How to compare two DNA sequences and return the identical nucleotides in a pair list
-4
I want to compare two DNA sequences and return the identical nucleotides in a pair list (position in sequence 1, position in sequence 2)
input:
a = [G, T, T, U, I, P]
b = [E, G, T, P]
output:
[[0,1], [1,2], [2,2], [5,3]]
python
edited Nov 11 '18 at 23:46
Jon Clements♦
98.2k19173218
asked Nov 11 '18 at 23:45
sir_Ouss
13
add a comment |
-4
I want to compare two DNA sequences and return the identical nucleotides in a pair list (position in sequence 1, position in sequence 2)
input:
a = [G, T, T, U, I, P]
b = [E, G, T, P]
output:
[[0,1], [1,2], [2,2], [5,3]]
python
edited Nov 11 '18 at 23:46
Jon Clements♦
98.2k19173218
asked Nov 11 '18 at 23:45
sir_Ouss
13
1
Are you after all pairs? So if you hada=['T', 'T', 'T']; b = ['T', 'T', 'T']you'd have 9 results?
– Jon Clements♦
Nov 11 '18 at 23:49
5
Did you write any code for this? You need to share the code and explain what exact issue you are facing in that
– Chetan Ranpariya
Nov 11 '18 at 23:50
add a comment |
-4
-4
-4
I want to compare two DNA sequences and return the identical nucleotides in a pair list (position in sequence 1, position in sequence 2)
input:
a = [G, T, T, U, I, P]
b = [E, G, T, P]
output:
[[0,1], [1,2], [2,2], [5,3]]
python
edited Nov 11 '18 at 23:46
Jon Clements♦
98.2k19173218
asked Nov 11 '18 at 23:45
sir_Ouss
13
I want to compare two DNA sequences and return the identical nucleotides in a pair list (position in sequence 1, position in sequence 2)
input:
a = [G, T, T, U, I, P]
b = [E, G, T, P]
output:
[[0,1], [1,2], [2,2], [5,3]]
python
python
edited Nov 11 '18 at 23:46
Jon Clements♦
98.2k19173218
asked Nov 11 '18 at 23:45
sir_Ouss
13
edited Nov 11 '18 at 23:46
Jon Clements♦
98.2k19173218
asked Nov 11 '18 at 23:45
sir_Ouss
13
edited Nov 11 '18 at 23:46
Jon Clements♦
98.2k19173218
edited Nov 11 '18 at 23:46
Jon Clements♦
98.2k19173218
edited Nov 11 '18 at 23:46
Jon Clements♦
98.2k19173218
98.2k19173218
asked Nov 11 '18 at 23:45
sir_Ouss
13
asked Nov 11 '18 at 23:45
sir_Ouss
13
asked Nov 11 '18 at 23:45
sir_Ouss
13
13
1
Are you after all pairs? So if you hada=['T', 'T', 'T']; b = ['T', 'T', 'T']you'd have 9 results?
– Jon Clements♦
Nov 11 '18 at 23:49
5
Did you write any code for this? You need to share the code and explain what exact issue you are facing in that
– Chetan Ranpariya
Nov 11 '18 at 23:50
add a comment |
1
Are you after all pairs? So if you hada=['T', 'T', 'T']; b = ['T', 'T', 'T']you'd have 9 results?
– Jon Clements♦
Nov 11 '18 at 23:49
5
Did you write any code for this? You need to share the code and explain what exact issue you are facing in that
– Chetan Ranpariya
Nov 11 '18 at 23:50
1
1
Are you after all pairs? So if you had
a=['T', 'T', 'T']; b = ['T', 'T', 'T'] you'd have 9 results?– Jon Clements♦
Nov 11 '18 at 23:49
Are you after all pairs? So if you had
a=['T', 'T', 'T']; b = ['T', 'T', 'T'] you'd have 9 results?– Jon Clements♦
Nov 11 '18 at 23:49
5
5
Did you write any code for this? You need to share the code and explain what exact issue you are facing in that
– Chetan Ranpariya
Nov 11 '18 at 23:50
Did you write any code for this? You need to share the code and explain what exact issue you are facing in that
– Chetan Ranpariya
Nov 11 '18 at 23:50
add a comment |
2 Answers
2
active
oldest
votes
1
You can do it with for loops:
a_s = ["G", "T", "T", "U", "I", "P"]
b_s = ["E", "G", "T", "P"]
d =
for i,a in enumerate(a_s):
for j,b in enumerate(b_s):
if a == b:
d.append([i,j])
print(d)
Out:
[[0, 1], [1, 2], [2, 2], [5, 3]]
Or you can do it in a single row:
a_s = ["G", "T", "T", "U", "I", "P"]
b_s = ["E", "G", "T", "P"]
print([[x, y] for x, av in enumerate(a_s) for y, bv in enumerate(b_s) if av == bv])
With the above, same output.
Note:
The first version is in most case more readable, the second is more concise. You can always chose any of both depending on the code context and the purpose of it.
answered Nov 11 '18 at 23:52
Geeocode
2,2261820
2
Or rolled into a list-comp:[[x, y] for x, av in enumerate(a) for y, bv in enumerate(b) if av == bv]
– Jon Clements♦
Nov 11 '18 at 23:52
@JonClements Sometimes more readable the open format, but can show him as well
– Geeocode
Nov 11 '18 at 23:54
Indeed... if you're starting out it's definitely more readable. However, it's useful for learning and for others in the future that come across this post to see the other version - doesn't hurt to show both
– Jon Clements♦
Nov 11 '18 at 23:57
thank you so much , that's exactly what i want
– sir_Ouss
Nov 12 '18 at 1:38
@sir_Ouss You'r welcome!
– Geeocode
Nov 12 '18 at 1:46
|
show 3 more comments
0
Two examples leveraging "product" from the "itertools" module.
The first is a traditional for loop that appends a list.
The second is a list comprehension equivalent.
from itertools import product
a = list('GTTUIP')
b = list('EGTP')
# Without a comprehension.
results =
for (x, a_s), (y, b_s) in product(enumerate(a), enumerate(b)):
if a_s == b_s:
results.append([x, y])
print(results)
# With a comprehension
results = [[x, y]
for (x, a_s), (y, b_s)
in product(enumerate(a), enumerate(b))
if a_s == b_s]
print(results)
OUT:
[[0, 1], [1, 2], [2, 2], [5, 3]]
[[0, 1], [1, 2], [2, 2], [5, 3]]
answered Nov 12 '18 at 0:12
DMfll
99011830
For what it's worth, usingitertools.productis roughly 10% faster. See this gist using thetimeitmodule. Each comprehension is run 1 million times: gist.github.com/bb4f02e391c7e15df947df6918e7bd93
– DMfll
Nov 12 '18 at 10:56
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
1
You can do it with for loops:
a_s = ["G", "T", "T", "U", "I", "P"]
b_s = ["E", "G", "T", "P"]
d =
for i,a in enumerate(a_s):
for j,b in enumerate(b_s):
if a == b:
d.append([i,j])
print(d)
Out:
[[0, 1], [1, 2], [2, 2], [5, 3]]
Or you can do it in a single row:
a_s = ["G", "T", "T", "U", "I", "P"]
b_s = ["E", "G", "T", "P"]
print([[x, y] for x, av in enumerate(a_s) for y, bv in enumerate(b_s) if av == bv])
With the above, same output.
Note:
The first version is in most case more readable, the second is more concise. You can always chose any of both depending on the code context and the purpose of it.
answered Nov 11 '18 at 23:52
Geeocode
2,2261820
2
Or rolled into a list-comp:[[x, y] for x, av in enumerate(a) for y, bv in enumerate(b) if av == bv]
– Jon Clements♦
Nov 11 '18 at 23:52
@JonClements Sometimes more readable the open format, but can show him as well
– Geeocode
Nov 11 '18 at 23:54
Indeed... if you're starting out it's definitely more readable. However, it's useful for learning and for others in the future that come across this post to see the other version - doesn't hurt to show both
– Jon Clements♦
Nov 11 '18 at 23:57
thank you so much , that's exactly what i want
– sir_Ouss
Nov 12 '18 at 1:38
@sir_Ouss You'r welcome!
– Geeocode
Nov 12 '18 at 1:46
|
show 3 more comments
1
You can do it with for loops:
a_s = ["G", "T", "T", "U", "I", "P"]
b_s = ["E", "G", "T", "P"]
d =
for i,a in enumerate(a_s):
for j,b in enumerate(b_s):
if a == b:
d.append([i,j])
print(d)
Out:
[[0, 1], [1, 2], [2, 2], [5, 3]]
Or you can do it in a single row:
a_s = ["G", "T", "T", "U", "I", "P"]
b_s = ["E", "G", "T", "P"]
print([[x, y] for x, av in enumerate(a_s) for y, bv in enumerate(b_s) if av == bv])
With the above, same output.
Note:
The first version is in most case more readable, the second is more concise. You can always chose any of both depending on the code context and the purpose of it.
answered Nov 11 '18 at 23:52
Geeocode
2,2261820
2
Or rolled into a list-comp:[[x, y] for x, av in enumerate(a) for y, bv in enumerate(b) if av == bv]
– Jon Clements♦
Nov 11 '18 at 23:52
@JonClements Sometimes more readable the open format, but can show him as well
– Geeocode
Nov 11 '18 at 23:54
Indeed... if you're starting out it's definitely more readable. However, it's useful for learning and for others in the future that come across this post to see the other version - doesn't hurt to show both
– Jon Clements♦
Nov 11 '18 at 23:57
thank you so much , that's exactly what i want
– sir_Ouss
Nov 12 '18 at 1:38
@sir_Ouss You'r welcome!
– Geeocode
Nov 12 '18 at 1:46
|
show 3 more comments
1
1
1
You can do it with for loops:
a_s = ["G", "T", "T", "U", "I", "P"]
b_s = ["E", "G", "T", "P"]
d =
for i,a in enumerate(a_s):
for j,b in enumerate(b_s):
if a == b:
d.append([i,j])
print(d)
Out:
[[0, 1], [1, 2], [2, 2], [5, 3]]
Or you can do it in a single row:
a_s = ["G", "T", "T", "U", "I", "P"]
b_s = ["E", "G", "T", "P"]
print([[x, y] for x, av in enumerate(a_s) for y, bv in enumerate(b_s) if av == bv])
With the above, same output.
Note:
The first version is in most case more readable, the second is more concise. You can always chose any of both depending on the code context and the purpose of it.
answered Nov 11 '18 at 23:52
Geeocode
2,2261820
You can do it with for loops:
a_s = ["G", "T", "T", "U", "I", "P"]
b_s = ["E", "G", "T", "P"]
d =
for i,a in enumerate(a_s):
for j,b in enumerate(b_s):
if a == b:
d.append([i,j])
print(d)
Out:
[[0, 1], [1, 2], [2, 2], [5, 3]]
Or you can do it in a single row:
a_s = ["G", "T", "T", "U", "I", "P"]
b_s = ["E", "G", "T", "P"]
print([[x, y] for x, av in enumerate(a_s) for y, bv in enumerate(b_s) if av == bv])
With the above, same output.
Note:
The first version is in most case more readable, the second is more concise. You can always chose any of both depending on the code context and the purpose of it.
answered Nov 11 '18 at 23:52
Geeocode
2,2261820
edited Nov 11 '18 at 23:58
answered Nov 11 '18 at 23:52
Geeocode
2,2261820
answered Nov 11 '18 at 23:52
Geeocode
2,2261820
answered Nov 11 '18 at 23:52
Geeocode
2,2261820
2,2261820
2
Or rolled into a list-comp:[[x, y] for x, av in enumerate(a) for y, bv in enumerate(b) if av == bv]
– Jon Clements♦
Nov 11 '18 at 23:52
@JonClements Sometimes more readable the open format, but can show him as well
– Geeocode
Nov 11 '18 at 23:54
Indeed... if you're starting out it's definitely more readable. However, it's useful for learning and for others in the future that come across this post to see the other version - doesn't hurt to show both
– Jon Clements♦
Nov 11 '18 at 23:57
thank you so much , that's exactly what i want
– sir_Ouss
Nov 12 '18 at 1:38
@sir_Ouss You'r welcome!
– Geeocode
Nov 12 '18 at 1:46
|
show 3 more comments
2
Or rolled into a list-comp:[[x, y] for x, av in enumerate(a) for y, bv in enumerate(b) if av == bv]
– Jon Clements♦
Nov 11 '18 at 23:52
@JonClements Sometimes more readable the open format, but can show him as well
– Geeocode
Nov 11 '18 at 23:54
Indeed... if you're starting out it's definitely more readable. However, it's useful for learning and for others in the future that come across this post to see the other version - doesn't hurt to show both
– Jon Clements♦
Nov 11 '18 at 23:57
thank you so much , that's exactly what i want
– sir_Ouss
Nov 12 '18 at 1:38
@sir_Ouss You'r welcome!
– Geeocode
Nov 12 '18 at 1:46
2
2
Or rolled into a list-comp:
[[x, y] for x, av in enumerate(a) for y, bv in enumerate(b) if av == bv]– Jon Clements♦
Nov 11 '18 at 23:52
Or rolled into a list-comp:
[[x, y] for x, av in enumerate(a) for y, bv in enumerate(b) if av == bv]– Jon Clements♦
Nov 11 '18 at 23:52
@JonClements Sometimes more readable the open format, but can show him as well
– Geeocode
Nov 11 '18 at 23:54
@JonClements Sometimes more readable the open format, but can show him as well
– Geeocode
Nov 11 '18 at 23:54
Indeed... if you're starting out it's definitely more readable. However, it's useful for learning and for others in the future that come across this post to see the other version - doesn't hurt to show both
– Jon Clements♦
Nov 11 '18 at 23:57
Indeed... if you're starting out it's definitely more readable. However, it's useful for learning and for others in the future that come across this post to see the other version - doesn't hurt to show both
– Jon Clements♦
Nov 11 '18 at 23:57
thank you so much , that's exactly what i want
– sir_Ouss
Nov 12 '18 at 1:38
thank you so much , that's exactly what i want
– sir_Ouss
Nov 12 '18 at 1:38
@sir_Ouss You'r welcome!
– Geeocode
Nov 12 '18 at 1:46
@sir_Ouss You'r welcome!
– Geeocode
Nov 12 '18 at 1:46
|
show 3 more comments
0
Two examples leveraging "product" from the "itertools" module.
The first is a traditional for loop that appends a list.
The second is a list comprehension equivalent.
from itertools import product
a = list('GTTUIP')
b = list('EGTP')
# Without a comprehension.
results =
for (x, a_s), (y, b_s) in product(enumerate(a), enumerate(b)):
if a_s == b_s:
results.append([x, y])
print(results)
# With a comprehension
results = [[x, y]
for (x, a_s), (y, b_s)
in product(enumerate(a), enumerate(b))
if a_s == b_s]
print(results)
OUT:
[[0, 1], [1, 2], [2, 2], [5, 3]]
[[0, 1], [1, 2], [2, 2], [5, 3]]
answered Nov 12 '18 at 0:12
DMfll
99011830
For what it's worth, usingitertools.productis roughly 10% faster. See this gist using thetimeitmodule. Each comprehension is run 1 million times: gist.github.com/bb4f02e391c7e15df947df6918e7bd93
– DMfll
Nov 12 '18 at 10:56
add a comment |
0
Two examples leveraging "product" from the "itertools" module.
The first is a traditional for loop that appends a list.
The second is a list comprehension equivalent.
from itertools import product
a = list('GTTUIP')
b = list('EGTP')
# Without a comprehension.
results =
for (x, a_s), (y, b_s) in product(enumerate(a), enumerate(b)):
if a_s == b_s:
results.append([x, y])
print(results)
# With a comprehension
results = [[x, y]
for (x, a_s), (y, b_s)
in product(enumerate(a), enumerate(b))
if a_s == b_s]
print(results)
OUT:
[[0, 1], [1, 2], [2, 2], [5, 3]]
[[0, 1], [1, 2], [2, 2], [5, 3]]
answered Nov 12 '18 at 0:12
DMfll
99011830
For what it's worth, usingitertools.productis roughly 10% faster. See this gist using thetimeitmodule. Each comprehension is run 1 million times: gist.github.com/bb4f02e391c7e15df947df6918e7bd93
– DMfll
Nov 12 '18 at 10:56
add a comment |
0
0
0
Two examples leveraging "product" from the "itertools" module.
The first is a traditional for loop that appends a list.
The second is a list comprehension equivalent.
from itertools import product
a = list('GTTUIP')
b = list('EGTP')
# Without a comprehension.
results =
for (x, a_s), (y, b_s) in product(enumerate(a), enumerate(b)):
if a_s == b_s:
results.append([x, y])
print(results)
# With a comprehension
results = [[x, y]
for (x, a_s), (y, b_s)
in product(enumerate(a), enumerate(b))
if a_s == b_s]
print(results)
OUT:
[[0, 1], [1, 2], [2, 2], [5, 3]]
[[0, 1], [1, 2], [2, 2], [5, 3]]
answered Nov 12 '18 at 0:12
DMfll
99011830
Two examples leveraging "product" from the "itertools" module.
The first is a traditional for loop that appends a list.
The second is a list comprehension equivalent.
from itertools import product
a = list('GTTUIP')
b = list('EGTP')
# Without a comprehension.
results =
for (x, a_s), (y, b_s) in product(enumerate(a), enumerate(b)):
if a_s == b_s:
results.append([x, y])
print(results)
# With a comprehension
results = [[x, y]
for (x, a_s), (y, b_s)
in product(enumerate(a), enumerate(b))
if a_s == b_s]
print(results)
OUT:
[[0, 1], [1, 2], [2, 2], [5, 3]]
[[0, 1], [1, 2], [2, 2], [5, 3]]
answered Nov 12 '18 at 0:12
DMfll
99011830
edited Nov 12 '18 at 0:25
answered Nov 12 '18 at 0:12
DMfll
99011830
answered Nov 12 '18 at 0:12
DMfll
99011830
answered Nov 12 '18 at 0:12
DMfll
99011830
99011830
For what it's worth, usingitertools.productis roughly 10% faster. See this gist using thetimeitmodule. Each comprehension is run 1 million times: gist.github.com/bb4f02e391c7e15df947df6918e7bd93
– DMfll
Nov 12 '18 at 10:56
add a comment |
For what it's worth, usingitertools.productis roughly 10% faster. See this gist using thetimeitmodule. Each comprehension is run 1 million times: gist.github.com/bb4f02e391c7e15df947df6918e7bd93
– DMfll
Nov 12 '18 at 10:56
For what it's worth, using
itertools.product is roughly 10% faster. See this gist using the timeit module. Each comprehension is run 1 million times: gist.github.com/bb4f02e391c7e15df947df6918e7bd93– DMfll
Nov 12 '18 at 10:56
For what it's worth, using
itertools.product is roughly 10% faster. See this gist using the timeit module. Each comprehension is run 1 million times: gist.github.com/bb4f02e391c7e15df947df6918e7bd93– DMfll
Nov 12 '18 at 10:56
add a comment |
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1
Are you after all pairs? So if you had
a=['T', 'T', 'T']; b = ['T', 'T', 'T']you'd have 9 results?– Jon Clements♦
Nov 11 '18 at 23:49
5
Did you write any code for this? You need to share the code and explain what exact issue you are facing in that
– Chetan Ranpariya
Nov 11 '18 at 23:50