如何循环和比较线对?

时间:2017-02-15 18:17:44

标签: python python-3.x loops file-handling

我有2个这样的文件。

档案1

e 1 0 ppp
e 3 1 rrr
e 3 2 rrr

文件2

e 1 0 rrr

我想分割每行代码来比较数组[3]如果行相同的rrr,它应该打印OK。我写这样的代码。

file1 = open('file1.txt', 'r', encoding="utf8")
file2 = open('file2.txt', 'r')

for line1 in file1:     #loop file1

    L1 = line1.split()   #split to array

    if(L1[0] == 'e'):
        print("line1 " + line1)

        for line2 in file2:  #loop file2

            L2 = line2.split()   #split to array

            if(L2[0] == 'e'):
                print("line2 " + line2)

                if(L2[3] == L1[3]):  #check rrr same

                    print("OK")

当我运行代码时,它打印出来:

line1 e 1 0 ppp
line2 e 1 0 rrr
line1 e 3 1 rrr
line1 e 3 2 rrr

代码是循环循环我想显示如下结果:

line1 e 1 0 ppp
line2 e 1 0 rrr
line1 e 3 1 rrr
line2 e 1 0 rrr
line1 e 3 2 rrr
line2 e 1 0 rrr

如何修改代码以比较文件的每一行?

2 个答案:

答案 0 :(得分:0)

基本上你需要跟踪两个文件的最后一行。如果任何文件较小,则在另一个文件之前完成迭代。 您需要将当前行与另一个文件中的上一行进行比较。

怎么样?

prev_f1_token = None
prev_f1_line = None

prev_f2_token = None
prev_f2_line = None

with open('file1.txt') as f1:
    with open('file2.txt') as f2:
        for line in f1:
            prev_f1_token = line[0]
            prev_f1_line = line
            for line in f2:
                prev_f2_token = line[0]
                prev_f2_line = line

            if(prev_f1_token == 'e'):
                print("line1 ", prev_f1_line)

            if (prev_f2_token == 'e'):
                print ("line2 ", prev_f2_line)

                if(prev_f2_line.split()[3] == prev_f1_line.split()[3]):  #check rrr same
                        print("OK")

答案 1 :(得分:0)

你编写的代码多于它所需要的代码,如果你的代码得到了很多缩进,那么继续编写代码并不总是一个好主意。这是我试图打印你想要的东西,我相信这可以缩短更多!

您可以在此处使用itertools.cycle (如果列表较短,则返回循环迭代器)

import itertools
f1, f2 = open('file1.txt', 'r'), open('file2.txt', 'r')
file1, file2 = f1.read().split(), f2.read().split()
if len(file1) > len(file2):
    file2 = itertools.cycle(file2)
else:
    file1 = itertools.cycle(file1)  
print '\n'.join(['line1 '+i+'\nline2 '+j for i,j in zip(file1, file2)])
f1.close();f2.close()