需要在python中比较1.5GB左右的非常大的文件

时间：2013-04-19 17:24:21

标签： python csv numpy pandas large-data-volumes 

"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，约为1.5Gb

我希望在另一个csv文件中输出类似这样的

"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

即如果第一次进入，我需要追加1，如果它发生第二次我需要追加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更多次我想要日期之间的差异，并记住日期没有排序所以我们必须针对特定的电子邮件地址对它们进行排序，我正在寻找python使用numpy或pandas库或任何其他库的解决方案处理这种类型的巨大数据，而不会给出超出限制的内存异常我有双核处理器与centos 6.3和ram为4GB

3 个答案:

答案 0 :(得分：8)

确保你有0.11，请阅读这些文档：http://pandas.pydata.org/pandas-docs/dev/io.html#hdf5-pytables，以及这些食谱：http://pandas.pydata.org/pandas-docs/dev/cookbook.html#hdfstore（特别是'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：

1）通过块从csv读取数据并附加到hdfstore 2）对商店进行迭代，创建另一个存储组合

基本上，我们从表中获取一个块，并与文件的其他部分中的块组合。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。有点像递归几乎减少了。

这应该是O（num_of_chunks ** 2）内存和计算时间 在你的情况下，chunksize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'])

答案 1 :(得分：7)

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的文件也没有问题。

答案 2 :(得分：4)

另一种可能的（系统管理员）方式，避免数据库和SQL查询以及运行时进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： -

  1. Convert the timestamp到秒（来自Epoch）并使用UNIX sort，使用电子邮件和这个新字段（即：sort -k2 -k4 -n -t, < converted_input_file > output_file）
  2. 初始化3变量，EMAIL，PREV_TIME和COUNT
  3. 对每行进行交互，如果遇到新邮件，请添加“1,0天”。更新PREV_TIME=timestamp，COUNT=1，EMAIL=new_email
  4. 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，增加COUNT = 1，更新PREV_TIME，添加“Count，Difference_in_days”
    • b）如果同一封电子邮件，同一时间戳：增加COUNT，请添加“COUNT,0天”
    • c）如果是新电子邮件，请从3开始。

    5. 替代1.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太大而无法随意排序，请将其拆分为较小的卡盘，使用电子邮件作为分割点。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

    output_file.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
      ...

    将输出通过管道传输到Perl，Python或AWK脚本以处理步骤2.到步骤4。

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