[英]merge multiple files by reading them simultaneously line by line?
我有3个文件:
文件1:
chrM 6423 5
chrM 6432 4
chrM 7575 1
chrM 7670 1
chrM 7933 1
chrM 7984 1
chrM 8123 1
chrM 9944 1
chrM 10434 1
chrM 10998 13
chrM 10999 19
chrM 11024 17
chrM 11025 29
chrM 11117 21
chrM 11118 42
chr1 197095350 2
chr1 197103061 1
chr1 197103582 1
chr1 197103615 1
chr1 197103810 3
chr1 197103885 2
chr1 197104256 1
chr1 197107467 4
chr1 197107480 5
chr1 197107498 6
chr1 197107528 10
chr1 197107805 1
chr1 197107806 1
chr1 197107813 1
chr1 197107814 1
chr1 197107839 1
chr1 197107840 1
chr1 197107855 1
chr1 197107856 1
chr1 197107877 1
chr1 197107878 1
chr1 197111511 1
chr1 197120122 1
chr1 197125503 1
chr1 197126978 1
chr1 197127070 1
chr1 197127084 1
chr1 197129731 2
chr1 197129758 2
chr1 197129765 1
chr1 197167632 2
chr1 197167652 2
chr1 197167668 2
chr1 197167682 2
chr1 197181417 1
chr1 197181973 3
chr1 197181975 3
chr1 197192150 0
文件2:
chrM 6423 5
chrM 6432 4
chrM 6582 1
chrM 6640 1
chrM 6643 1
chrM 7140 1
chrM 10998 7
chrM 10999 8
chrM 11024 10
chrM 11025 13
chrM 11117 12
chrM 11118 33
chr1 197095157 2
chr1 197095185 2
chr1 197098860 1
chr1 197105061 1
chr1 197107422 1
chr1 197107436 1
chr1 197107467 3
chr1 197107480 4
chr1 197107498 3
chr1 197107528 4
chr1 197107805 2
chr1 197107813 2
chr1 197107839 1
chr1 197108557 1
chr1 197108591 1
chr1 197108596 1
chr1 197108617 1
chr1 197108651 1
chr1 197139308 1
chr1 197139335 1
chr1 197143403 1
chr1 197143442 1
chr1 197145546 1
chr1 197148715 1
chr1 197148723 1
chr1 197148731 1
chr1 197148761 1
chr1 197153190 1
chr1 197166831 1
chr1 197166847 2
chr1 197166922 2
chr1 197166950 1
chr1 197166954 1
chr1 197167041 1
chr1 197167778 1
chr1 197167791 1
chr1 197167834 1
chr1 197167857 2
chr1 197167860 2
chr1 197167865 1
chr1 197167867 1
chr1 197167871 1
chr1 197167935 2
chr1 197167946 2
chr1 197167948 2
chr1 197167951 2
chr1 197167974 1
chr1 197167980 1
chr1 197168142 1
chr1 197168163 1
chr1 197168195 1
chr1 197168210 1
chr1 197169548 1
chr1 197169580 1
chr1 197169609 1
chr1 197183318 1
chr1 197183404 1
chr1 197184910 1
chr1 197184937 1
chr1 197186368 1
chr1 197191991 1
chr1 197192031 1
chr1 197192047 1
chr1 197192097 1
chr1 197192106 1
chr1 197192125 1
chr1 197192150 1
文件3:
chrM 6423 2
chrM 6432 1
chrM 6766 1
chrM 6785 1
chrM 10075 1
chrM 10084 1
chrM 10998 7
chrM 10999 8
chrM 11024 7
chrM 11025 14
chrM 11117 8
chr1 197095943 1
chr1 197096144 1
chr1 197104061 1
chr1 197104257 1
chr1 197107805 2
chr1 197122470 1
chr1 197123085 1
chr1 197123093 1
chr1 197126978 1
chr1 197142562 1
chr1 197157076 1
chr1 197157101 2
chr1 197162035 4
chr1 197167431 1
chr1 197167470 1
chr1 197167535 1
chr1 197167652 1
chr1 197167668 1
chr1 197167682 1
chr1 197167715 1
chr1 197167734 1
chr1 197167755 1
chr1 197168107 2
chr1 197168113 2
chr1 197172198 1
chr1 197172211 1
chr1 197172221 1
chr1 197172271 1
chr1 197175787 1
chr1 197175806 1
chr1 197175822 1
chr1 197192150 0
结果文件应如下所示:
6423 chrM 2 5 5
6432 chrM 1 4 4
6582 chrM 1
197093370 chr1 1
197093385 chr1 1
197094791 chr1 1
197094813 chr1 1
197094855 chr1 1
197094857 chr1 1
197095157 chr1 2
197095185 chr1 2
197095350 chr1 2
197095943 chr1 1
197096
现在我的代码可以正常工作了,但是在while循环中有一个issu,在合并了许多记录后,几乎在合并文件的末尾停止了对该文件的写操作,而只是写了197096 ....最后调用):文件“”,第4行,在IndexError中:列表索引超出范围
我认为这个错误与while循环有关。我不知道为什么会这样。我也在更改我的代码,如下所示:
看看她来的问题:您可以在结果文件中清楚地看到,在这种情况下,正在发生一些事情,即从单个文件读取后,代码无法从所有文件中读取公共值,并且在这种情况下,它没有给出应有的7575 7140之后
我有多个大文件,如果它们对于第2列都具有相同的值,我想逐行读取它们并将它们合并在一起,为此我使用了将所有第二列val放在列表中然后发现其中的最小值。 从文件中写入最小值的记录(保存在mycover中的第3列)将显示最小值的值记录到新文件中。 然后在my_newfile[]
跟踪读取的文件以从文件中读取下一行,并删除已写入文件的记录。
希望足以理解。 我不知道如何重复该过程,直到所有文件都到达末尾,以便从所有文件中读取所有记录。 我的代码如下:
import sys
import glob
import errno
path = '*Sorted_Coverage.txt'
filenames = glob.glob(path)
files = [open(i, "r") for i in filenames]
p=1
mylist=[]
mychr=[]
mycover=[]
new_mychr=[]
new_mycover=[]
new_mylist=[]
myfile=[]
new_myfile=[]
ab=""
g=1
result_f = open('MERGING_water_onlyselected.txt', 'a')
for j in files:
line = j.readline()
parts = line.split()
mychr.append(parts[0])
mycover.append(parts[2])
mylist.append(parts[1])
myfile.append(j)
mylist=map(int,mylist)
minval = min(mylist)
ind = [i for i, v in enumerate(mylist) if v == minval]
not_ind = [i for i, v in enumerate(mylist) if v != minval]
w=""
j=0
for j in xrange(len(ind)): # writing records to file with minimum value
if(j==0):
ab = (str(mylist[ind[j]])+'\t'+mychr[ind[j]]+'\t'+mycover[ind[j]])
else:
ab=ab+'\t'+mycover[ind[j]]
#smallest written on file
result_f.writelines(ab+'\n')
ab=""
for i in ind:
new_myfile.append(myfile[i])
#removing the records by index which have been used from mylists .
for i in sorted(ind, reverse=True):
del mylist[i]
del mycover[i]
del mychr[i]
del myfile[i]
#how to iterate the following code from all records of all files till the end of each file
while(True):
for i in xrange(len(new_myfile)):
print len(new_myfile)
myfile.append(new_myfile[i])
line = new_myfile[i].readline()
parts = line.split()
mychr.append(parts[0])
mycover.append(parts[2])
mylist.append(parts[1])
new_myfile=[]
mylist=map(int, mylist)
minval = min(mylist)
print minval
print("list values:")
print mylist
ind = [i for i, v in enumerate(mylist) if v == minval]
not_ind = [i for i, v in enumerate(mylist) if v != minval]
k=0
ab=""
for j in xrange(len(ind)): # writing records to file with minimum value
if(j==0):
ab = (str(mylist[ind[j]])+'\t'+str(mychr[ind[j]])+'\t'+str(mycover[ind[j]]))
k=k+1
else:
ab=ab+'\t'+str(mycover[ind[j]])
k=k+1
#smallest written on file
result_f.writelines(ab+'\n')
ab=""
for i in ind:
new_myfile.append(myfile[i])
#removing the records by index which have been used from mylists .
for i in sorted(ind, reverse=True):
del mylist[i]
del mycover[i]
del mychr[i]
del myfile[i]
result_f.close()
我一直在寻找解决方案很多天,但仍然找不到任何解决方案。 我不知道该代码是否可以进一步改进,因为我对python很陌生。
如果有人可以提供帮助,我将不胜感激。
这是一个非常简单的方法。 我不知道它在大文件上的表现如何(请参阅下面的评论)。
我假设所有文件都已经针对第二列进行了排序 。 另外,我假设第一列签名('chrM','chr1')对于第二列中的固定值保持不变(我在下面将其称为“ id”)。
该算法很简单:
从每个文件读取一行(我称读取行为“项”)
选择一个具有最小“ id”(任意一个)的“ item”,并将其与“ current_item”进行比较:
如果两者都具有相同的ID:将它们组合在一起:将“ current_item”写入文件并将其替换为“ item”
从与读取“ item”相同的文件中读取一行(如果还有任何一行)
从1.开始重复,直到读取所有文件中的所有行。
import glob
import numpy as np
path = './file[0-9]*'
filenames = glob.glob(path)
files = [open(i, "r") for i in filenames]
output_file = open('output_file', mode = 'a')
# last_ids[i] = last id number read from files[i]
# I choose np.array because of function np.argmin
last_ids = np.ones(shape = len(files)) * np.inf
last_items = [None] *len(files)
# Note: When we hit EOF in a file, the corresponding entries from "files", "last_items", and "last_ids" will be deleted
for i in range(len(files)):
line = files[i].readline()
if line:
item = line.strip().split()
last_ids[i] = int(item[1])
last_items[i] = item
# Find an item with the smallest id
pos = np.argmin(last_ids)
current_item = last_items[pos]
# Inverting positions, so that id is first
current_item[0], current_item[1] = current_item[1], current_item[0]
while True:
# Read next item from the corresponding file
line = files[pos].readline()
if line:
item = line.strip().split()
last_ids[pos] = int(item[1])
last_items[pos] = item
else:
# EOF in files[pos], so delete it from the lists
files[pos].close()
del(files[pos])
del(last_items[pos])
last_ids = np.delete(last_ids, pos)
if last_ids.size == 0:
# No more files to read from
break
# Find an item with the smallest id
pos = np.argmin(last_ids)
if last_items[pos][1] == current_item[0]:
# combine:
current_item.append(last_items[pos][2])
else:
# write current to file and replace:
output_file.write(' '.join(current_item) + '\n')
current_item = last_items[pos]
current_item[0], current_item[1] = current_item[1], current_item[0]
# The last item to write:
output_file.write(' '.join(current_item) + '\n')
output_file.close()
如果所有文件都足够小以适合内存,那么以下代码肯定会更短。 是否更快可能取决于数据。 (请参阅下面的评论。)
import glob
import pandas as pd
path = './file[0-9]*'
filenames = glob.glob(path)
df_list = []
# Read in all files and concatenate to a single data frame:
for file in filenames:
df_list.append(pd.read_csv(file, header = None, sep = '\s+'))
df = pd.concat(df_list)
# changing type for convenience:
df[2] = df[2].astype(str)
# sorting here is not necessary:
# df = df.sort_values(by = 1)
df2 = df.groupby(by = 1).aggregate({0:'first', 2: lambda x: ' '.join(x)})
df2.to_csv('output_file', header = None)
# (Columns in 'output_file' are separated by commas. )
我在1000-10000行的几个输入文件上测试了这两种解决方案。 通常,基本解决方案速度更快(有时是另一种解决方案速度的两倍)。 但这取决于数据的结构。 如果有很多重复的“ id”,那么熊猫可能会更具优势(相差很小)。
我认为这两种方法都可以与带有选项chunksize
或iterator
的pd.read_csv
结合使用。 这样,我们就可以读入更大的数据块并对其进行操作(而不是单行)。 但是我现在不确定它是否会导致更快的代码。
如果失败(并且没有人找到更好的方法),则可以考虑在Amazon Web Services上运行map reduce算法。 在一开始有一些工作可以解决所有设置,但是map-reduce算法对于这类问题非常简单。
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