如何从文件中读取字节 [英] How to Read Bytes from a file
问题描述
看起来这很容易,但我正在画一个空白。
我想做的是能够以二进制模式打开任何文件,并阅读
一次一个字节(8位),然后计算
那个字节的1位数。
I得到了这个,但它给了我一些字符串,我不确定如何准确到达字节/位级别。
f1 = file(''somefile'',''rb'')
而1:
abyte = f1.read(1)
提前感谢您的帮助。
-Greg
It seems like this would be easy but I''m drawing a blank.
What I want to do is be able to open any file in binary mode, and read
in one byte (8 bits) at a time and then count the number of 1 bits in
that byte.
I got as far as this but it is giving me strings and I''m not sure how
to accurately get to the byte/bit level.
f1=file(''somefile'',''rb'')
while 1:
abyte=f1.read(1)
Thanks in advance for any help.
-Greg
推荐答案
gr********@gmail.com < gr ******** @ gmail.comwrote:
gr********@gmail.com <gr********@gmail.comwrote:
看起来这很容易,但我正在画一个空白。
什么我想做的是能够以二进制模式打开任何文件,并一次读取一个字节(8位)的
然后凑这个字节中的1比特数是多少。
我得到了这个但它给了我字符串我不知道怎么样
准确到达字节/位级别。
f1 = file(''somefile'',''rb'')
而1:
abyte = f1.read(1)
It seems like this would be easy but I''m drawing a blank.
What I want to do is be able to open any file in binary mode, and read
in one byte (8 bits) at a time and then count the number of 1 bits in
that byte.
I got as far as this but it is giving me strings and I''m not sure how
to accurately get to the byte/bit level.
f1=file(''somefile'',''rb'')
while 1:
abyte=f1.read(1)
你应该在循环前准备char的映射编号
的1位字符:
m = {}
范围内的c(256):
m [c] = countones(c)
然后将m [abyte]的值总结为一个运行总计(从
$ b中断当''不是abyte''时,$ b循环,即你正在读取0字节,即使
要求1 - 告诉你罚款完成,记得关闭
it。
完成countones功能的一个简单方法:
def countones(x):
断言x> = 0
c = 0
而x:
c + =(x& 1)
x>> = 1
返回c
你不想经常打电话,从那里以前的建议
只需要256次来准备映射。
如果下载并安装gmpy,你可以使用gmpy.popcount作为快速
计数器的实现:-)。
Alex
You should probaby prepare before the loop a mapping from char to number
of 1 bits in that char:
m = {}
for c in range(256):
m[c] = countones(c)
and then sum up the values of m[abyte] into a running total (break from
the loop when ''not abyte'', i.e. you''re reading 0 bytes even though
asking for 1 -- that tells you the fine is finished, remember to close
it).
A trivial way to do the countones function:
def countones(x):
assert x>=0
c = 0
while x:
c += (x&1)
x >>= 1
return c
you just don''t want to call it too often, whence the previous advice to
call it just 256 times to prep a mapping.
If you download and install gmpy you can use gmpy.popcount as a fast
implementation of countones:-).
Alex
Alex Martelli写道:
Alex Martelli wrote:
你应该在循环之前准备一个从char到数字的映射
的1比特:
m = { }
范围内的c(256):
m [c] = countones(c)
You should probaby prepare before the loop a mapping from char to number
of 1 bits in that char:
m = {}
for c in range(256):
m[c] = countones(c)
Wouldn 列表更有效率吗?
m = [x(256)中c的countones(c)
Wouldn''t a list be more efficient?
m = [countones(c) for c in xrange(256)]
On 3月1日上午7:52,gregpin ... @ gmail.com < gregpin ... @ gmail.com>
写道:
On Mar 1, 7:52 am, "gregpin...@gmail.com" <gregpin...@gmail.com>
wrote:
看起来这很容易但我正在画画空白。
我想要做的是能够以二进制模式打开任何文件,并在一个字节(8位)中读取
时间,然后计算
那个字节的1位数。
我得到了这个,但它给了我一些字符串,我是不确定
如何准确到达字节/位级别。
f1 = file(''somefile'',''rb'')
而1:
abyte = f1.read(1)
It seems like this would be easy but I''m drawing a blank.
What I want to do is be able to open any file in binary mode, and read
in one byte (8 bits) at a time and then count the number of 1 bits in
that byte.
I got as far as this but it is giving me strings and I''m not sure how
to accurately get to the byte/bit level.
f1=file(''somefile'',''rb'')
while 1:
abyte=f1.read(1)
import struct
buf = open(''somefile'',''rb'')。read()
count1 = lambda x:(x& 1)+(x& 2> 0)+( x& 4> 0)+(x& 8> 0)+(x& 16> 0)+(x& 32> 0)+
(x& 64> 0)+(x& 128> 0)
byteOnes = map(count1,struct.unpack(''B''* len(buf),buf))
byteOnes [ n]是数字是n字节n中的数字。
import struct
buf = open(''somefile'',''rb'').read()
count1 = lambda x: (x&1)+(x&2>0)+(x&4>0)+(x&8>0)+(x&16>0)+(x&32>0)+
(x&64>0)+(x&128>0)
byteOnes = map(count1,struct.unpack(''B''*len(buf),buf))
byteOnes[n] is number is number of ones in byte n.
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