split an array into a list of arrays
How can I split a 2D array by a grouping variable, and return a list of arrays please (also the order is important).
To show expected outcome, the equivalent in R can be done as
> (A = matrix(c("a", "b", "a", "c", "b", "d"), nr=3, byrow=TRUE)) # input
[,1] [,2]
[1,] "a" "b"
[2,] "a" "c"
[3,] "b" "d"
> (split.data.frame(A, A[,1])) # output
$a
[,1] [,2]
[1,] "a" "b"
[2,] "a" "c"
$b
[,1] [,2]
[1,] "b" "d"
EDIT: To clarify: I'd like to split the array/matrix, A into a list of multiple arrays based on the unique values in the first column. That is, split A into one array where the first column has an a, and another array where the first column has a b.
I have tried Python equivalent of R "split"-function but this gives three arrays
import numpy as np
import itertools
A = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
b = a[:,0]
def split(x, f):
return list(itertools.compress(x, f)), list(itertools.compress(x, (not i for i in f)))
split(A, b)
([array(['a', 'b'], dtype='<U1'),
array(['a', 'c'], dtype='<U1'),
array(['b', 'd'], dtype='<U1')],
)
And also numpy.split, using np.split(A, b), but which needs integers. I though I may be able to use How to convert strings into integers in Python? to convert the letters to integers, but even if I pass integers, it doesn't split as expected
c = np.transpose(np.array([1,1,2]))
np.split(A, c) # returns 4 arrays
Can this be done? thanks
EDIT: please note that this is a small example, and the number of groups may be greater than two and they may not be ordered.
python arrays numpy
asked Nov 14 '18 at 18:42
user2957945
1,14011022
add a comment |
How can I split a 2D array by a grouping variable, and return a list of arrays please (also the order is important).
To show expected outcome, the equivalent in R can be done as
> (A = matrix(c("a", "b", "a", "c", "b", "d"), nr=3, byrow=TRUE)) # input
[,1] [,2]
[1,] "a" "b"
[2,] "a" "c"
[3,] "b" "d"
> (split.data.frame(A, A[,1])) # output
$a
[,1] [,2]
[1,] "a" "b"
[2,] "a" "c"
$b
[,1] [,2]
[1,] "b" "d"
EDIT: To clarify: I'd like to split the array/matrix, A into a list of multiple arrays based on the unique values in the first column. That is, split A into one array where the first column has an a, and another array where the first column has a b.
I have tried Python equivalent of R "split"-function but this gives three arrays
import numpy as np
import itertools
A = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
b = a[:,0]
def split(x, f):
return list(itertools.compress(x, f)), list(itertools.compress(x, (not i for i in f)))
split(A, b)
([array(['a', 'b'], dtype='<U1'),
array(['a', 'c'], dtype='<U1'),
array(['b', 'd'], dtype='<U1')],
)
And also numpy.split, using np.split(A, b), but which needs integers. I though I may be able to use How to convert strings into integers in Python? to convert the letters to integers, but even if I pass integers, it doesn't split as expected
c = np.transpose(np.array([1,1,2]))
np.split(A, c) # returns 4 arrays
Can this be done? thanks
EDIT: please note that this is a small example, and the number of groups may be greater than two and they may not be ordered.
python arrays numpy
asked Nov 14 '18 at 18:42
user2957945
1,14011022
Not sure I understand your expected output @user2957945
– RafaelC
Nov 14 '18 at 18:45
okay, thanks @RafaelC -- I'll clarify
– user2957945
Nov 14 '18 at 18:46
add a comment |
How can I split a 2D array by a grouping variable, and return a list of arrays please (also the order is important).
To show expected outcome, the equivalent in R can be done as
> (A = matrix(c("a", "b", "a", "c", "b", "d"), nr=3, byrow=TRUE)) # input
[,1] [,2]
[1,] "a" "b"
[2,] "a" "c"
[3,] "b" "d"
> (split.data.frame(A, A[,1])) # output
$a
[,1] [,2]
[1,] "a" "b"
[2,] "a" "c"
$b
[,1] [,2]
[1,] "b" "d"
EDIT: To clarify: I'd like to split the array/matrix, A into a list of multiple arrays based on the unique values in the first column. That is, split A into one array where the first column has an a, and another array where the first column has a b.
I have tried Python equivalent of R "split"-function but this gives three arrays
import numpy as np
import itertools
A = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
b = a[:,0]
def split(x, f):
return list(itertools.compress(x, f)), list(itertools.compress(x, (not i for i in f)))
split(A, b)
([array(['a', 'b'], dtype='<U1'),
array(['a', 'c'], dtype='<U1'),
array(['b', 'd'], dtype='<U1')],
)
And also numpy.split, using np.split(A, b), but which needs integers. I though I may be able to use How to convert strings into integers in Python? to convert the letters to integers, but even if I pass integers, it doesn't split as expected
c = np.transpose(np.array([1,1,2]))
np.split(A, c) # returns 4 arrays
Can this be done? thanks
EDIT: please note that this is a small example, and the number of groups may be greater than two and they may not be ordered.
python arrays numpy
asked Nov 14 '18 at 18:42
user2957945
1,14011022
How can I split a 2D array by a grouping variable, and return a list of arrays please (also the order is important).
To show expected outcome, the equivalent in R can be done as
> (A = matrix(c("a", "b", "a", "c", "b", "d"), nr=3, byrow=TRUE)) # input
[,1] [,2]
[1,] "a" "b"
[2,] "a" "c"
[3,] "b" "d"
> (split.data.frame(A, A[,1])) # output
$a
[,1] [,2]
[1,] "a" "b"
[2,] "a" "c"
$b
[,1] [,2]
[1,] "b" "d"
EDIT: To clarify: I'd like to split the array/matrix, A into a list of multiple arrays based on the unique values in the first column. That is, split A into one array where the first column has an a, and another array where the first column has a b.
I have tried Python equivalent of R "split"-function but this gives three arrays
import numpy as np
import itertools
A = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
b = a[:,0]
def split(x, f):
return list(itertools.compress(x, f)), list(itertools.compress(x, (not i for i in f)))
split(A, b)
([array(['a', 'b'], dtype='<U1'),
array(['a', 'c'], dtype='<U1'),
array(['b', 'd'], dtype='<U1')],
)
And also numpy.split, using np.split(A, b), but which needs integers. I though I may be able to use How to convert strings into integers in Python? to convert the letters to integers, but even if I pass integers, it doesn't split as expected
c = np.transpose(np.array([1,1,2]))
np.split(A, c) # returns 4 arrays
Can this be done? thanks
EDIT: please note that this is a small example, and the number of groups may be greater than two and they may not be ordered.
python arrays numpy
python arrays numpy
asked Nov 14 '18 at 18:42
user2957945
1,14011022
asked Nov 14 '18 at 18:42
user2957945
1,14011022
edited Nov 14 '18 at 18:57
asked Nov 14 '18 at 18:42
user2957945
1,14011022
asked Nov 14 '18 at 18:42
user2957945
1,14011022
asked Nov 14 '18 at 18:42
user2957945
1,14011022
1,14011022
Not sure I understand your expected output @user2957945
– RafaelC
Nov 14 '18 at 18:45
okay, thanks @RafaelC -- I'll clarify
– user2957945
Nov 14 '18 at 18:46
add a comment |
Not sure I understand your expected output @user2957945
– RafaelC
Nov 14 '18 at 18:45
okay, thanks @RafaelC -- I'll clarify
– user2957945
Nov 14 '18 at 18:46
Not sure I understand your expected output @user2957945
– RafaelC
Nov 14 '18 at 18:45
Not sure I understand your expected output @user2957945
– RafaelC
Nov 14 '18 at 18:45
okay, thanks @RafaelC -- I'll clarify
– user2957945
Nov 14 '18 at 18:46
okay, thanks @RafaelC -- I'll clarify
– user2957945
Nov 14 '18 at 18:46
add a comment |
2 Answers
2
active
oldest
votes
You can use pandas:
import pandas as pd
import numpy as np
a = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
listofdfs = {}
for n,g in pd.DataFrame(a).groupby(0):
listofdfs[n] = g
listofdfs['a'].values
Output:
array([['a', 'b'],
['a', 'c']], dtype=object)
And,
listofdfs['b'].values
Output:
array([['b', 'd']], dtype=object)
Or, you could use itertools groupby:
import numpy as np
from itertools import groupby
l = [np.stack(list(g)) for k, g in groupby(a, lambda x: x[0])]
l[0]
Output:
array([['a', 'b'],
['a', 'c']], dtype='<U1')
And,
l[1]
Output:
array([['b', 'd']], dtype='<U1')
answered Nov 14 '18 at 19:08
Scott Boston
52k72955
1
Great, thanks Scott, that looks good. I'd considered coercing to a dataframe but I thought there may be array tools -- but this is good.
– user2957945
Nov 14 '18 at 19:14
@user2957945 I did add an update.
– Scott Boston
Nov 14 '18 at 19:27
brilliant, thank you very much. I'm trapesing through stackoverflow.com/questions/773/…, so your edit gives me the output for my understanding to work towards
– user2957945
Nov 14 '18 at 19:30
add a comment |
If I understand your question, you can do simple slicing, as in:
a = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
x,y=a[:2,:],a[2,:]
x
array([['a', 'b'],
['a', 'c']], dtype='<U1')
y
array(['b', 'd'], dtype='<U1')
answered Nov 14 '18 at 18:49
G. Anderson
1,06829
Hi G.Anderson, thank you for your answer. This would fail fora = np.array([["a", "b"], ["b", "d"], ["a", "c"], ["b", "d"]]), or if there were more groups. Apologies maybe my example was to minimal.
– user2957945
Nov 14 '18 at 18:55
I see. I answered before you edited about grouping the splits based on value. Perhaps this answer might help?
– G. Anderson
Nov 14 '18 at 19:07
Thanks. That looks promising-- I'm just trying to tweak it to my example.
– user2957945
Nov 14 '18 at 19:13
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 use pandas:
import pandas as pd
import numpy as np
a = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
listofdfs = {}
for n,g in pd.DataFrame(a).groupby(0):
listofdfs[n] = g
listofdfs['a'].values
Output:
array([['a', 'b'],
['a', 'c']], dtype=object)
And,
listofdfs['b'].values
Output:
array([['b', 'd']], dtype=object)
Or, you could use itertools groupby:
import numpy as np
from itertools import groupby
l = [np.stack(list(g)) for k, g in groupby(a, lambda x: x[0])]
l[0]
Output:
array([['a', 'b'],
['a', 'c']], dtype='<U1')
And,
l[1]
Output:
array([['b', 'd']], dtype='<U1')
answered Nov 14 '18 at 19:08
Scott Boston
52k72955
1
Great, thanks Scott, that looks good. I'd considered coercing to a dataframe but I thought there may be array tools -- but this is good.
– user2957945
Nov 14 '18 at 19:14
@user2957945 I did add an update.
– Scott Boston
Nov 14 '18 at 19:27
brilliant, thank you very much. I'm trapesing through stackoverflow.com/questions/773/…, so your edit gives me the output for my understanding to work towards
– user2957945
Nov 14 '18 at 19:30
add a comment |
You can use pandas:
import pandas as pd
import numpy as np
a = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
listofdfs = {}
for n,g in pd.DataFrame(a).groupby(0):
listofdfs[n] = g
listofdfs['a'].values
Output:
array([['a', 'b'],
['a', 'c']], dtype=object)
And,
listofdfs['b'].values
Output:
array([['b', 'd']], dtype=object)
Or, you could use itertools groupby:
import numpy as np
from itertools import groupby
l = [np.stack(list(g)) for k, g in groupby(a, lambda x: x[0])]
l[0]
Output:
array([['a', 'b'],
['a', 'c']], dtype='<U1')
And,
l[1]
Output:
array([['b', 'd']], dtype='<U1')
answered Nov 14 '18 at 19:08
Scott Boston
52k72955
1
Great, thanks Scott, that looks good. I'd considered coercing to a dataframe but I thought there may be array tools -- but this is good.
– user2957945
Nov 14 '18 at 19:14
@user2957945 I did add an update.
– Scott Boston
Nov 14 '18 at 19:27
brilliant, thank you very much. I'm trapesing through stackoverflow.com/questions/773/…, so your edit gives me the output for my understanding to work towards
– user2957945
Nov 14 '18 at 19:30
add a comment |
You can use pandas:
import pandas as pd
import numpy as np
a = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
listofdfs = {}
for n,g in pd.DataFrame(a).groupby(0):
listofdfs[n] = g
listofdfs['a'].values
Output:
array([['a', 'b'],
['a', 'c']], dtype=object)
And,
listofdfs['b'].values
Output:
array([['b', 'd']], dtype=object)
Or, you could use itertools groupby:
import numpy as np
from itertools import groupby
l = [np.stack(list(g)) for k, g in groupby(a, lambda x: x[0])]
l[0]
Output:
array([['a', 'b'],
['a', 'c']], dtype='<U1')
And,
l[1]
Output:
array([['b', 'd']], dtype='<U1')
answered Nov 14 '18 at 19:08
Scott Boston
52k72955
You can use pandas:
import pandas as pd
import numpy as np
a = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
listofdfs = {}
for n,g in pd.DataFrame(a).groupby(0):
listofdfs[n] = g
listofdfs['a'].values
Output:
array([['a', 'b'],
['a', 'c']], dtype=object)
And,
listofdfs['b'].values
Output:
array([['b', 'd']], dtype=object)
Or, you could use itertools groupby:
import numpy as np
from itertools import groupby
l = [np.stack(list(g)) for k, g in groupby(a, lambda x: x[0])]
l[0]
Output:
array([['a', 'b'],
['a', 'c']], dtype='<U1')
And,
l[1]
Output:
array([['b', 'd']], dtype='<U1')
answered Nov 14 '18 at 19:08
Scott Boston
52k72955
edited Nov 14 '18 at 19:24
answered Nov 14 '18 at 19:08
Scott Boston
52k72955
answered Nov 14 '18 at 19:08
Scott Boston
52k72955
answered Nov 14 '18 at 19:08
Scott Boston
52k72955
52k72955
1
Great, thanks Scott, that looks good. I'd considered coercing to a dataframe but I thought there may be array tools -- but this is good.
– user2957945
Nov 14 '18 at 19:14
@user2957945 I did add an update.
– Scott Boston
Nov 14 '18 at 19:27
brilliant, thank you very much. I'm trapesing through stackoverflow.com/questions/773/…, so your edit gives me the output for my understanding to work towards
– user2957945
Nov 14 '18 at 19:30
add a comment |
1
Great, thanks Scott, that looks good. I'd considered coercing to a dataframe but I thought there may be array tools -- but this is good.
– user2957945
Nov 14 '18 at 19:14
@user2957945 I did add an update.
– Scott Boston
Nov 14 '18 at 19:27
brilliant, thank you very much. I'm trapesing through stackoverflow.com/questions/773/…, so your edit gives me the output for my understanding to work towards
– user2957945
Nov 14 '18 at 19:30
1
1
Great, thanks Scott, that looks good. I'd considered coercing to a dataframe but I thought there may be array tools -- but this is good.
– user2957945
Nov 14 '18 at 19:14
Great, thanks Scott, that looks good. I'd considered coercing to a dataframe but I thought there may be array tools -- but this is good.
– user2957945
Nov 14 '18 at 19:14
@user2957945 I did add an update.
– Scott Boston
Nov 14 '18 at 19:27
@user2957945 I did add an update.
– Scott Boston
Nov 14 '18 at 19:27
brilliant, thank you very much. I'm trapesing through stackoverflow.com/questions/773/…, so your edit gives me the output for my understanding to work towards
– user2957945
Nov 14 '18 at 19:30
brilliant, thank you very much. I'm trapesing through stackoverflow.com/questions/773/…, so your edit gives me the output for my understanding to work towards
– user2957945
Nov 14 '18 at 19:30
add a comment |
If I understand your question, you can do simple slicing, as in:
a = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
x,y=a[:2,:],a[2,:]
x
array([['a', 'b'],
['a', 'c']], dtype='<U1')
y
array(['b', 'd'], dtype='<U1')
answered Nov 14 '18 at 18:49
G. Anderson
1,06829
Hi G.Anderson, thank you for your answer. This would fail fora = np.array([["a", "b"], ["b", "d"], ["a", "c"], ["b", "d"]]), or if there were more groups. Apologies maybe my example was to minimal.
– user2957945
Nov 14 '18 at 18:55
I see. I answered before you edited about grouping the splits based on value. Perhaps this answer might help?
– G. Anderson
Nov 14 '18 at 19:07
Thanks. That looks promising-- I'm just trying to tweak it to my example.
– user2957945
Nov 14 '18 at 19:13
add a comment |
If I understand your question, you can do simple slicing, as in:
a = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
x,y=a[:2,:],a[2,:]
x
array([['a', 'b'],
['a', 'c']], dtype='<U1')
y
array(['b', 'd'], dtype='<U1')
answered Nov 14 '18 at 18:49
G. Anderson
1,06829
Hi G.Anderson, thank you for your answer. This would fail fora = np.array([["a", "b"], ["b", "d"], ["a", "c"], ["b", "d"]]), or if there were more groups. Apologies maybe my example was to minimal.
– user2957945
Nov 14 '18 at 18:55
I see. I answered before you edited about grouping the splits based on value. Perhaps this answer might help?
– G. Anderson
Nov 14 '18 at 19:07
Thanks. That looks promising-- I'm just trying to tweak it to my example.
– user2957945
Nov 14 '18 at 19:13
add a comment |
If I understand your question, you can do simple slicing, as in:
a = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
x,y=a[:2,:],a[2,:]
x
array([['a', 'b'],
['a', 'c']], dtype='<U1')
y
array(['b', 'd'], dtype='<U1')
answered Nov 14 '18 at 18:49
G. Anderson
1,06829
If I understand your question, you can do simple slicing, as in:
a = np.array([["a", "b"], ["a", "c"], ["b", "d"]])
x,y=a[:2,:],a[2,:]
x
array([['a', 'b'],
['a', 'c']], dtype='<U1')
y
array(['b', 'd'], dtype='<U1')
answered Nov 14 '18 at 18:49
G. Anderson
1,06829
answered Nov 14 '18 at 18:49
G. Anderson
1,06829
answered Nov 14 '18 at 18:49
G. Anderson
1,06829
answered Nov 14 '18 at 18:49
G. Anderson
1,06829
1,06829
Hi G.Anderson, thank you for your answer. This would fail fora = np.array([["a", "b"], ["b", "d"], ["a", "c"], ["b", "d"]]), or if there were more groups. Apologies maybe my example was to minimal.
– user2957945
Nov 14 '18 at 18:55
I see. I answered before you edited about grouping the splits based on value. Perhaps this answer might help?
– G. Anderson
Nov 14 '18 at 19:07
Thanks. That looks promising-- I'm just trying to tweak it to my example.
– user2957945
Nov 14 '18 at 19:13
add a comment |
Hi G.Anderson, thank you for your answer. This would fail fora = np.array([["a", "b"], ["b", "d"], ["a", "c"], ["b", "d"]]), or if there were more groups. Apologies maybe my example was to minimal.
– user2957945
Nov 14 '18 at 18:55
I see. I answered before you edited about grouping the splits based on value. Perhaps this answer might help?
– G. Anderson
Nov 14 '18 at 19:07
Thanks. That looks promising-- I'm just trying to tweak it to my example.
– user2957945
Nov 14 '18 at 19:13
Hi G.Anderson, thank you for your answer. This would fail for
a = np.array([["a", "b"], ["b", "d"], ["a", "c"], ["b", "d"]]), or if there were more groups. Apologies maybe my example was to minimal.– user2957945
Nov 14 '18 at 18:55
Hi G.Anderson, thank you for your answer. This would fail for
a = np.array([["a", "b"], ["b", "d"], ["a", "c"], ["b", "d"]]), or if there were more groups. Apologies maybe my example was to minimal.– user2957945
Nov 14 '18 at 18:55
I see. I answered before you edited about grouping the splits based on value. Perhaps this answer might help?
– G. Anderson
Nov 14 '18 at 19:07
I see. I answered before you edited about grouping the splits based on value. Perhaps this answer might help?
– G. Anderson
Nov 14 '18 at 19:07
Thanks. That looks promising-- I'm just trying to tweak it to my example.
– user2957945
Nov 14 '18 at 19:13
Thanks. That looks promising-- I'm just trying to tweak it to my example.
– user2957945
Nov 14 '18 at 19:13
add a comment |
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Not sure I understand your expected output @user2957945
– RafaelC
Nov 14 '18 at 18:45
okay, thanks @RafaelC -- I'll clarify
– user2957945
Nov 14 '18 at 18:46