遍历文件

Question

我有100个文件在文件夹命名为1.htm - 100.htm。我运行这段代码从文件中提取一些信息，并将提取的信息放在另一个文件final.txt中。目前，我必须手动运行100个文件的程序。我需要构建一个循环，可以运行程序100次，读取每个文件一次。(请详细解释我需要在我的代码中做的确切编辑)

下面是文件6.htm: 的代码

import glob
import BeautifulSoup
from BeautifulSoup import BeautifulSoup

fo = open("6.htm", "r")
bo = open("output.txt" ,"w")
f = open("final.txt","a+")
htmltext = fo.read()
soup = BeautifulSoup(htmltext)
#print len(urls)
table = soup.findAll('table')
rows = table[0].findAll('tr');
for tr in rows:
    cols = tr.findAll('td')
    for td in cols:
        text = str(td.find(text=True)) + ';;;'
        if(text!=" ;;;"):
            bo.write(text);
            bo.write('n');
fo.close()
bo.close()
b= open("output.txt", "r")
for j in range (1,5):
str=b.readline();
for j in range(1, 15):
str=b.readline();
c=str.split(";;;")
#print c[1]
if(c[0]=="APD ID:"):
    f.write(c[1])
    f.write("#")
if(c[0]=="Name/Class:"):
    f.write(c[1])
    f.write("#")
if(c[0]=="Source:"):
    f.write(c[1])
    f.write("#")
if(c[0]=="Sequence:"):
    f.write(c[1])
    f.write("#")
if(c[0]=="Length:"):
    f.write(c[1])
    f.write("#")
if(c[0]=="Net charge:"):
    f.write(c[1])
    f.write("#")
if(c[0]=="Hydrophobic residue%:"):
    f.write(c[1])
    f.write("#")
if(c[0]=="Boman Index:"):
    f.write(c[1])
    f.write("#")
f.write('n');
b.close();
f.close();

f.close();
print "End"

Answer 1

import os
f = open("final.txt","a+")
for root, folders, files in os.walk('./path/to/html_files/'):
    for fileName in files:
        fo = open(os.path.abspath(root + '/' + fileName, "r")
        ...

然后剩下的代码放到那里。

也考虑(最佳实践)

with open(os.path.abspath(root + '/' + fileName, "r") as fo:
    ...

所以你不要忘记关闭那些文件句柄，因为在你的操作系统中允许打开的文件句柄数量是有限的，这将确保你不会错误地填充它。

让你的代码看起来像这样:

import os
with open("final.txt","a+") as f:
    for root, folders, files in os.walk('./path/to/html_files/'):
        for fileName in files:
            with open(os.path.abspath(root + '/' + fileName, "r") as fo:
                ...

同时NEVER替换全局变量名，如str:

str=b.readline();

在代码行末尾也不需要;，这是Python..我们以舒适的方式编写代码!

最后但并非最不重要的……

if(c[0]=="APD ID:"):
if(c[0]=="Name/Class:"):
if(c[0]=="Source:"):
if(c[0]=="Sequence:"):
if(c[0]=="Length:"):
if(c[0]=="Net charge:"):
if(c[0]=="Hydrophobic residue%:"):
if(c[0]=="Boman Index:"):

应:

if(c[0]=="APD ID:"):
elif(c[0]=="Name/Class:"):
elif(c[0]=="Source:"):
elif(c[0]=="Sequence:"):
elif(c[0]=="Length:"):
elif(c[0]=="Net charge:"):
elif(c[0]=="Hydrophobic residue%:"):
elif(c[0]=="Boman Index:"):

除非你一路上修改了c，当然你不会。所以开关!

该死的，我只是不断发现更多关于这段代码的可怕的事情(你显然是从所有星系的例子中复制粘贴的…):

您可以将上述所有if/elif/else压缩成一个if块:

if(c[0] in ("APD ID:", "Name/Class:", "Source:", "Sequence:", "Length:", "Net charge:", "Hydrophobic residue%:", "Boman Index:")):
    f.write(c[1])
    f.write("#")

还有，跳过(…)围绕你的if块，再一次…这是Python..我们以一种舒适的方式编程:

if c[0] in ("APD ID:", "Name/Class:", "Source:", "Sequence:", "Length:", "Net charge:", "Hydrophobic residue%:", "Boman Index:"):
    f.write(c[1])
    f.write("#")

Answer 2

可能是这样的结构:

# declare main files
bo = open("output.txt" ,"w")
f = open("final.txt","a+")
#loop over range ii = [1,100]
for ii in range(1,101):
    fo = open(str(ii) + ".htm", "r")
    # Run program like normal
    ...
    ...
    ...
    fo.close()
f.close()
bo.close()

Answer 3

Listdir列出特定目录下的所有文件。

正如@Torxed所指出的，最佳实践是使用with子句(这样文件句柄就关闭了)。

你可以像这样查找。htm文件:

import os
# Creates a list of 1-100.htm file names
filenames = map(lambda x: str(x) + ".htm", range(1,101))
for file in os.listdir("/mydir"):
    if (file in filenames):
        # Do your logic here.