在Python中跳過多行

Question

因此我試圖跳過.txt文件中的多行，然後使用csv閱讀器創建CSV文件。 18行需要跳過。這可以完成工作，但我確信有一個簡單的方法可以跳過18行而不是使用next（）18次。

import csv 
import os 

my_file_name = os.path.abspath('LensBank.txt') 
cleaned_file = "LensBankClean.csv" 
with open(my_file_name, 'r', newline='') as infile, open(cleaned_file, 'w',newline='') as outfile: 
    writer = csv.writer(outfile) 
    cr = csv.reader(infile, delimiter=',') 

    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    next(cr) 
    writer.writerow(next(cr)) 

    for line in (r[:20] for r in cr): 
    writer.writerow(line) 

這適用於我，但我怎麼會清理代碼到一個更簡單的版本。謝謝！

Answer 1

使用：

for skip in range(18): 
    next(cr) 

Answer 2

for i in range(18): 
    next(cr) 

使用for循環。或者你可以使用itertools.dropwhile

for line in (r[:20] for i, r in itertools.dropwhile(lambda x: x[1] < 18 , enumerate(cr))): 

Answer 3

這是奇怪的是，你用一個for循環之下，但沒有考慮它了同樣的問題。

你的代碼可以很容易的東西來替代這樣

for i in range(18): 
    next(cr) 
writer.writerow(next(cr)) 

這將調用next（CR）18次，事後致電writer.writerow

Answer 4

這個怎麼樣，

import csv 

# read a csv file into a list of lists 
with open(in_file, 'r') as f_in: 
    lists = [row for row in csv.reader(f_in, delimiter=',')] 

# write a list of lists to a csv file 
with open(out_file, 'w') as f_out: 
    writer = csv.writer(f_out) 
    writer.writerows(lists[18:]) # skip the first 18 lines 

---

正如@PatrickHaugh所述，上述解決方案對大文件無效。以下是大文件的解決方案。

with open(in_file,'r') as f_in, open(out_file,'w') as f_out: 
    # skip the first n lines 
    for _ in range(18): 
     next(f_in) 
    for line in f_in: 
     f_out.write(line)