Make every four elements into a single number (element) - python
So how to the title description?
So the list:
l=[1,2,3,4,5,6,7,8]
Desired output:
[1234,5678]
So merge the every four packs of elements!
I am thinking about a list comprehension, but it's not a working well (it is, it's below my question)
python list merge element
asked Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
add a comment |
So how to the title description?
So the list:
l=[1,2,3,4,5,6,7,8]
Desired output:
[1234,5678]
So merge the every four packs of elements!
I am thinking about a list comprehension, but it's not a working well (it is, it's below my question)
python list merge element
asked Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
add a comment |
So how to the title description?
So the list:
l=[1,2,3,4,5,6,7,8]
Desired output:
[1234,5678]
So merge the every four packs of elements!
I am thinking about a list comprehension, but it's not a working well (it is, it's below my question)
python list merge element
asked Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
So how to the title description?
So the list:
l=[1,2,3,4,5,6,7,8]
Desired output:
[1234,5678]
So merge the every four packs of elements!
I am thinking about a list comprehension, but it's not a working well (it is, it's below my question)
python list merge element
python list merge element
asked Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
asked Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
asked Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
asked Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
asked Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
16.6k51543
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You can get the elements from slicing index of the range, we can do that because of the step argument (last one), that make that sequence into an integer (number).
Note: i am ordering them by rank, so (best on top, and worst at bottom)
Option 1:
list comprehension:
>>> [int(''.join(map(str,l[i:i+4]))) for i in range(0,len(l),4)]
[1234, 5678]
>>>
Option 2:
map:
>>> list(map(lambda i: int(''.join(map(str,l[i:i+4]))),range(0,len(l),4)))
[1234, 5678]
>>>
answered Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
1
Note: Option 1 is the better option, both on readability and performance; any time amapwould require the use oflambdathat the equivalent listcomp/genexpr can inline, themapwill be slower. The use ofmapin option 1 is fine (becausestris an existing function implemented in C which couldn't be avoided in an equivalent listcomp anyway).
– ShadowRanger
Nov 13 '18 at 0:13
@ShadowRanger Yes sir, That's how i order them :-), i know that :D
– U9-Forward
Nov 13 '18 at 0:16
add a comment |
Assume the length of l is the multiple of 4 and each element of l is an integer (1 to 9)
Here is another option without using string.
[1000*l[4*i]+100*l[4*i+1]+10*l[4*i+2]+l[4*i+3] for i in range(len(l)//4)]
answered Nov 13 '18 at 0:17
Banghua ZhaoBanghua Zhao
1,2871721
Ouch, Little unreadable, and inefficient..., but thanks for it.
– U9-Forward
Nov 13 '18 at 0:19
Note:range(len(l)/4)should berange(len(l)//4), use//not/.
– U9-Forward
Nov 13 '18 at 0:20
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can get the elements from slicing index of the range, we can do that because of the step argument (last one), that make that sequence into an integer (number).
Note: i am ordering them by rank, so (best on top, and worst at bottom)
Option 1:
list comprehension:
>>> [int(''.join(map(str,l[i:i+4]))) for i in range(0,len(l),4)]
[1234, 5678]
>>>
Option 2:
map:
>>> list(map(lambda i: int(''.join(map(str,l[i:i+4]))),range(0,len(l),4)))
[1234, 5678]
>>>
answered Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
1
Note: Option 1 is the better option, both on readability and performance; any time amapwould require the use oflambdathat the equivalent listcomp/genexpr can inline, themapwill be slower. The use ofmapin option 1 is fine (becausestris an existing function implemented in C which couldn't be avoided in an equivalent listcomp anyway).
– ShadowRanger
Nov 13 '18 at 0:13
@ShadowRanger Yes sir, That's how i order them :-), i know that :D
– U9-Forward
Nov 13 '18 at 0:16
add a comment |
You can get the elements from slicing index of the range, we can do that because of the step argument (last one), that make that sequence into an integer (number).
Note: i am ordering them by rank, so (best on top, and worst at bottom)
Option 1:
list comprehension:
>>> [int(''.join(map(str,l[i:i+4]))) for i in range(0,len(l),4)]
[1234, 5678]
>>>
Option 2:
map:
>>> list(map(lambda i: int(''.join(map(str,l[i:i+4]))),range(0,len(l),4)))
[1234, 5678]
>>>
answered Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
1
Note: Option 1 is the better option, both on readability and performance; any time amapwould require the use oflambdathat the equivalent listcomp/genexpr can inline, themapwill be slower. The use ofmapin option 1 is fine (becausestris an existing function implemented in C which couldn't be avoided in an equivalent listcomp anyway).
– ShadowRanger
Nov 13 '18 at 0:13
@ShadowRanger Yes sir, That's how i order them :-), i know that :D
– U9-Forward
Nov 13 '18 at 0:16
add a comment |
You can get the elements from slicing index of the range, we can do that because of the step argument (last one), that make that sequence into an integer (number).
Note: i am ordering them by rank, so (best on top, and worst at bottom)
Option 1:
list comprehension:
>>> [int(''.join(map(str,l[i:i+4]))) for i in range(0,len(l),4)]
[1234, 5678]
>>>
Option 2:
map:
>>> list(map(lambda i: int(''.join(map(str,l[i:i+4]))),range(0,len(l),4)))
[1234, 5678]
>>>
answered Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
You can get the elements from slicing index of the range, we can do that because of the step argument (last one), that make that sequence into an integer (number).
Note: i am ordering them by rank, so (best on top, and worst at bottom)
Option 1:
list comprehension:
>>> [int(''.join(map(str,l[i:i+4]))) for i in range(0,len(l),4)]
[1234, 5678]
>>>
Option 2:
map:
>>> list(map(lambda i: int(''.join(map(str,l[i:i+4]))),range(0,len(l),4)))
[1234, 5678]
>>>
answered Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
edited Nov 13 '18 at 0:15
answered Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
answered Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
answered Nov 13 '18 at 0:02
U9-ForwardU9-Forward
16.6k51543
16.6k51543
1
Note: Option 1 is the better option, both on readability and performance; any time amapwould require the use oflambdathat the equivalent listcomp/genexpr can inline, themapwill be slower. The use ofmapin option 1 is fine (becausestris an existing function implemented in C which couldn't be avoided in an equivalent listcomp anyway).
– ShadowRanger
Nov 13 '18 at 0:13
@ShadowRanger Yes sir, That's how i order them :-), i know that :D
– U9-Forward
Nov 13 '18 at 0:16
add a comment |
1
Note: Option 1 is the better option, both on readability and performance; any time amapwould require the use oflambdathat the equivalent listcomp/genexpr can inline, themapwill be slower. The use ofmapin option 1 is fine (becausestris an existing function implemented in C which couldn't be avoided in an equivalent listcomp anyway).
– ShadowRanger
Nov 13 '18 at 0:13
@ShadowRanger Yes sir, That's how i order them :-), i know that :D
– U9-Forward
Nov 13 '18 at 0:16
1
1
Note: Option 1 is the better option, both on readability and performance; any time a
map would require the use of lambda that the equivalent listcomp/genexpr can inline, the map will be slower. The use of map in option 1 is fine (because str is an existing function implemented in C which couldn't be avoided in an equivalent listcomp anyway).– ShadowRanger
Nov 13 '18 at 0:13
Note: Option 1 is the better option, both on readability and performance; any time a
map would require the use of lambda that the equivalent listcomp/genexpr can inline, the map will be slower. The use of map in option 1 is fine (because str is an existing function implemented in C which couldn't be avoided in an equivalent listcomp anyway).– ShadowRanger
Nov 13 '18 at 0:13
@ShadowRanger Yes sir, That's how i order them :-), i know that :D
– U9-Forward
Nov 13 '18 at 0:16
@ShadowRanger Yes sir, That's how i order them :-), i know that :D
– U9-Forward
Nov 13 '18 at 0:16
add a comment |
Assume the length of l is the multiple of 4 and each element of l is an integer (1 to 9)
Here is another option without using string.
[1000*l[4*i]+100*l[4*i+1]+10*l[4*i+2]+l[4*i+3] for i in range(len(l)//4)]
answered Nov 13 '18 at 0:17
Banghua ZhaoBanghua Zhao
1,2871721
Ouch, Little unreadable, and inefficient..., but thanks for it.
– U9-Forward
Nov 13 '18 at 0:19
Note:range(len(l)/4)should berange(len(l)//4), use//not/.
– U9-Forward
Nov 13 '18 at 0:20
add a comment |
Assume the length of l is the multiple of 4 and each element of l is an integer (1 to 9)
Here is another option without using string.
[1000*l[4*i]+100*l[4*i+1]+10*l[4*i+2]+l[4*i+3] for i in range(len(l)//4)]
answered Nov 13 '18 at 0:17
Banghua ZhaoBanghua Zhao
1,2871721
Ouch, Little unreadable, and inefficient..., but thanks for it.
– U9-Forward
Nov 13 '18 at 0:19
Note:range(len(l)/4)should berange(len(l)//4), use//not/.
– U9-Forward
Nov 13 '18 at 0:20
add a comment |
Assume the length of l is the multiple of 4 and each element of l is an integer (1 to 9)
Here is another option without using string.
[1000*l[4*i]+100*l[4*i+1]+10*l[4*i+2]+l[4*i+3] for i in range(len(l)//4)]
answered Nov 13 '18 at 0:17
Banghua ZhaoBanghua Zhao
1,2871721
Assume the length of l is the multiple of 4 and each element of l is an integer (1 to 9)
Here is another option without using string.
[1000*l[4*i]+100*l[4*i+1]+10*l[4*i+2]+l[4*i+3] for i in range(len(l)//4)]
answered Nov 13 '18 at 0:17
Banghua ZhaoBanghua Zhao
1,2871721
edited Nov 13 '18 at 0:24
answered Nov 13 '18 at 0:17
Banghua ZhaoBanghua Zhao
1,2871721
answered Nov 13 '18 at 0:17
Banghua ZhaoBanghua Zhao
1,2871721
answered Nov 13 '18 at 0:17
Banghua ZhaoBanghua Zhao
1,2871721
1,2871721
Ouch, Little unreadable, and inefficient..., but thanks for it.
– U9-Forward
Nov 13 '18 at 0:19
Note:range(len(l)/4)should berange(len(l)//4), use//not/.
– U9-Forward
Nov 13 '18 at 0:20
add a comment |
Ouch, Little unreadable, and inefficient..., but thanks for it.
– U9-Forward
Nov 13 '18 at 0:19
Note:range(len(l)/4)should berange(len(l)//4), use//not/.
– U9-Forward
Nov 13 '18 at 0:20
Ouch, Little unreadable, and inefficient..., but thanks for it.
– U9-Forward
Nov 13 '18 at 0:19
Ouch, Little unreadable, and inefficient..., but thanks for it.
– U9-Forward
Nov 13 '18 at 0:19
Note:
range(len(l)/4) should be range(len(l)//4), use // not /.– U9-Forward
Nov 13 '18 at 0:20
Note:
range(len(l)/4) should be range(len(l)//4), use // not /.– U9-Forward
Nov 13 '18 at 0:20
add a comment |
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