Python - Replacing repeating elements in a list with unique elements from another lists
Multi tool use
2
I'm trying to assign unique colour values to each repeating element in a list. For this I have two different lists, one being the list which contains the repeating values (eg. labels = [1, 5, 6, 7, 8, 12, 13, 17]).
The second list (or list of list) contains the hue information itself:
group_pal = seaborn.husl_palette(len(set(labels)), s=.45).
The key here is that the length of group_pal is always the number of unique elements in labels.
Is there a python one liner to perform this operation ?
Input:
labels = [0, 0, 0, 1, 1, 2, 3]
group_pal = sns.husl_palette(len(set(labels)), s=.45)
Desired output:
labels = mapingMacro(labels, group_pal)
labels:
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.7256380093027939, 0.5865684184445076, 0.45124969098702544]
[0.7256380093027939, 0.5865684184445076, 0.45124969098702544]
[0.601243246823196, 0.6281411529879642, 0.44959498566071004]
[0.46712078684915886, 0.6454760674453914, 0.6277122757100324]
python list dictionary matplotlib seaborn
edited Dec 28 '18 at 3:16
meW
1,701114
asked Dec 28 '18 at 3:03
Siddharth
10411
|
show 5 more comments
2
I'm trying to assign unique colour values to each repeating element in a list. For this I have two different lists, one being the list which contains the repeating values (eg. labels = [1, 5, 6, 7, 8, 12, 13, 17]).
The second list (or list of list) contains the hue information itself:
group_pal = seaborn.husl_palette(len(set(labels)), s=.45).
The key here is that the length of group_pal is always the number of unique elements in labels.
Is there a python one liner to perform this operation ?
Input:
labels = [0, 0, 0, 1, 1, 2, 3]
group_pal = sns.husl_palette(len(set(labels)), s=.45)
Desired output:
labels = mapingMacro(labels, group_pal)
labels:
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.7256380093027939, 0.5865684184445076, 0.45124969098702544]
[0.7256380093027939, 0.5865684184445076, 0.45124969098702544]
[0.601243246823196, 0.6281411529879642, 0.44959498566071004]
[0.46712078684915886, 0.6454760674453914, 0.6277122757100324]
python list dictionary matplotlib seaborn
edited Dec 28 '18 at 3:16
meW
1,701114
asked Dec 28 '18 at 3:03
Siddharth
10411
So you just want[group_pal[i] for i in labels]?
– Jack Moody
Dec 28 '18 at 3:23
@JackMoody That work for now, but with a listlabelslike even:[3,3,3,2,2,1,0]won't work as expected.
– U9-Forward
Dec 28 '18 at 3:36
@U9-Forward I’m a bit confused. Why wouldn’t it work for that list?
– Jack Moody
Dec 28 '18 at 3:40
@JackMoody Because your simply getting the values with the element as the index, so if the first one is the last one, it wouldn't work.
– U9-Forward
Dec 28 '18 at 3:41
@JackMoody Try it. and see
– U9-Forward
Dec 28 '18 at 3:42
|
show 5 more comments
2
2
2
I'm trying to assign unique colour values to each repeating element in a list. For this I have two different lists, one being the list which contains the repeating values (eg. labels = [1, 5, 6, 7, 8, 12, 13, 17]).
The second list (or list of list) contains the hue information itself:
group_pal = seaborn.husl_palette(len(set(labels)), s=.45).
The key here is that the length of group_pal is always the number of unique elements in labels.
Is there a python one liner to perform this operation ?
Input:
labels = [0, 0, 0, 1, 1, 2, 3]
group_pal = sns.husl_palette(len(set(labels)), s=.45)
Desired output:
labels = mapingMacro(labels, group_pal)
labels:
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.7256380093027939, 0.5865684184445076, 0.45124969098702544]
[0.7256380093027939, 0.5865684184445076, 0.45124969098702544]
[0.601243246823196, 0.6281411529879642, 0.44959498566071004]
[0.46712078684915886, 0.6454760674453914, 0.6277122757100324]
python list dictionary matplotlib seaborn
edited Dec 28 '18 at 3:16
meW
1,701114
asked Dec 28 '18 at 3:03
Siddharth
10411
I'm trying to assign unique colour values to each repeating element in a list. For this I have two different lists, one being the list which contains the repeating values (eg. labels = [1, 5, 6, 7, 8, 12, 13, 17]).
The second list (or list of list) contains the hue information itself:
group_pal = seaborn.husl_palette(len(set(labels)), s=.45).
The key here is that the length of group_pal is always the number of unique elements in labels.
Is there a python one liner to perform this operation ?
Input:
labels = [0, 0, 0, 1, 1, 2, 3]
group_pal = sns.husl_palette(len(set(labels)), s=.45)
Desired output:
labels = mapingMacro(labels, group_pal)
labels:
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723]
[0.7256380093027939, 0.5865684184445076, 0.45124969098702544]
[0.7256380093027939, 0.5865684184445076, 0.45124969098702544]
[0.601243246823196, 0.6281411529879642, 0.44959498566071004]
[0.46712078684915886, 0.6454760674453914, 0.6277122757100324]
python list dictionary matplotlib seaborn
python list dictionary matplotlib seaborn
edited Dec 28 '18 at 3:16
meW
1,701114
asked Dec 28 '18 at 3:03
Siddharth
10411
edited Dec 28 '18 at 3:16
meW
1,701114
asked Dec 28 '18 at 3:03
Siddharth
10411
edited Dec 28 '18 at 3:16
meW
1,701114
edited Dec 28 '18 at 3:16
meW
1,701114
edited Dec 28 '18 at 3:16
meW
1,701114
1,701114
asked Dec 28 '18 at 3:03
Siddharth
10411
asked Dec 28 '18 at 3:03
Siddharth
10411
asked Dec 28 '18 at 3:03
Siddharth
10411
10411
So you just want[group_pal[i] for i in labels]?
– Jack Moody
Dec 28 '18 at 3:23
@JackMoody That work for now, but with a listlabelslike even:[3,3,3,2,2,1,0]won't work as expected.
– U9-Forward
Dec 28 '18 at 3:36
@U9-Forward I’m a bit confused. Why wouldn’t it work for that list?
– Jack Moody
Dec 28 '18 at 3:40
@JackMoody Because your simply getting the values with the element as the index, so if the first one is the last one, it wouldn't work.
– U9-Forward
Dec 28 '18 at 3:41
@JackMoody Try it. and see
– U9-Forward
Dec 28 '18 at 3:42
|
show 5 more comments
So you just want[group_pal[i] for i in labels]?
– Jack Moody
Dec 28 '18 at 3:23
@JackMoody That work for now, but with a listlabelslike even:[3,3,3,2,2,1,0]won't work as expected.
– U9-Forward
Dec 28 '18 at 3:36
@U9-Forward I’m a bit confused. Why wouldn’t it work for that list?
– Jack Moody
Dec 28 '18 at 3:40
@JackMoody Because your simply getting the values with the element as the index, so if the first one is the last one, it wouldn't work.
– U9-Forward
Dec 28 '18 at 3:41
@JackMoody Try it. and see
– U9-Forward
Dec 28 '18 at 3:42
So you just want
[group_pal[i] for i in labels]?– Jack Moody
Dec 28 '18 at 3:23
So you just want
[group_pal[i] for i in labels]?– Jack Moody
Dec 28 '18 at 3:23
@JackMoody That work for now, but with a list
labels like even: [3,3,3,2,2,1,0] won't work as expected.– U9-Forward
Dec 28 '18 at 3:36
@JackMoody That work for now, but with a list
labels like even: [3,3,3,2,2,1,0] won't work as expected.– U9-Forward
Dec 28 '18 at 3:36
@U9-Forward I’m a bit confused. Why wouldn’t it work for that list?
– Jack Moody
Dec 28 '18 at 3:40
@U9-Forward I’m a bit confused. Why wouldn’t it work for that list?
– Jack Moody
Dec 28 '18 at 3:40
@JackMoody Because your simply getting the values with the element as the index, so if the first one is the last one, it wouldn't work.
– U9-Forward
Dec 28 '18 at 3:41
@JackMoody Because your simply getting the values with the element as the index, so if the first one is the last one, it wouldn't work.
– U9-Forward
Dec 28 '18 at 3:41
@JackMoody Try it. and see
– U9-Forward
Dec 28 '18 at 3:42
@JackMoody Try it. and see
– U9-Forward
Dec 28 '18 at 3:42
|
show 5 more comments
1 Answer
1
active
oldest
votes
0
Use:
group_pal = [i for x,y in zip(sns.husl_palette(len(set(labels)), s=.45),
sorted(set(labels),key=labels.index))
for i in np.repeat([x],labels.count(y),axis=0).tolist()]
And now:
print(group_pal)
Is:
[[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.601243246823196, 0.6281411529879642, 0.44959498566071004],
[0.601243246823196, 0.6281411529879642, 0.44959498566071004],
[0.46712078684915886, 0.6454760674453914, 0.6277122757100324],
[0.6254162090818173, 0.5854245228463807, 0.7893617517727602]]
edited Dec 28 '18 at 3:25
James
13.1k11532
answered Dec 28 '18 at 3:23
U9-Forward
13.3k21237
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
Use:
group_pal = [i for x,y in zip(sns.husl_palette(len(set(labels)), s=.45),
sorted(set(labels),key=labels.index))
for i in np.repeat([x],labels.count(y),axis=0).tolist()]
And now:
print(group_pal)
Is:
[[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.601243246823196, 0.6281411529879642, 0.44959498566071004],
[0.601243246823196, 0.6281411529879642, 0.44959498566071004],
[0.46712078684915886, 0.6454760674453914, 0.6277122757100324],
[0.6254162090818173, 0.5854245228463807, 0.7893617517727602]]
edited Dec 28 '18 at 3:25
James
13.1k11532
answered Dec 28 '18 at 3:23
U9-Forward
13.3k21237
add a comment |
0
Use:
group_pal = [i for x,y in zip(sns.husl_palette(len(set(labels)), s=.45),
sorted(set(labels),key=labels.index))
for i in np.repeat([x],labels.count(y),axis=0).tolist()]
And now:
print(group_pal)
Is:
[[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.601243246823196, 0.6281411529879642, 0.44959498566071004],
[0.601243246823196, 0.6281411529879642, 0.44959498566071004],
[0.46712078684915886, 0.6454760674453914, 0.6277122757100324],
[0.6254162090818173, 0.5854245228463807, 0.7893617517727602]]
edited Dec 28 '18 at 3:25
James
13.1k11532
answered Dec 28 '18 at 3:23
U9-Forward
13.3k21237
add a comment |
0
0
0
Use:
group_pal = [i for x,y in zip(sns.husl_palette(len(set(labels)), s=.45),
sorted(set(labels),key=labels.index))
for i in np.repeat([x],labels.count(y),axis=0).tolist()]
And now:
print(group_pal)
Is:
[[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.601243246823196, 0.6281411529879642, 0.44959498566071004],
[0.601243246823196, 0.6281411529879642, 0.44959498566071004],
[0.46712078684915886, 0.6454760674453914, 0.6277122757100324],
[0.6254162090818173, 0.5854245228463807, 0.7893617517727602]]
edited Dec 28 '18 at 3:25
James
13.1k11532
answered Dec 28 '18 at 3:23
U9-Forward
13.3k21237
Use:
group_pal = [i for x,y in zip(sns.husl_palette(len(set(labels)), s=.45),
sorted(set(labels),key=labels.index))
for i in np.repeat([x],labels.count(y),axis=0).tolist()]
And now:
print(group_pal)
Is:
[[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.8167028311697733, 0.5345122109266688, 0.5750280113923723],
[0.601243246823196, 0.6281411529879642, 0.44959498566071004],
[0.601243246823196, 0.6281411529879642, 0.44959498566071004],
[0.46712078684915886, 0.6454760674453914, 0.6277122757100324],
[0.6254162090818173, 0.5854245228463807, 0.7893617517727602]]
edited Dec 28 '18 at 3:25
James
13.1k11532
answered Dec 28 '18 at 3:23
U9-Forward
13.3k21237
edited Dec 28 '18 at 3:25
James
13.1k11532
edited Dec 28 '18 at 3:25
James
13.1k11532
edited Dec 28 '18 at 3:25
James
13.1k11532
13.1k11532
answered Dec 28 '18 at 3:23
U9-Forward
13.3k21237
answered Dec 28 '18 at 3:23
U9-Forward
13.3k21237
answered Dec 28 '18 at 3:23
U9-Forward
13.3k21237
13.3k21237
add a comment |
add a comment |
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8k7353DE70cvJaQFHV N207IlFTWbE7Ie j4kBIhaqiEvOqL6m,o43dj
So you just want
[group_pal[i] for i in labels]?– Jack Moody
Dec 28 '18 at 3:23
@JackMoody That work for now, but with a list
labelslike even:[3,3,3,2,2,1,0]won't work as expected.– U9-Forward
Dec 28 '18 at 3:36
@U9-Forward I’m a bit confused. Why wouldn’t it work for that list?
– Jack Moody
Dec 28 '18 at 3:40
@JackMoody Because your simply getting the values with the element as the index, so if the first one is the last one, it wouldn't work.
– U9-Forward
Dec 28 '18 at 3:41
@JackMoody Try it. and see
– U9-Forward
Dec 28 '18 at 3:42