我有一个简单的问题。导航到罚款中的某一行，然后删除所有内容。我使用合适的file.truncate（）调用。但是，下面两段代码表现不同。
1）
with open(file, "a+b", 1) as f:
#Navigate to the MARKER
while True:
line = f.readline()
if MARKER in line:
f.truncate()
f.write(stuff)
break
2）
with open(file, "a+b", 1) as f:
#Navigate to the MARKER
for line in f:
if MARKER in line:
f.truncate()
f.write(stuff)
break
（1）表现如预期。但是在（2）的情况下，找到 MARKER后截断的几行的文件。我推测有一些缓冲正在进行，但正如你所看到的，我明确地将缓冲行为定义为open（）调用的“行缓冲”。
有什么想法？我想使用更直观的“for line in file”语法......
答案 0 :(得分：3)
线索似乎在this bit of Python's C source中 - Python 2.7似乎为for line in file:使用了8 KB的预读缓冲区。
答案 1 :(得分：2)
通常，for x in y类型的语句要求y不要在循环内更改。你违反了合同。
答案 2 :(得分：2)
来自Python documentation，5。内置类型/ 5.9。文件对象：
为了使for循环最多 有效的循环方式 文件的行（非常常见 操作），next（）方法使用a 隐藏的预读缓冲区。
顺便说一句：一般不鼓励使用关键字（例如file）作为变量名称。
答案 3 :(得分：0)
这是因为'a'模式：
一
开放追加（最后写作 的文件）。如果是，则创建该文件 不存在。 流是 位于文件末尾。
A +
开放阅读和追加 （在文件末尾写）。该文件是 如果它不存在则创建。该 用于阅读的初始文件位置是 在文件的开头，但是 输出始终附加到文件末尾。
我的上述答案是错误的。
我已经知道循环文件的行使用缓冲区的预读，但我相信 truncate（）会触发文件的指针移动到文件的末尾，因为据我所知，截断文件包括编写一个称为EOF的小字节序列，意味着文件结束，而'a'模式总是激发文件末尾的写入，无论位置是什么在写作之前的文件指针。
嗯，不是那样的，我应该通过执行代码来验证。所以我的答案值得投票。
但是在这种情况下，没有任何解释的贬低是破旧和令人沮丧的，在这种情况下，这个答案中的错误并不明显。
以下代码显示在 truncate（）的操作之前，文件的指针未移动到文件的末尾。
为了清楚起见，文件'fileA'由长度为100个字符的行组成（'\ r \ n包含），结尾就像那样（'\ r \ n'在这里不可见）：
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00000100
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00000200
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00000300
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00000400
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00000500
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00000600
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00000700
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00000800
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00000900
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00001000
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00001100
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00001200
....ffffffgggggggggghhhhhhhhhhiiiiiiiiii00001300
............................
代码：
print '\n===================== 1 ==================\n'
from os.path import getsize
# length of ecr is 90 :
ecr = 10*'a' + 10*'b' + 10*'c' + 10*'d' + 10*'e' +\
10*'f' + 10*'g' + 10*'h' + 10*'i'
# creation of a file whose length exceeds the reading buffer's size
with open('fileA.txt','wb') as f:
for i in xrange(100,53001,100): # 530 turns of iteration
f.write(ecr + str(i).zfill(8) + '\r\n')
# Length of each written line is 100 :
# 90 (ecr) + 8 (str(i).zfill(8)) + 2 ('\r\n')
# File's length will be 53000
print 'size of fileA before truncating : ',getsize('fileA.txt')
# truncating file at uncontroled position
with open('fileA.txt','a+b') as g:
for line in g:
if '00000800' in line:
print repr(line[78:]),' g.tell()==',g.tell()
# at this point, 800 characters should have been read
# in the file if there wasn't a reading buffer
g.truncate()
print 'size of fileA after truncating : ',getsize('fileA.txt')
结果
===================== 1 ==================
size of fileA before truncating : 53000
'hhiiiiiiiiii00000800\r\n' g.tell()== 8192
size of fileA after truncating : 8192
因此，AKX和Fenisko在唤起缓冲区时是正确的（但是他们没有比我更多地测试这个假设）因为'a'模式下文件的打开对的操作没有影响截断（）即可。我认为这是文档的以下摘录中的上限句子所说的：
直到现在，我从未理解过这句话。file.truncate（[size]）截断 文件的大小。如果是可选尺寸 参数存在，文件是 截断到（最多）该大小。该 size默认为当前位置。 目前的文件位置不是 CHANGED
正如AKX所指出的那样，缓冲区的大小为 8192 ....一读。
但是对于下一个读数，缓冲区显然是 10240 个字符：
print '\n=================== 2 ====================\n'
from os.path import getsize
# length of ecr is 90 :
ecr = 10*'a' + 10*'b' + 10*'c' + 10*'d' + 10*'e' +\
10*'f' + 10*'g' + 10*'h' + 10*'i'
# creation of a file whose length exceeds the reading buffer's size
with open('fileA.txt','wb') as f:
for i in xrange(100,53001,100): # 530 turns of iteration
f.write(ecr + str(i).zfill(8) + '\r\n')
# length of each written line is 100
# file's length will be 53000
print 'size of fileA before truncating : ',getsize('fileA.txt')
# truncating file at uncontroled position
with open('fileA.txt','a+b') as g:
for line in g:
if '00008100' in line:
print repr(line[78:]),' g.tell()==',g.tell()
# at this point, 800 characters should have been read
# in the file if there wasn't a reading buffer
g.truncate()
print 'size of fileA after truncating : ',getsize('fileA.txt')
# -----------
print
# creation of a file whose length exceeds the reading buffer's size
with open('fileA.txt','wb') as f:
for i in xrange(100,53001,100): # 530 turns of iteration
f.write(ecr + str(i).zfill(8) + '\r\n')
# length of each written line is 100
# file's length will be 53000
print 'size of fileA before truncating : ',getsize('fileA.txt')
# truncating file at uncontroled position
with open('fileA.txt','a+b') as g:
for line in g:
if '00008200' in line:
print repr(line[78:]),' g.tell()==',g.tell()
# at this point, 800 characters should have been read
# in the file if there wasn't a reading buffer
g.truncate()
print 'size of fileA after truncating : ',getsize('fileA.txt')
# -----------
print
# creation of a file whose length exceeds the reading buffer's size
with open('fileA.txt','wb') as f:
for i in xrange(100,53001,100): # 530 turns of iteration
f.write(ecr + str(i).zfill(8) + '\r\n')
# length of each written line is 100
# file's length will be 53000
print 'size of fileA before truncating : ',getsize('fileA.txt')
# truncating file at uncontroled position
with open('fileA.txt','a+b') as g:
for line in g:
if '00018400' in line:
print repr(line[78:]),' g.tell()==',g.tell()
# at this point, 800 characters should have been read
# in the file if there wasn't a reading buffer
g.truncate()
print 'size of fileA after truncating : ',getsize('fileA.txt')
# -----------
print
# creation of a file whose length exceeds the reading buffer's size
with open('fileA.txt','wb') as f:
for i in xrange(100,53001,100): # 530 turns of iteration
f.write(ecr + str(i).zfill(8) + '\r\n')
# length of each written line is 100
# file's length will be 53000
print 'size of fileA before truncating : ',getsize('fileA.txt')
# truncating file at uncontroled position
with open('fileA.txt','a+b') as g:
for line in g:
if '00018500' in line:
print repr(line[78:]),' g.tell()==',g.tell()
# at this point, 800 characters should have been read
# in the file if there wasn't a reading buffer
g.truncate()
print 'size of fileA after truncating : ',getsize('fileA.txt')
结果
=================== 2 ====================
size of fileA before truncating : 53000
'hhiiiiiiiiii00008100\r\n' g.tell()== 8192
size of fileA after truncating : 8192
size of fileA before truncating : 53000
'hhiiiiiiiiii00008200\r\n' g.tell()== 18432
size of fileA after truncating : 18432
size of fileA before truncating : 53000
'hhiiiiiiiiii00018400\r\n' g.tell()== 18432
size of fileA after truncating : 18432
size of fileA before truncating : 53000
'hhiiiiiiiiii00018500\r\n' g.tell()== 28672
size of fileA after truncating : 28672
顺便说一句， truncate（）不会关闭文件：
print '\n=================== 3 ====================\n'
from os.path import getsize
# length of ecr is 90 :
ecr = 10*'a' + 10*'b' + 10*'c' + 10*'d' + 10*'e' +\
10*'f' + 10*'g' + 10*'h' + 10*'i'
# creation of a file whose length exceeds the reading buffer's size
with open('fileA.txt','wb') as f:
for i in xrange(100,53001,100): # 530 turns of iteration
f.write(ecr + str(i).zfill(8) + '\r\n')
# length of each written line is 100
# file's length will be 53000
print 'size of fileA before truncating : ',getsize('fileA.txt')
with open('fileA.txt','a+b') as g:
for line in g:
if '00000200' in line:
print repr(line[78:]),' g.tell()==',g.tell()
# at this point, 800 characters should have been read
# if there wasn't a buffer
g.truncate()
g.seek(6000,0)
k = 0
for li in g:
k+=1
print 'k==',k,' ',repr(li[-32:])
if k==7:
break
print 'size of fileA after truncating : ',getsize('fileA.txt')
结果
=================== 3 ====================
size of fileA before truncating : 53000
'hhiiiiiiiiii00000200\r\n' g.tell()== 8192
k== 1 'gghhhhhhhhhhiiiiiiiiii00006100\r\n'
k== 2 'gghhhhhhhhhhiiiiiiiiii00006200\r\n'
k== 3 'gghhhhhhhhhhiiiiiiiiii00006300\r\n'
k== 4 'gghhhhhhhhhhiiiiiiiiii00006400\r\n'
k== 5 'gghhhhhhhhhhiiiiiiiiii00006500\r\n'
k== 6 'gghhhhhhhhhhiiiiiiiiii00006600\r\n'
k== 7 'gghhhhhhhhhhiiiiiiiiii00006700\r\n'
size of fileA after truncating : 8192
但如果在 truncate（）之后放置了写入指令，则程序的行为就会变得不连贯。试试吧。