需要比较python中1.5GB左右的非常大的文件

3条回答

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1楼 · 编辑于 2024-10-01 09:33:43

另一种可能的（系统管理）方法，避免数据库和SQL查询以及运行时进程和硬件资源的大量需求。在

Convert the timestamp到秒（从Epoch开始），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{}
在每一行进行交互，如果遇到新邮件，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，增量计数=1，更新上一个时间，添加“COUNT，Difference_in_days”
b）如果相同的电子邮件，相同的时间戳：increment COUNT，添加“COUNT，0 day”
c）如果是新邮件，从3开始。在

替代1。是添加一个新的字段时间戳，并在打印出行时将其删除。在

注意：如果1.5GB太大，无法一次性排序，请将其拆分为更小的chuck，使用电子邮件作为拆分点。你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公司名称：

"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
...

您可以将输出通过管道传输到Perl、Python或AWK脚本来处理步骤2。到4。在

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2楼 · 编辑于 2024-10-01 09:33:43

使用内置的sqlite3数据库：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。在

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3楼 · 编辑于 2024-10-01 09:33:43

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

这里有一个似乎有效的解决方案。工作流程如下：

1）从csv中按块读取数据并附加到HDF存储 2）对存储区的迭代，它创建另一个执行合并器的存储区

本质上，我们从表中获取一个块，并与文件其他部分的一个块进行组合。combiner函数不会减少，而是计算该块中所有元素之间的函数（以天为单位的差异），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。在

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'])

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