Creating new matrix based on dictionary values and a list
up vote
0
down vote
favorite
The following is a simplified example:
val = [10,23,45,31,78,43,1,67,82]
indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]
samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3
Say I have a dictionary (samp) that has two groups: Group 0 and Group 1.
The dictionary has indicies for values in the vector val.
I want to pull out all of the values in val based on the given group in the dictionary by creating a 8 X 2 matrix,
Where i have the group and the value in two columns in order by index so it looks like this:
val group
10 2
23 0
45 2
31 0
87 0
43 1
1 1
67 0
82 1
How do I go about doing this?
python pandas dictionary numpy-ndarray
asked Nov 10 at 4:59
Sheila
84561327
add a comment |
up vote
0
down vote
favorite
The following is a simplified example:
val = [10,23,45,31,78,43,1,67,82]
indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]
samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3
Say I have a dictionary (samp) that has two groups: Group 0 and Group 1.
The dictionary has indicies for values in the vector val.
I want to pull out all of the values in val based on the given group in the dictionary by creating a 8 X 2 matrix,
Where i have the group and the value in two columns in order by index so it looks like this:
val group
10 2
23 0
45 2
31 0
87 0
43 1
1 1
67 0
82 1
How do I go about doing this?
python pandas dictionary numpy-ndarray
asked Nov 10 at 4:59
Sheila
84561327
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The following is a simplified example:
val = [10,23,45,31,78,43,1,67,82]
indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]
samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3
Say I have a dictionary (samp) that has two groups: Group 0 and Group 1.
The dictionary has indicies for values in the vector val.
I want to pull out all of the values in val based on the given group in the dictionary by creating a 8 X 2 matrix,
Where i have the group and the value in two columns in order by index so it looks like this:
val group
10 2
23 0
45 2
31 0
87 0
43 1
1 1
67 0
82 1
How do I go about doing this?
python pandas dictionary numpy-ndarray
asked Nov 10 at 4:59
Sheila
84561327
The following is a simplified example:
val = [10,23,45,31,78,43,1,67,82]
indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]
samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3
Say I have a dictionary (samp) that has two groups: Group 0 and Group 1.
The dictionary has indicies for values in the vector val.
I want to pull out all of the values in val based on the given group in the dictionary by creating a 8 X 2 matrix,
Where i have the group and the value in two columns in order by index so it looks like this:
val group
10 2
23 0
45 2
31 0
87 0
43 1
1 1
67 0
82 1
How do I go about doing this?
python pandas dictionary numpy-ndarray
python pandas dictionary numpy-ndarray
asked Nov 10 at 4:59
Sheila
84561327
asked Nov 10 at 4:59
Sheila
84561327
edited Nov 10 at 5:06
asked Nov 10 at 4:59
Sheila
84561327
asked Nov 10 at 4:59
Sheila
84561327
asked Nov 10 at 4:59
Sheila
84561327
84561327
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
One way to get this
[(j, next(k for k,v in samp.items() if i in v)) for i,j in enumerate(val)]
Output:
[(10, 2),
(23, 0),
(45, 2),
(31, 1),
(78, 0),
(43, 0),
(1, 1),
(67, 1),
(82, 0)]
answered Nov 10 at 5:26
Transhuman
2,6111411
add a comment |
up vote
0
down vote
Here is a solution without using pandas that outputs a (8,2) numpy matrix:
val = [10,23,45,31,78,43,1,67,82]
indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]
indices = [indx, indx2, indx3]
def get_group(x):
for i,indx_arr in enumerate(indices):
if x in indx_arr:
return i
pairs = [(v,get_group(i)) for i,v in enumerate(val)]
np.asarray(pairs)
array([[10, 2],
[23, 0],
[45, 2],
[31, 1],
[78, 0],
[43, 0],
[ 1, 1],
[67, 1],
[82, 0]])
answered Nov 10 at 5:19
Dani G
427411
add a comment |
up vote
0
down vote
Use dictionary comprehension to reverse the key, value pairs in dictionary and then use map:
df = pd.DataFrame(val,columns=['val'])
d = {value1:key for key,value in samp.items() for value1 in value}
df['group'] = df.index.map(d)
print(df)
val group
0 10 2
1 23 0
2 45 2
3 31 1
4 78 0
5 43 0
6 1 1
7 67 1
print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}
What if the values are numpy arrays:
indx = np.array([1,4,5,8])
indx2 = np.array([3,6,7])
indx3 = np.array([0,2])
samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3
print(samp)
{0: array([1, 4, 5, 8]), 1: array([3, 6, 7]), 2: array([0, 2])}
d = {value1:key for key,value in samp.items() for value1 in value}
print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}
answered Nov 10 at 5:09
Sandeep Kadapa
5,642427
Thanks @Sandeep. What if the values in the dictonaries are numpy arrays? I noticed that .items() is not supported for np.arrays
– Sheila
Nov 10 at 5:18
@Sheila It would work even if the values are numpy arrays check the update.
– Sandeep Kadapa
Nov 10 at 5:20
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
up vote
1
down vote
accepted
One way to get this
[(j, next(k for k,v in samp.items() if i in v)) for i,j in enumerate(val)]
Output:
[(10, 2),
(23, 0),
(45, 2),
(31, 1),
(78, 0),
(43, 0),
(1, 1),
(67, 1),
(82, 0)]
answered Nov 10 at 5:26
Transhuman
2,6111411
add a comment |
up vote
1
down vote
accepted
One way to get this
[(j, next(k for k,v in samp.items() if i in v)) for i,j in enumerate(val)]
Output:
[(10, 2),
(23, 0),
(45, 2),
(31, 1),
(78, 0),
(43, 0),
(1, 1),
(67, 1),
(82, 0)]
answered Nov 10 at 5:26
Transhuman
2,6111411
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
One way to get this
[(j, next(k for k,v in samp.items() if i in v)) for i,j in enumerate(val)]
Output:
[(10, 2),
(23, 0),
(45, 2),
(31, 1),
(78, 0),
(43, 0),
(1, 1),
(67, 1),
(82, 0)]
answered Nov 10 at 5:26
Transhuman
2,6111411
One way to get this
[(j, next(k for k,v in samp.items() if i in v)) for i,j in enumerate(val)]
Output:
[(10, 2),
(23, 0),
(45, 2),
(31, 1),
(78, 0),
(43, 0),
(1, 1),
(67, 1),
(82, 0)]
answered Nov 10 at 5:26
Transhuman
2,6111411
answered Nov 10 at 5:26
Transhuman
2,6111411
answered Nov 10 at 5:26
Transhuman
2,6111411
answered Nov 10 at 5:26
Transhuman
2,6111411
2,6111411
add a comment |
add a comment |
up vote
0
down vote
Here is a solution without using pandas that outputs a (8,2) numpy matrix:
val = [10,23,45,31,78,43,1,67,82]
indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]
indices = [indx, indx2, indx3]
def get_group(x):
for i,indx_arr in enumerate(indices):
if x in indx_arr:
return i
pairs = [(v,get_group(i)) for i,v in enumerate(val)]
np.asarray(pairs)
array([[10, 2],
[23, 0],
[45, 2],
[31, 1],
[78, 0],
[43, 0],
[ 1, 1],
[67, 1],
[82, 0]])
answered Nov 10 at 5:19
Dani G
427411
add a comment |
up vote
0
down vote
Here is a solution without using pandas that outputs a (8,2) numpy matrix:
val = [10,23,45,31,78,43,1,67,82]
indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]
indices = [indx, indx2, indx3]
def get_group(x):
for i,indx_arr in enumerate(indices):
if x in indx_arr:
return i
pairs = [(v,get_group(i)) for i,v in enumerate(val)]
np.asarray(pairs)
array([[10, 2],
[23, 0],
[45, 2],
[31, 1],
[78, 0],
[43, 0],
[ 1, 1],
[67, 1],
[82, 0]])
answered Nov 10 at 5:19
Dani G
427411
add a comment |
up vote
0
down vote
up vote
0
down vote
Here is a solution without using pandas that outputs a (8,2) numpy matrix:
val = [10,23,45,31,78,43,1,67,82]
indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]
indices = [indx, indx2, indx3]
def get_group(x):
for i,indx_arr in enumerate(indices):
if x in indx_arr:
return i
pairs = [(v,get_group(i)) for i,v in enumerate(val)]
np.asarray(pairs)
array([[10, 2],
[23, 0],
[45, 2],
[31, 1],
[78, 0],
[43, 0],
[ 1, 1],
[67, 1],
[82, 0]])
answered Nov 10 at 5:19
Dani G
427411
Here is a solution without using pandas that outputs a (8,2) numpy matrix:
val = [10,23,45,31,78,43,1,67,82]
indx = [1,4,5,8]
indx2 = [3,6,7]
indx3 = [0,2]
indices = [indx, indx2, indx3]
def get_group(x):
for i,indx_arr in enumerate(indices):
if x in indx_arr:
return i
pairs = [(v,get_group(i)) for i,v in enumerate(val)]
np.asarray(pairs)
array([[10, 2],
[23, 0],
[45, 2],
[31, 1],
[78, 0],
[43, 0],
[ 1, 1],
[67, 1],
[82, 0]])
answered Nov 10 at 5:19
Dani G
427411
answered Nov 10 at 5:19
Dani G
427411
answered Nov 10 at 5:19
Dani G
427411
answered Nov 10 at 5:19
Dani G
427411
427411
add a comment |
add a comment |
up vote
0
down vote
Use dictionary comprehension to reverse the key, value pairs in dictionary and then use map:
df = pd.DataFrame(val,columns=['val'])
d = {value1:key for key,value in samp.items() for value1 in value}
df['group'] = df.index.map(d)
print(df)
val group
0 10 2
1 23 0
2 45 2
3 31 1
4 78 0
5 43 0
6 1 1
7 67 1
print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}
What if the values are numpy arrays:
indx = np.array([1,4,5,8])
indx2 = np.array([3,6,7])
indx3 = np.array([0,2])
samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3
print(samp)
{0: array([1, 4, 5, 8]), 1: array([3, 6, 7]), 2: array([0, 2])}
d = {value1:key for key,value in samp.items() for value1 in value}
print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}
answered Nov 10 at 5:09
Sandeep Kadapa
5,642427
Thanks @Sandeep. What if the values in the dictonaries are numpy arrays? I noticed that .items() is not supported for np.arrays
– Sheila
Nov 10 at 5:18
@Sheila It would work even if the values are numpy arrays check the update.
– Sandeep Kadapa
Nov 10 at 5:20
add a comment |
up vote
0
down vote
Use dictionary comprehension to reverse the key, value pairs in dictionary and then use map:
df = pd.DataFrame(val,columns=['val'])
d = {value1:key for key,value in samp.items() for value1 in value}
df['group'] = df.index.map(d)
print(df)
val group
0 10 2
1 23 0
2 45 2
3 31 1
4 78 0
5 43 0
6 1 1
7 67 1
print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}
What if the values are numpy arrays:
indx = np.array([1,4,5,8])
indx2 = np.array([3,6,7])
indx3 = np.array([0,2])
samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3
print(samp)
{0: array([1, 4, 5, 8]), 1: array([3, 6, 7]), 2: array([0, 2])}
d = {value1:key for key,value in samp.items() for value1 in value}
print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}
answered Nov 10 at 5:09
Sandeep Kadapa
5,642427
Thanks @Sandeep. What if the values in the dictonaries are numpy arrays? I noticed that .items() is not supported for np.arrays
– Sheila
Nov 10 at 5:18
@Sheila It would work even if the values are numpy arrays check the update.
– Sandeep Kadapa
Nov 10 at 5:20
add a comment |
up vote
0
down vote
up vote
0
down vote
Use dictionary comprehension to reverse the key, value pairs in dictionary and then use map:
df = pd.DataFrame(val,columns=['val'])
d = {value1:key for key,value in samp.items() for value1 in value}
df['group'] = df.index.map(d)
print(df)
val group
0 10 2
1 23 0
2 45 2
3 31 1
4 78 0
5 43 0
6 1 1
7 67 1
print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}
What if the values are numpy arrays:
indx = np.array([1,4,5,8])
indx2 = np.array([3,6,7])
indx3 = np.array([0,2])
samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3
print(samp)
{0: array([1, 4, 5, 8]), 1: array([3, 6, 7]), 2: array([0, 2])}
d = {value1:key for key,value in samp.items() for value1 in value}
print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}
answered Nov 10 at 5:09
Sandeep Kadapa
5,642427
Use dictionary comprehension to reverse the key, value pairs in dictionary and then use map:
df = pd.DataFrame(val,columns=['val'])
d = {value1:key for key,value in samp.items() for value1 in value}
df['group'] = df.index.map(d)
print(df)
val group
0 10 2
1 23 0
2 45 2
3 31 1
4 78 0
5 43 0
6 1 1
7 67 1
print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}
What if the values are numpy arrays:
indx = np.array([1,4,5,8])
indx2 = np.array([3,6,7])
indx3 = np.array([0,2])
samp = {}
samp[0] = indx
samp[1] = indx2
samp[2] = indx3
print(samp)
{0: array([1, 4, 5, 8]), 1: array([3, 6, 7]), 2: array([0, 2])}
d = {value1:key for key,value in samp.items() for value1 in value}
print(d)
{1: 0, 4: 0, 5: 0, 8: 0, 3: 1, 6: 1, 7: 1, 0: 2, 2: 2}
answered Nov 10 at 5:09
Sandeep Kadapa
5,642427
edited Nov 10 at 5:21
answered Nov 10 at 5:09
Sandeep Kadapa
5,642427
answered Nov 10 at 5:09
Sandeep Kadapa
5,642427
answered Nov 10 at 5:09
Sandeep Kadapa
5,642427
5,642427
Thanks @Sandeep. What if the values in the dictonaries are numpy arrays? I noticed that .items() is not supported for np.arrays
– Sheila
Nov 10 at 5:18
@Sheila It would work even if the values are numpy arrays check the update.
– Sandeep Kadapa
Nov 10 at 5:20
add a comment |
Thanks @Sandeep. What if the values in the dictonaries are numpy arrays? I noticed that .items() is not supported for np.arrays
– Sheila
Nov 10 at 5:18
@Sheila It would work even if the values are numpy arrays check the update.
– Sandeep Kadapa
Nov 10 at 5:20
Thanks @Sandeep. What if the values in the dictonaries are numpy arrays? I noticed that .items() is not supported for np.arrays
– Sheila
Nov 10 at 5:18
Thanks @Sandeep. What if the values in the dictonaries are numpy arrays? I noticed that .items() is not supported for np.arrays
– Sheila
Nov 10 at 5:18
@Sheila It would work even if the values are numpy arrays check the update.
– Sandeep Kadapa
Nov 10 at 5:20
@Sheila It would work even if the values are numpy arrays check the update.
– Sandeep Kadapa
Nov 10 at 5:20
add a comment |
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