def getitems(iterable, items):
  items = list(items) # get a list from any iterable and make our own copy
                      # since we modify it
  if items:
    items.sort()
    for n, v in enumerate(iterable):
      if n == items[0]:
        yield v
        items.pop(0)
        if not items:
          break

print list(getitems(open("/usr/share/dict/words"), [25, 29]))
# ['Abelson\n', 'Abernathy\n']
# note that index 25 is the 26th item

为了提供另一种解决方案:

import linecache
linecache.getline('Sample.txt', Number_of_Line)

我希望这是快速和简单的:)

您可以使用一种最简单的逻辑将字符串拆分为数组或List。

f = open('filepath')
r = f.read()
s = r.split("\n")
n = [linenumber1, linenumber2] # [26, 29] in your 
                               #case
for x in n:
  print(s[x-1])
f.close()

如果你想看第7行

line = open("file.txt", "r").readlines()[7]

从某一行读出:

n = 4   # for reading from 5th line
with open("write.txt",'r') as t:
     for i,line in enumerate(t):
         if i >= n:             # i == n-1 for nth line
            print(line)

不要使用阅读线!

我的解决方案是:

with open(filename) as f:
    specify = [26, 30]
    results = list(
        map(lambda line: line[1],
            filter(lambda line: line[0] in specify,
                   enumerate(f))
            )
    )

对6.5G文件进行如下测试:

import time

filename = 'a.txt'
start = time.time()
with open(filename, 'w') as f:
    for i in range(10_000_000):
        f.write(f'{str(i)*100}\n')       
end1 = time.time()

with open(filename) as f:
    specify = [26, 30]
    results = list(
        map(lambda line: line[1],
            filter(lambda line: line[0] in specify,
                   enumerate(f))
            )
    )
end2 = time.time()
print(f'write time: {end1-start}')
print(f'read time: {end2-end1}')
# write time: 14.38945460319519
# read time: 8.380386352539062