For Binary seperation, which would be better:using list or divmod?
I'm trying to divide the binary string from the end, by constant length:for example, '1001011000' by 3->['1','001','011','000'].
starting with the number 600,
def bin_divby(dec,leng):
n = 0
mid_res = ''
res=list()
for nums in bin(dec)[2:][::-1]:
n+=1
mid_res+=nums
if not n%leng:
res+=[mid_res[::-1]]
mid_res=''
if n%leng:
res+=[mid_res[::-1]]
return res[::-1]
(I'm not sure but code were kind of like this)
Using for loop with few variable would make it work, but I'm curious if using divmod would make it much faster. or would base conversion would be better? I thought using Binary form would be effective than re-calculating once more by divmod, but well as you see, I'm using much variable and functions as well. For repeating thousands of calculations, which would be better?
python binary base
edited Nov 11 at 20:13
Willem Van Onsem
143k16135227
asked Nov 11 at 20:12
ILoveG11
325
add a comment |
I'm trying to divide the binary string from the end, by constant length:for example, '1001011000' by 3->['1','001','011','000'].
starting with the number 600,
def bin_divby(dec,leng):
n = 0
mid_res = ''
res=list()
for nums in bin(dec)[2:][::-1]:
n+=1
mid_res+=nums
if not n%leng:
res+=[mid_res[::-1]]
mid_res=''
if n%leng:
res+=[mid_res[::-1]]
return res[::-1]
(I'm not sure but code were kind of like this)
Using for loop with few variable would make it work, but I'm curious if using divmod would make it much faster. or would base conversion would be better? I thought using Binary form would be effective than re-calculating once more by divmod, but well as you see, I'm using much variable and functions as well. For repeating thousands of calculations, which would be better?
python binary base
edited Nov 11 at 20:13
Willem Van Onsem
143k16135227
asked Nov 11 at 20:12
ILoveG11
325
add a comment |
I'm trying to divide the binary string from the end, by constant length:for example, '1001011000' by 3->['1','001','011','000'].
starting with the number 600,
def bin_divby(dec,leng):
n = 0
mid_res = ''
res=list()
for nums in bin(dec)[2:][::-1]:
n+=1
mid_res+=nums
if not n%leng:
res+=[mid_res[::-1]]
mid_res=''
if n%leng:
res+=[mid_res[::-1]]
return res[::-1]
(I'm not sure but code were kind of like this)
Using for loop with few variable would make it work, but I'm curious if using divmod would make it much faster. or would base conversion would be better? I thought using Binary form would be effective than re-calculating once more by divmod, but well as you see, I'm using much variable and functions as well. For repeating thousands of calculations, which would be better?
python binary base
edited Nov 11 at 20:13
Willem Van Onsem
143k16135227
asked Nov 11 at 20:12
ILoveG11
325
I'm trying to divide the binary string from the end, by constant length:for example, '1001011000' by 3->['1','001','011','000'].
starting with the number 600,
def bin_divby(dec,leng):
n = 0
mid_res = ''
res=list()
for nums in bin(dec)[2:][::-1]:
n+=1
mid_res+=nums
if not n%leng:
res+=[mid_res[::-1]]
mid_res=''
if n%leng:
res+=[mid_res[::-1]]
return res[::-1]
(I'm not sure but code were kind of like this)
Using for loop with few variable would make it work, but I'm curious if using divmod would make it much faster. or would base conversion would be better? I thought using Binary form would be effective than re-calculating once more by divmod, but well as you see, I'm using much variable and functions as well. For repeating thousands of calculations, which would be better?
python binary base
python binary base
edited Nov 11 at 20:13
Willem Van Onsem
143k16135227
asked Nov 11 at 20:12
ILoveG11
325
edited Nov 11 at 20:13
Willem Van Onsem
143k16135227
asked Nov 11 at 20:12
ILoveG11
325
edited Nov 11 at 20:13
Willem Van Onsem
143k16135227
edited Nov 11 at 20:13
Willem Van Onsem
143k16135227
edited Nov 11 at 20:13
Willem Van Onsem
143k16135227
143k16135227
asked Nov 11 at 20:12
ILoveG11
325
asked Nov 11 at 20:12
ILoveG11
325
asked Nov 11 at 20:12
ILoveG11
325
325
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
We can use list comprehension for this:
def bin_divby(dec, leng):
bn = bin(dec)[:1:-1]
return [bn[i:i+leng][::-1] for i in range(0, len(bn), leng)][::-1]
we can slightly improve efficiency by reversing the range(..) object:
def bin_divby(dec, leng):
bn = bin(dec)[:1:-1]
n = len(bn) - 1
return [bn[i:i+leng][::-1] for i in range(n - n%leng, -leng, -leng)]
So here we first obtain the binary string in reverse, and then we iterate over that string, and each time slice the string. The end result is reversed.
This produces the expected:
>>> bin_divby(0b1001011000, 1)
['1', '0', '0', '1', '0', '1', '1', '0', '0', '0']
>>> bin_divby(0b1001011000, 2)
['10', '01', '01', '10', '00']
>>> bin_divby(0b1001011000, 3)
['1', '001', '011', '000']
>>> bin_divby(0b1001011000, 4)
['10', '0101', '1000']
>>> bin_divby(0b1001011000, 5)
['10010', '11000']
>>> bin_divby(0b1001011000, 6)
['1001', '011000']
>>> bin_divby(0b1001011000, 7)
['100', '1011000']
answered Nov 11 at 20:22
Willem Van Onsem
143k16135227
It's being three times more faster than original! May I ask about 'struct' : would unpacking directly from it would be faster?
– ILoveG11
Nov 12 at 5:46
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
We can use list comprehension for this:
def bin_divby(dec, leng):
bn = bin(dec)[:1:-1]
return [bn[i:i+leng][::-1] for i in range(0, len(bn), leng)][::-1]
we can slightly improve efficiency by reversing the range(..) object:
def bin_divby(dec, leng):
bn = bin(dec)[:1:-1]
n = len(bn) - 1
return [bn[i:i+leng][::-1] for i in range(n - n%leng, -leng, -leng)]
So here we first obtain the binary string in reverse, and then we iterate over that string, and each time slice the string. The end result is reversed.
This produces the expected:
>>> bin_divby(0b1001011000, 1)
['1', '0', '0', '1', '0', '1', '1', '0', '0', '0']
>>> bin_divby(0b1001011000, 2)
['10', '01', '01', '10', '00']
>>> bin_divby(0b1001011000, 3)
['1', '001', '011', '000']
>>> bin_divby(0b1001011000, 4)
['10', '0101', '1000']
>>> bin_divby(0b1001011000, 5)
['10010', '11000']
>>> bin_divby(0b1001011000, 6)
['1001', '011000']
>>> bin_divby(0b1001011000, 7)
['100', '1011000']
answered Nov 11 at 20:22
Willem Van Onsem
143k16135227
It's being three times more faster than original! May I ask about 'struct' : would unpacking directly from it would be faster?
– ILoveG11
Nov 12 at 5:46
add a comment |
We can use list comprehension for this:
def bin_divby(dec, leng):
bn = bin(dec)[:1:-1]
return [bn[i:i+leng][::-1] for i in range(0, len(bn), leng)][::-1]
we can slightly improve efficiency by reversing the range(..) object:
def bin_divby(dec, leng):
bn = bin(dec)[:1:-1]
n = len(bn) - 1
return [bn[i:i+leng][::-1] for i in range(n - n%leng, -leng, -leng)]
So here we first obtain the binary string in reverse, and then we iterate over that string, and each time slice the string. The end result is reversed.
This produces the expected:
>>> bin_divby(0b1001011000, 1)
['1', '0', '0', '1', '0', '1', '1', '0', '0', '0']
>>> bin_divby(0b1001011000, 2)
['10', '01', '01', '10', '00']
>>> bin_divby(0b1001011000, 3)
['1', '001', '011', '000']
>>> bin_divby(0b1001011000, 4)
['10', '0101', '1000']
>>> bin_divby(0b1001011000, 5)
['10010', '11000']
>>> bin_divby(0b1001011000, 6)
['1001', '011000']
>>> bin_divby(0b1001011000, 7)
['100', '1011000']
answered Nov 11 at 20:22
Willem Van Onsem
143k16135227
It's being three times more faster than original! May I ask about 'struct' : would unpacking directly from it would be faster?
– ILoveG11
Nov 12 at 5:46
add a comment |
We can use list comprehension for this:
def bin_divby(dec, leng):
bn = bin(dec)[:1:-1]
return [bn[i:i+leng][::-1] for i in range(0, len(bn), leng)][::-1]
we can slightly improve efficiency by reversing the range(..) object:
def bin_divby(dec, leng):
bn = bin(dec)[:1:-1]
n = len(bn) - 1
return [bn[i:i+leng][::-1] for i in range(n - n%leng, -leng, -leng)]
So here we first obtain the binary string in reverse, and then we iterate over that string, and each time slice the string. The end result is reversed.
This produces the expected:
>>> bin_divby(0b1001011000, 1)
['1', '0', '0', '1', '0', '1', '1', '0', '0', '0']
>>> bin_divby(0b1001011000, 2)
['10', '01', '01', '10', '00']
>>> bin_divby(0b1001011000, 3)
['1', '001', '011', '000']
>>> bin_divby(0b1001011000, 4)
['10', '0101', '1000']
>>> bin_divby(0b1001011000, 5)
['10010', '11000']
>>> bin_divby(0b1001011000, 6)
['1001', '011000']
>>> bin_divby(0b1001011000, 7)
['100', '1011000']
answered Nov 11 at 20:22
Willem Van Onsem
143k16135227
We can use list comprehension for this:
def bin_divby(dec, leng):
bn = bin(dec)[:1:-1]
return [bn[i:i+leng][::-1] for i in range(0, len(bn), leng)][::-1]
we can slightly improve efficiency by reversing the range(..) object:
def bin_divby(dec, leng):
bn = bin(dec)[:1:-1]
n = len(bn) - 1
return [bn[i:i+leng][::-1] for i in range(n - n%leng, -leng, -leng)]
So here we first obtain the binary string in reverse, and then we iterate over that string, and each time slice the string. The end result is reversed.
This produces the expected:
>>> bin_divby(0b1001011000, 1)
['1', '0', '0', '1', '0', '1', '1', '0', '0', '0']
>>> bin_divby(0b1001011000, 2)
['10', '01', '01', '10', '00']
>>> bin_divby(0b1001011000, 3)
['1', '001', '011', '000']
>>> bin_divby(0b1001011000, 4)
['10', '0101', '1000']
>>> bin_divby(0b1001011000, 5)
['10010', '11000']
>>> bin_divby(0b1001011000, 6)
['1001', '011000']
>>> bin_divby(0b1001011000, 7)
['100', '1011000']
answered Nov 11 at 20:22
Willem Van Onsem
143k16135227
edited Nov 11 at 20:31
answered Nov 11 at 20:22
Willem Van Onsem
143k16135227
answered Nov 11 at 20:22
Willem Van Onsem
143k16135227
answered Nov 11 at 20:22
Willem Van Onsem
143k16135227
143k16135227
It's being three times more faster than original! May I ask about 'struct' : would unpacking directly from it would be faster?
– ILoveG11
Nov 12 at 5:46
add a comment |
It's being three times more faster than original! May I ask about 'struct' : would unpacking directly from it would be faster?
– ILoveG11
Nov 12 at 5:46
It's being three times more faster than original! May I ask about 'struct' : would unpacking directly from it would be faster?
– ILoveG11
Nov 12 at 5:46
It's being three times more faster than original! May I ask about 'struct' : would unpacking directly from it would be faster?
– ILoveG11
Nov 12 at 5:46
add a comment |
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