpy 或 pandas 库或任何其他可以处理这种类型的海量数据而不会放弃的库在 python 中寻找解决方案绑定内存异常我有双核处理器，centos 6.3，内存为 4GB

score 8 · Accepted Answer

确保你有 0.11，阅读这些文档：http ://pandas.pydata.org/pandas-docs/dev/io.html#hdf5-pytables和这些食谱： http: //pandas.pydata.org/pandas- docs/dev/cookbook.html#hdfstore（尤其是“合并数百万行”

这是一个似乎有效的解决方案。这是工作流程：

按块从 csv 读取数据并附加到 hdfstore
遍历存储，这将创建另一个执行组合器的存储

本质上，我们从表中取出一个块，并与文件其他每个部分的一个块组合在一起。组合器函数不会减少，而是计算该块中所有元素之间的函数（以天为单位的差异），消除重复项，并在每次循环后获取最新数据。几乎有点像递归减少。

在您的情况下，这应该是 O(num_of_chunks**2) 内存和计算时间块大小可以说是 1m（或更多）

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'])

score 7 · Accepted Answer

使用内置的sqlite3数据库：您可以根据需要插入数据、排序和分组，使用大于可用 RAM 的文件没有问题。

score 4 · Accepted Answer

另一种可能的（系统管理员）方式，避免数据库和 SQL 查询以及对运行时进程和硬件资源的大量要求。

更新 20/04添加了更多代码和简化方法：-

将时间戳转换为秒（来自 Epoch）并使用 UNIX sort，使用 email 和这个新字段（即sort -k2 -k4 -n -t, < converted_input_file > output_file：）
初始化 3 个变量EMAIL,PREV_TIME和COUNT
遍历每一行，如果遇到新电子邮件，添加“1,0 天”。更新PREV_TIME=timestamp，，，COUNT=1_EMAIL=new_email
下一行：3 种可能的情况
- a) 如果相同的电子邮件，不同的时间戳：计算天数，递增 COUNT=1，更新 PREV_TIME，添加“Count, Difference_in_days”
- b）如果相同的电子邮件，相同的时间戳：增加 COUNT，添加“COUNT，0 天”
- c) 如果是新邮件，从 3 开始。

1. 的替代方法是添加一个新字段 TIMESTAMP 并在打印出该行时将其删除。

注意：如果 1.5GB 太大而无法一次排序，请将其拆分为更小的卡盘，使用电子邮件作为拆分点。您可以在不同的机器上并行运行这些块

/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

输出文件.txt：

"DF","00000000@11111.COM","FLTINT1000130394756","26JUL2010","B2C","6799.2",1280102400 "DF","0001HARISH@GMAIL.COM","NF252022031180","09DEC2010"," B2C","3439",1291852800 "DF","0001HARISH@GMAIL.COM","NF251742087846","12DEC2010","B2C","1000",1292112000 "DF","0001HARISH@GMAIL.COM"," NF251352240086","22DEC2010","B2C","4006",1292976000
...

您将输出通过管道传输到 Perl、Python 或 AWK 脚本以处理步骤 2 到 4。

python - 需要在 python 中比较 1.5GB 左右的非常大的文件

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