[已解决]新手python求助，逐行比较，输出不同类型，急急急。。。。

poppy章鱼

求大佬帮忙，第二次发帖，急急急。。。。

数据如下，第一列是顺序排列的
scaffold_1..12548467    6
scaffold_1..12548468    6
scaffold_1..12548469    6
scaffold_1..12548470    6
scaffold_1..12548471    6
scaffold_1..12548472    6
scaffold_1..12548473    5
scaffold_1..12548474    5
scaffold_1..12548475    6
scaffold_1..12548476    6
scaffold_1..12548477    6
scaffold_1..12548478    5
scaffold_1..12548479    4
scaffold_1..12548480    4
scaffold_1..12548481    4
scaffold_1..12548482    4
scaffold_1..12548483    4




想要的结果输出格式
scaffold_1..12548467        scaffold_1..12548472        6        6
scaffold_1..12548473        scaffold_1..12548474        5        2
scaffold_1..12548475         scaffold_1..12548477        6        3
scaffold_1..12548478                                                5        1
scaffold_1..12548479        scaffold_1..12548483        4        5


##
逐行读取, 比如原数据中第二列为6，总共连续出现了6行第二列都是6，,就输出一行，格式为: start,end,6,频次  

然后后面再出现6 ，又重新输出，start，end,6,频次


如果第二列是1，则只输出start



感激不尽~~~

最佳答案

月排行榜 / 总排行榜

hrpzcf

2021-11-12 14:20:49

1. # coding: utf-8
   
2. 
   
3. # string = """scaffold_1..12548467    6
   
4. # scaffold_1..12548468    6
   
5. # scaffold_1..12548469    6
   
6. # scaffold_1..12548470    6
   
7. # scaffold_1..12548471    6
   
8. # scaffold_1..12548472    6
   
9. # scaffold_1..12548473    5
   
10. # scaffold_1..12548474    5
    
11. # scaffold_1..12548475    6
    
12. # scaffold_1..12548476    6
    
13. # scaffold_1..12548477    6
    
14. # scaffold_1..12548478    5
    
15. # scaffold_1..12548479    4
    
16. # scaffold_1..12548480    4
    
17. # scaffold_1..12548481    4
    
18. # scaffold_1..12548482    4
    
19. # scaffold_1..12548483    4"""
    
20. 
    
21. 
    
22. with open("data.txt", "rt", encoding="utf-8") as f:
    
23.     # slist = string.splitlines()
    
24.     slist = f.read().splitlines()
    
25.     freq = 1
    
26.     sp = slist[0].split()
    
27.     start, num = sp[0], sp[-1]
    
28.     end = start
    
29.     for s in slist[1:]:
    
30.         if s[-1] == num:
    
31.             freq += 1
    
32.             end = s.split()[0]
    
33.         else:
    
34.             if freq == 1:
    
35.                 end = " " * len(end)
    
36.             print("{} {} {} {}".format(start, end, num, freq))
    
37.             freq = 1
    
38.             sp = s.split()
    
39.             start, num = sp[0], sp[-1]
    
40.             end = start
    
41.     else:
    
42.         print("{} {} {} {}".format(start, end, num, freq))
    

复制代码

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hrpzcf

1. # coding: utf-8
   
2. 
   
3. # string = """scaffold_1..12548467    6
   
4. # scaffold_1..12548468    6
   
5. # scaffold_1..12548469    6
   
6. # scaffold_1..12548470    6
   
7. # scaffold_1..12548471    6
   
8. # scaffold_1..12548472    6
   
9. # scaffold_1..12548473    5
   
10. # scaffold_1..12548474    5
    
11. # scaffold_1..12548475    6
    
12. # scaffold_1..12548476    6
    
13. # scaffold_1..12548477    6
    
14. # scaffold_1..12548478    5
    
15. # scaffold_1..12548479    4
    
16. # scaffold_1..12548480    4
    
17. # scaffold_1..12548481    4
    
18. # scaffold_1..12548482    4
    
19. # scaffold_1..12548483    4"""
    
20. 
    
21. 
    
22. with open("data.txt", "rt", encoding="utf-8") as f:
    
23.     # slist = string.splitlines()
    
24.     slist = f.read().splitlines()
    
25.     freq = 1
    
26.     sp = slist[0].split()
    
27.     start, num = sp[0], sp[-1]
    
28.     end = start
    
29.     for s in slist[1:]:
    
30.         if s[-1] == num:
    
31.             freq += 1
    
32.             end = s.split()[0]
    
33.         else:
    
34.             if freq == 1:
    
35.                 end = " " * len(end)
    
36.             print("{} {} {} {}".format(start, end, num, freq))
    
37.             freq = 1
    
38.             sp = s.split()
    
39.             start, num = sp[0], sp[-1]
    
40.             end = start
    
41.     else:
    
42.         print("{} {} {} {}".format(start, end, num, freq))
    

复制代码

poppy章鱼

hrpzcf 发表于 2021-11-12 14:20

谢谢您，非常感谢。我在本地测试是对了。但是我的原文件较大，所以我改了一下输入方式，结果报错了。您有帮我解答下吗？感激不尽，祝好。
#!/usr/bin/python3  
# -*- coding: utf-8 -*-
import sys

args=sys.argv
input1=args[1]

with open(input1, 'r') as f:
    slist = f.splitlines()
    freq = 1
    sp = slist[0].split()
    start, num = sp[0], sp[-1]
    end = start
    for s in slist[1:]:
        if s[-1] == num:
            freq += 1
            end = s.split()[0]
        else:
            if freq == 1:
                end = " " * len(end)
            print("{} {} {} {}".format(start, end, num, freq))
            freq = 1
            s = s.split()
            start, num = s[0], s[-1]
            end = start
    else:
        print("{} {} {} {}".format(start, end, num, freq))


报错信息：
python3 coordinate.count3.py ./05.txt > 05.new.txt
Traceback (most recent call last):
  File "coordinate.count3.py", line 9, in <module>
    slist = f.splitlines()
AttributeError: 'file' object has no attribute 'splitlines'

hrpzcf

poppy章鱼 发表于 2021-11-12 14:42
谢谢您，非常感谢。我在本地测试是对了。但是我的原文件较大，所以我改了一下输入方式，结果报错了。您有 ...

slist = f.splitlines()写错了，改成
slist = f.read().splitlines()

但是如果文件比可用内存还大的话就麻烦了

poppy章鱼

hrpzcf 发表于 2021-11-12 14:45
slist = f.splitlines()写错了，改成
slist = f.read().splitlines()

谢谢您，跑成功了，我在服务器上跑，没事。我学习下你的脚本，祝万事胜意。