blocks|key|2561099|text|使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2561100|In+[1]:+hex(777007543)
Out[1]:+'0x2e5031b7'|code-block|syntax|javascript|2561101|您应该能够从这里重新格式化它。|2561102|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Use this:

<pre><code>In [1]: hex(777007543)
Out[1]: '0x2e5031b7'
</code></pre>

you should be able to reformat it from here.

blocks|key|4157969|text|Struct模块将给出实际的字节数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4157970|>>>+struct.pack('L',777007543)
'.P1\xb7'|code-block|syntax|javascript|4157971|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Struct module will give you the actual bytes:

<pre><code>&gt;&gt;&gt; struct.pack('L',777007543)
'.P1\xb7'
</code></pre>

blocks|key|493982|text|使用struct+module|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|493983|In+[6]:+import+struct
In+[14]:+map(hex,struct.unpack('>4B',struct.pack('>L',777007543)))
Out[14]:+['0x2e',+'0x50',+'0x31',+'0xb7']|code-block|syntax|javascript|493984|或者，如果大写很重要，|493985|In+[17]:+map('0x{0:X}'.format,struct.unpack('>4B',struct.pack('>L',777007543)))
Out[17]:+['0x2E',+'0x50',+'0x31',+'0xB7']|493986|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/struct.html#module-struct^0|2|D|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

Using the <a href="http://docs.python.org/library/struct.html#module-struct" rel="nofollow">struct module</a>:

<pre><code>In [6]: import struct
In [14]: map(hex,struct.unpack('&gt;4B',struct.pack('&gt;L',777007543)))
Out[14]: ['0x2e', '0x50', '0x31', '0xb7']
</code></pre>

or, if capitalization is important,

<pre><code>In [17]: map('0x{0:X}'.format,struct.unpack('&gt;4B',struct.pack('&gt;L',777007543)))
Out[17]: ['0x2E', '0x50', '0x31', '0xB7']
</code></pre>

blocks|key|4158037|text|我能想到的最简单的方法是在列表理解中使用struct+module：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4158038|import+struct
print+[hex(ord(b))+for+b+in+struct.pack('>L',777007543)]
#+['0x2e',+'0x50',+'0x31',+'0xb7']|code-block|syntax|javascript|4158039|获取大写的十六进制数字有点复杂，但也不是那么糟糕：|4158040|import+string
import+struct
xlate+=+string.maketrans('abcdef',+'ABCDEF')

print+[hex(ord(b)).translate(xlate)+for+b+in+struct.pack('>L',777007543)]
#+['0x2E',+'0x50',+'0x31',+'0xB7']|4158041|更新|BOLD|4158042|因为从你的评论中听起来你可能在使用Python+3-尽管你的问题没有"python-3.x“标签-而且现在大多数人都在使用更高的版本，这里的代码演示了如何在两个版本中都能工作(产生大写的十六进制字母)：|4158043|import+struct
import+sys

if+sys.version_info+<+(3,):++#+Python+2?
++++def+hexfmt(val):
++++++++return+'0x{:02X}'.format(ord(val))
else:
++++def+hexfmt(val):
++++++++return+'0x{:02X}'.format(val)

byte_list+=+[hexfmt(b)+for+b+in+struct.pack('>L',+777007543)]
print(byte_list)++#+->+['0x2E',+'0x50',+'0x31',+'0xB7']|4158044|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/struct.html#module-struct^0|K|6|K|D|0|0|0|0|0|0|2|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@$9|16|A|17|1|18]]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1C|8|@$9|1D|A|1E|B|Q]]|D|@]|E|$]]|$1|R|3|S|5|6|7|1F|8|@]|D|@]|E|$]]|$1|T|3|U|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|V|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|W|$X|$5|Y|Z|10|E|$11|12]]]]

The simplest way I can think of is to use the <a href="http://docs.python.org/library/struct.html#module-struct" rel="nofollow noreferrer"><code>struct</code> module</a> from within a list comprehension:

<pre><code>import struct
print [hex(ord(b)) for b in struct.pack('&gt;L',777007543)]
# ['0x2e', '0x50', '0x31', '0xb7']
</code></pre>

It's a little bit more complicated to get uppercase hex digits, but not that bad:

<pre><code>import string
import struct
xlate = string.maketrans('abcdef', 'ABCDEF')

print [hex(ord(b)).translate(xlate) for b in struct.pack('&gt;L',777007543)]
# ['0x2E', '0x50', '0x31', '0xB7']
</code></pre>

Update

Since from your comments it sounds like you may be using Python 3 — even though your question doesn't have a "python-3.x" tag — and that fact that nowadays most folks are using the later version, here's code illustrating how to do it that will work in both versions (producing uppercase hexadecimal letters):

<pre><code>import struct
import sys

if sys.version_info &lt; (3,): # Python 2?
 def hexfmt(val):
 return '0x{:02X}'.format(ord(val))
else:
 def hexfmt(val):
 return '0x{:02X}'.format(val)

byte_list = [hexfmt(b) for b in struct.pack('&gt;L', 777007543)]
print(byte_list) # -&gt; ['0x2E', '0x50', '0x31', '0xB7']
</code></pre>

blocks|key|4158057|text|对于这个简单的问题，字符串%2B结构实在是言过其实。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4158058|>>>+x=777007543
>>>+[hex(0xff&x>>8*i)+for+i+in+3,2,1,0]
['0x2e',+'0x50',+'0x31',+'0xb7']
>>>+[hex(0xff&x>>8*i).upper()+for+i+in+3,2,1,0]
['0X2E',+'0X50',+'0X31',+'0XB7']|code-block|syntax|javascript|4158059|下面是使用ipython的快速比较|4158060|In+[1]:+import+struct

In+[2]:+x=777007543

In+[3]:+timeit+[hex(ord(b))+for+b+in+struct.pack('>L',x)]
100000+loops,+best+of+3:+2.06+us+per+loop

In+[4]:+timeit+[hex(0xff&x>>8*i)+for+i+in+3,2,1,0]
1000000+loops,+best+of+3:+1.35+us+per+loop

In+[5]:+timeit+[hex(0xff&x>>i)+for+i+in+24,16,8,0]
1000000+loops,+best+of+3:+1.15+us+per+loop|4158061|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Strings + struct are really overkill for this simple problem

<pre><code>&gt;&gt;&gt; x=777007543
&gt;&gt;&gt; [hex(0xff&amp;x&gt;&gt;8*i) for i in 3,2,1,0]
['0x2e', '0x50', '0x31', '0xb7']
&gt;&gt;&gt; [hex(0xff&amp;x&gt;&gt;8*i).upper() for i in 3,2,1,0]
['0X2E', '0X50', '0X31', '0XB7']
</code></pre>

Here is a quick comparison using ipython

<pre><code>In [1]: import struct

In [2]: x=777007543

In [3]: timeit [hex(ord(b)) for b in struct.pack('&gt;L',x)]
100000 loops, best of 3: 2.06 us per loop

In [4]: timeit [hex(0xff&amp;x&gt;&gt;8*i) for i in 3,2,1,0]
1000000 loops, best of 3: 1.35 us per loop

In [5]: timeit [hex(0xff&amp;x&gt;&gt;i) for i in 24,16,8,0]
1000000 loops, best of 3: 1.15 us per loop
</code></pre>

How do you turn a long unsigned int into a list of four bytes in hexidecimal?

Example...

777007543 = 0x2E 0x50 0x31 0xB7

How do you convert a decimal number into a list of bytes in python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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如何将一个长的无符号整型转换为十六进制的四个字节的列表？例如。777007543 = 0x2E 0x50 0x31 0xB7

问在python中如何将十进制数转换为字节列表？
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问在python中如何将十进制数转换为字节列表？EN