如何在Python中打印文件的特定行

Question

我有一個.txt文件，我想打印第3,7,11,15行3, 7, 11, 15,...

因此，在打印第三行之后，我想在之后每隔4行打印一次。

我開始看模數運算符：

#Open the file
with open('file.txt') as file:

  #Iterate through lines
  for i, line in enumerate(file):

      #Choose every third line in a file
      if i % 3 == 0:
          print(line)

  #Close the file when you're done
  file.close()

但這種方法打印每三行。 如果i % 3 == 1則打印第1,4,7,10,13行等。

Answer 1

而不是使用模數，只需使用加法，使用要顯示的第一行啟動它，然后向其中添加4

next_line = 2  # Line 3 is index 2
for i, line in enumerate(file):

    if i == next_line:
        print(line)
        next_line = next_line + 4

Answer 2

你的代碼幾乎沒有問題，除了模數：你希望除法的余數為4。

with open('file.txt') as file:
  for i, line in enumerate(file):
      if i % 4 == 3:
          print(line)

請注意，您不需要明確地close您的文件結尾：這就是with意為，它可以確保你的文件被關閉，無論發生什么情況。

Answer 3

因此，您希望每隔四次發生一次事情，這意味着模數4.嘗試將if更改為if i % 4 == N: N數字很​​好。

順便說一句，當使用with語句時，你不必調用close() ，它會自動執行。

Answer 4

怎么樣：

# Fetch all lines from the file
lines = open('20 - Modular OS - lang_en_vs2.srt').readlines()

# Print the 3rd line
print(lines[2])

# throw away the first 3 lines, so the modulo (below) works ok
for i in range(3):
    del(lines[0])

# print every 4th line after that
for (i in range(len(lines)):
    if (i > 0 and i % 4 == 0):
        print(lines[i])

將每一行讀入數組。 輸出第3行。 然后我們需要每四行，所以通過刪除前三個元素，很容易簡單地對模4（“％4”）進行測試並輸出該行。

Answer 5

x = 0
with open('file.txt') as file:

  #Iterate through lines
  for i, line in enumerate(file):
      x += 1
      #Choose every third line in a file
      if x == 4:
          print(line)
          x = 0

  #Close the file when you're done
  file.close()

結果

>>> i = 0
>>> for x in range(0, 100):
...     i += 1
...     if i is 4:
...         print(x)
...         i = 0

3 7 11 15 19 23 27 31 35 39 43 47 51 55 59 63 67 71 75 79 83 87 91 95 99

Answer 6

file = open('file.txt')
print(file[2])
#Iterate through lines
for i in file:
  #Choose every third line in a file, beginning with 4
  if i % 4 == 0:
      print(i+3)
  elif i % 4 == 0:
      print(i)

這可行，但不是超級優雅。