You can use Convert.ToHexString
starting with .NET 5.
There's also a method for the reverse operation: Convert.FromHexString
.
For older versions of .NET you can either use:
public static string ByteArrayToString(byte[] ba)
{
StringBuilder hex = new StringBuilder(ba.Length * 2);
foreach (byte b in ba)
hex.AppendFormat("{0:x2}", b);
return hex.ToString();
}
or:
public static string ByteArrayToString(byte[] ba)
{
return BitConverter.ToString(ba).Replace("-","");
}
There are even more variants of doing it, for example here.
The reverse conversion would go like this:
public static byte[] StringToByteArray(String hex)
{
int NumberChars = hex.Length;
byte[] bytes = new byte[NumberChars / 2];
for (int i = 0; i < NumberChars; i += 2)
bytes[i / 2] = Convert.ToByte(hex.Substring(i, 2), 16);
return bytes;
}
Using Substring
is the best option in combination with Convert.ToByte
. See this answer for more information. If you need better performance, you must avoid Convert.ToByte
before you can drop SubString
.
Given a list of lists t
,
flat_list = [item for sublist in t for item in sublist]
which means:
flat_list = []
for sublist in t:
for item in sublist:
flat_list.append(item)
is faster than the shortcuts posted so far. (t
is the list to flatten.)
Here is the corresponding function:
def flatten(t):
return [item for sublist in t for item in sublist]
As evidence, you can use the timeit
module in the standard library:
$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in t for item in sublist]'
10000 loops, best of 3: 143 usec per loop
$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(t, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s't=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,t)'
1000 loops, best of 3: 1.1 msec per loop
Explanation: the shortcuts based on +
(including the implied use in sum
) are, of necessity, O(T**2)
when there are T sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have T sublists of k items each: the first k items are copied back and forth T-1 times, the second k items T-2 times, and so on; total number of copies is k times the sum of x for x from 1 to T excluded, i.e., k * (T**2)/2
.
The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.
Best Answer
You've got a workable idea, but the
#flatten!
is in the wrong place -- it flattens its receiver, so you could use it to turn[1, 2, ['foo', 'bar']]
into[1,2,'foo','bar']
.I'm doubtless forgetting some approaches, but you can concatenate:
or prepend/append:
or splice:
or append and flatten: