How do I replace a value in a matrix which corresponding to another matrix?
For example
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
And I want to get a matrix c:
c = [0.5, 0.8, 0, 0.11, 0, 0]
That's like if the i in a = ww for ww,ee in n for n in b, then replace with ee else 0
I try some if and else command and here is my code
for n in b:
for t,y in n:
for tt in a:
mmm = [y if t == ''.join(tt) else ''.join(tt)]
print(mmm)
But it failed. How should I code for this situation?
python python-3.x numpy if-statement replace
edited Nov 18 '18 at 17:55
Austin
10.2k3828
asked Nov 18 '18 at 17:37
wayne64001wayne64001
475
add a comment |
For example
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
And I want to get a matrix c:
c = [0.5, 0.8, 0, 0.11, 0, 0]
That's like if the i in a = ww for ww,ee in n for n in b, then replace with ee else 0
I try some if and else command and here is my code
for n in b:
for t,y in n:
for tt in a:
mmm = [y if t == ''.join(tt) else ''.join(tt)]
print(mmm)
But it failed. How should I code for this situation?
python python-3.x numpy if-statement replace
edited Nov 18 '18 at 17:55
Austin
10.2k3828
asked Nov 18 '18 at 17:37
wayne64001wayne64001
475
3
Shouldn't the last value incbe0.23?
– Austin
Nov 18 '18 at 17:49
[dict(sum(b,)).get(int(i),0) for i in a]
– Onyambu
Nov 18 '18 at 21:48
add a comment |
For example
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
And I want to get a matrix c:
c = [0.5, 0.8, 0, 0.11, 0, 0]
That's like if the i in a = ww for ww,ee in n for n in b, then replace with ee else 0
I try some if and else command and here is my code
for n in b:
for t,y in n:
for tt in a:
mmm = [y if t == ''.join(tt) else ''.join(tt)]
print(mmm)
But it failed. How should I code for this situation?
python python-3.x numpy if-statement replace
edited Nov 18 '18 at 17:55
Austin
10.2k3828
asked Nov 18 '18 at 17:37
wayne64001wayne64001
475
For example
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
And I want to get a matrix c:
c = [0.5, 0.8, 0, 0.11, 0, 0]
That's like if the i in a = ww for ww,ee in n for n in b, then replace with ee else 0
I try some if and else command and here is my code
for n in b:
for t,y in n:
for tt in a:
mmm = [y if t == ''.join(tt) else ''.join(tt)]
print(mmm)
But it failed. How should I code for this situation?
python python-3.x numpy if-statement replace
python python-3.x numpy if-statement replace
edited Nov 18 '18 at 17:55
Austin
10.2k3828
asked Nov 18 '18 at 17:37
wayne64001wayne64001
475
edited Nov 18 '18 at 17:55
Austin
10.2k3828
asked Nov 18 '18 at 17:37
wayne64001wayne64001
475
edited Nov 18 '18 at 17:55
Austin
10.2k3828
edited Nov 18 '18 at 17:55
Austin
10.2k3828
edited Nov 18 '18 at 17:55
Austin
10.2k3828
10.2k3828
asked Nov 18 '18 at 17:37
wayne64001wayne64001
475
asked Nov 18 '18 at 17:37
wayne64001wayne64001
475
asked Nov 18 '18 at 17:37
wayne64001wayne64001
475
475
3
Shouldn't the last value incbe0.23?
– Austin
Nov 18 '18 at 17:49
[dict(sum(b,)).get(int(i),0) for i in a]
– Onyambu
Nov 18 '18 at 21:48
add a comment |
3
Shouldn't the last value incbe0.23?
– Austin
Nov 18 '18 at 17:49
[dict(sum(b,)).get(int(i),0) for i in a]
– Onyambu
Nov 18 '18 at 21:48
3
3
Shouldn't the last value in
c be 0.23 ?– Austin
Nov 18 '18 at 17:49
Shouldn't the last value in
c be 0.23 ?– Austin
Nov 18 '18 at 17:49
[dict(sum(b,)).get(int(i),0) for i in a]– Onyambu
Nov 18 '18 at 21:48
[dict(sum(b,)).get(int(i),0) for i in a]– Onyambu
Nov 18 '18 at 21:48
add a comment |
3 Answers
3
active
oldest
votes
chain + dict + list comprehension
Your b mapping is a list of lists, you can flatten this into an iterable of tuples via chain.from_iterable. Then feed to dict to create an efficient mapping.
Finally, use a list comprehension with dict.get for the desired result. Just remember to convert the values of a from str to int.
from itertools import chain
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
b_dict = dict(chain.from_iterable(b))
c = [b_dict.get(i, 0) for i in map(int, a)]
print(c)
[0.5, 0.8, 0, 0.11, 0, 0.23]
answered Nov 18 '18 at 18:16
jppjpp
100k2161111
add a comment |
This iterates through list a comparing it's value with first value of tuples in b list. This appends the second value of tuple to output list if the first value of tuple matches with the value in a:
from itertools import chain
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
b = list(chain.from_iterable(b))
lst =
for x in a:
for y, z in b:
if y == int(x):
lst.append(z)
break
else:
lst.append(0)
print(lst)
# [0.5, 0.8, 0, 0.11, 0, 0.23]
answered Nov 18 '18 at 17:51
AustinAustin
10.2k3828
add a comment |
You can convert your double-list of mappings into a lookup dictionary and use a list-comp:
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
# convert b to a dictionary:
d = {str(k):v for tup in b for k,v in tup} # stringify the lookup key value
print(d)
# apply the lookup to a's values
result = [d.get(k,0) for k in a]
print(result)
Output:
# the lookup-dictionary
{'1': 0.5, '2': 0.8, '4': 0.11, '6': 0.23}
# result of list c,omprehension
[0.5, 0.8, 0, 0.11, 0, 0.23]
Related:
- dict.get(key[, default])
- Why dict.get(key) instead of dict[key]?
answered Nov 18 '18 at 18:16
Patrick ArtnerPatrick Artner
23.6k62443
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
chain + dict + list comprehension
Your b mapping is a list of lists, you can flatten this into an iterable of tuples via chain.from_iterable. Then feed to dict to create an efficient mapping.
Finally, use a list comprehension with dict.get for the desired result. Just remember to convert the values of a from str to int.
from itertools import chain
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
b_dict = dict(chain.from_iterable(b))
c = [b_dict.get(i, 0) for i in map(int, a)]
print(c)
[0.5, 0.8, 0, 0.11, 0, 0.23]
answered Nov 18 '18 at 18:16
jppjpp
100k2161111
add a comment |
chain + dict + list comprehension
Your b mapping is a list of lists, you can flatten this into an iterable of tuples via chain.from_iterable. Then feed to dict to create an efficient mapping.
Finally, use a list comprehension with dict.get for the desired result. Just remember to convert the values of a from str to int.
from itertools import chain
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
b_dict = dict(chain.from_iterable(b))
c = [b_dict.get(i, 0) for i in map(int, a)]
print(c)
[0.5, 0.8, 0, 0.11, 0, 0.23]
answered Nov 18 '18 at 18:16
jppjpp
100k2161111
add a comment |
chain + dict + list comprehension
Your b mapping is a list of lists, you can flatten this into an iterable of tuples via chain.from_iterable. Then feed to dict to create an efficient mapping.
Finally, use a list comprehension with dict.get for the desired result. Just remember to convert the values of a from str to int.
from itertools import chain
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
b_dict = dict(chain.from_iterable(b))
c = [b_dict.get(i, 0) for i in map(int, a)]
print(c)
[0.5, 0.8, 0, 0.11, 0, 0.23]
answered Nov 18 '18 at 18:16
jppjpp
100k2161111
chain + dict + list comprehension
Your b mapping is a list of lists, you can flatten this into an iterable of tuples via chain.from_iterable. Then feed to dict to create an efficient mapping.
Finally, use a list comprehension with dict.get for the desired result. Just remember to convert the values of a from str to int.
from itertools import chain
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
b_dict = dict(chain.from_iterable(b))
c = [b_dict.get(i, 0) for i in map(int, a)]
print(c)
[0.5, 0.8, 0, 0.11, 0, 0.23]
answered Nov 18 '18 at 18:16
jppjpp
100k2161111
answered Nov 18 '18 at 18:16
jppjpp
100k2161111
answered Nov 18 '18 at 18:16
jppjpp
100k2161111
answered Nov 18 '18 at 18:16
jppjpp
100k2161111
100k2161111
add a comment |
add a comment |
This iterates through list a comparing it's value with first value of tuples in b list. This appends the second value of tuple to output list if the first value of tuple matches with the value in a:
from itertools import chain
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
b = list(chain.from_iterable(b))
lst =
for x in a:
for y, z in b:
if y == int(x):
lst.append(z)
break
else:
lst.append(0)
print(lst)
# [0.5, 0.8, 0, 0.11, 0, 0.23]
answered Nov 18 '18 at 17:51
AustinAustin
10.2k3828
add a comment |
This iterates through list a comparing it's value with first value of tuples in b list. This appends the second value of tuple to output list if the first value of tuple matches with the value in a:
from itertools import chain
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
b = list(chain.from_iterable(b))
lst =
for x in a:
for y, z in b:
if y == int(x):
lst.append(z)
break
else:
lst.append(0)
print(lst)
# [0.5, 0.8, 0, 0.11, 0, 0.23]
answered Nov 18 '18 at 17:51
AustinAustin
10.2k3828
add a comment |
This iterates through list a comparing it's value with first value of tuples in b list. This appends the second value of tuple to output list if the first value of tuple matches with the value in a:
from itertools import chain
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
b = list(chain.from_iterable(b))
lst =
for x in a:
for y, z in b:
if y == int(x):
lst.append(z)
break
else:
lst.append(0)
print(lst)
# [0.5, 0.8, 0, 0.11, 0, 0.23]
answered Nov 18 '18 at 17:51
AustinAustin
10.2k3828
This iterates through list a comparing it's value with first value of tuples in b list. This appends the second value of tuple to output list if the first value of tuple matches with the value in a:
from itertools import chain
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
b = list(chain.from_iterable(b))
lst =
for x in a:
for y, z in b:
if y == int(x):
lst.append(z)
break
else:
lst.append(0)
print(lst)
# [0.5, 0.8, 0, 0.11, 0, 0.23]
answered Nov 18 '18 at 17:51
AustinAustin
10.2k3828
edited Nov 18 '18 at 18:01
answered Nov 18 '18 at 17:51
AustinAustin
10.2k3828
answered Nov 18 '18 at 17:51
AustinAustin
10.2k3828
answered Nov 18 '18 at 17:51
AustinAustin
10.2k3828
10.2k3828
add a comment |
add a comment |
You can convert your double-list of mappings into a lookup dictionary and use a list-comp:
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
# convert b to a dictionary:
d = {str(k):v for tup in b for k,v in tup} # stringify the lookup key value
print(d)
# apply the lookup to a's values
result = [d.get(k,0) for k in a]
print(result)
Output:
# the lookup-dictionary
{'1': 0.5, '2': 0.8, '4': 0.11, '6': 0.23}
# result of list c,omprehension
[0.5, 0.8, 0, 0.11, 0, 0.23]
Related:
- dict.get(key[, default])
- Why dict.get(key) instead of dict[key]?
answered Nov 18 '18 at 18:16
Patrick ArtnerPatrick Artner
23.6k62443
add a comment |
You can convert your double-list of mappings into a lookup dictionary and use a list-comp:
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
# convert b to a dictionary:
d = {str(k):v for tup in b for k,v in tup} # stringify the lookup key value
print(d)
# apply the lookup to a's values
result = [d.get(k,0) for k in a]
print(result)
Output:
# the lookup-dictionary
{'1': 0.5, '2': 0.8, '4': 0.11, '6': 0.23}
# result of list c,omprehension
[0.5, 0.8, 0, 0.11, 0, 0.23]
Related:
- dict.get(key[, default])
- Why dict.get(key) instead of dict[key]?
answered Nov 18 '18 at 18:16
Patrick ArtnerPatrick Artner
23.6k62443
add a comment |
You can convert your double-list of mappings into a lookup dictionary and use a list-comp:
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
# convert b to a dictionary:
d = {str(k):v for tup in b for k,v in tup} # stringify the lookup key value
print(d)
# apply the lookup to a's values
result = [d.get(k,0) for k in a]
print(result)
Output:
# the lookup-dictionary
{'1': 0.5, '2': 0.8, '4': 0.11, '6': 0.23}
# result of list c,omprehension
[0.5, 0.8, 0, 0.11, 0, 0.23]
Related:
- dict.get(key[, default])
- Why dict.get(key) instead of dict[key]?
answered Nov 18 '18 at 18:16
Patrick ArtnerPatrick Artner
23.6k62443
You can convert your double-list of mappings into a lookup dictionary and use a list-comp:
a = ['1', '2', '3', '4', '5', '6']
b = [[(1, 0.5), (2, 0.8)], [(4, 0.11), (6, 0.23)]]
# convert b to a dictionary:
d = {str(k):v for tup in b for k,v in tup} # stringify the lookup key value
print(d)
# apply the lookup to a's values
result = [d.get(k,0) for k in a]
print(result)
Output:
# the lookup-dictionary
{'1': 0.5, '2': 0.8, '4': 0.11, '6': 0.23}
# result of list c,omprehension
[0.5, 0.8, 0, 0.11, 0, 0.23]
Related:
- dict.get(key[, default])
- Why dict.get(key) instead of dict[key]?
answered Nov 18 '18 at 18:16
Patrick ArtnerPatrick Artner
23.6k62443
answered Nov 18 '18 at 18:16
Patrick ArtnerPatrick Artner
23.6k62443
answered Nov 18 '18 at 18:16
Patrick ArtnerPatrick Artner
23.6k62443
answered Nov 18 '18 at 18:16
Patrick ArtnerPatrick Artner
23.6k62443
23.6k62443
add a comment |
add a comment |
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3
Shouldn't the last value in
cbe0.23?– Austin
Nov 18 '18 at 17:49
[dict(sum(b,)).get(int(i),0) for i in a]– Onyambu
Nov 18 '18 at 21:48