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scatter plot with different color
2019 Community Moderator ElectionWhat is the difference between @staticmethod and @classmethod?Difference between append vs. extend list methods in PythonPrint in terminal with colors?Difference between __str__ and __repr__?How to put the legend out of the plotSave plot to image file instead of displaying it using Matplotlibset bin colors in 2d histogram (polyochromatic plots)How to make IPython notebook matplotlib plot inlineLimited color choice in matplotlib?How do I create a line plot in matplotlib with different colors along the line, dependent on categorical variables?
I am new to matplotlib and trying to plot this liner regression with customized color for a specific independent variable:
colors=['red','blue','green','black']
X=array([[1000],[2000],[3000],[4500]]
y=array([[200000],[200000],[200000],[200000]]
plt.scatter(X, y, color = colors[0])
plt.plot(X, lin_reg.predict(X), color = 'blue')
plt.xlabel('X')
plt.ylabel('y')
plt.show()
I need to set the color to black when X==3000 so I am using np.where:
colors_z=(np.where(X==3000,colors[4],colors[0]))
plt.scatter(X, y, color = colors_z)
But I am getting color error. any Idea what I am doing wrong? Thanks
python matplotlib
asked Mar 6 at 15:22
user3570022user3570022
164
add a comment |
I am new to matplotlib and trying to plot this liner regression with customized color for a specific independent variable:
colors=['red','blue','green','black']
X=array([[1000],[2000],[3000],[4500]]
y=array([[200000],[200000],[200000],[200000]]
plt.scatter(X, y, color = colors[0])
plt.plot(X, lin_reg.predict(X), color = 'blue')
plt.xlabel('X')
plt.ylabel('y')
plt.show()
I need to set the color to black when X==3000 so I am using np.where:
colors_z=(np.where(X==3000,colors[4],colors[0]))
plt.scatter(X, y, color = colors_z)
But I am getting color error. any Idea what I am doing wrong? Thanks
python matplotlib
asked Mar 6 at 15:22
user3570022user3570022
164
add a comment |
I am new to matplotlib and trying to plot this liner regression with customized color for a specific independent variable:
colors=['red','blue','green','black']
X=array([[1000],[2000],[3000],[4500]]
y=array([[200000],[200000],[200000],[200000]]
plt.scatter(X, y, color = colors[0])
plt.plot(X, lin_reg.predict(X), color = 'blue')
plt.xlabel('X')
plt.ylabel('y')
plt.show()
I need to set the color to black when X==3000 so I am using np.where:
colors_z=(np.where(X==3000,colors[4],colors[0]))
plt.scatter(X, y, color = colors_z)
But I am getting color error. any Idea what I am doing wrong? Thanks
python matplotlib
asked Mar 6 at 15:22
user3570022user3570022
164
I am new to matplotlib and trying to plot this liner regression with customized color for a specific independent variable:
colors=['red','blue','green','black']
X=array([[1000],[2000],[3000],[4500]]
y=array([[200000],[200000],[200000],[200000]]
plt.scatter(X, y, color = colors[0])
plt.plot(X, lin_reg.predict(X), color = 'blue')
plt.xlabel('X')
plt.ylabel('y')
plt.show()
I need to set the color to black when X==3000 so I am using np.where:
colors_z=(np.where(X==3000,colors[4],colors[0]))
plt.scatter(X, y, color = colors_z)
But I am getting color error. any Idea what I am doing wrong? Thanks
python matplotlib
python matplotlib
asked Mar 6 at 15:22
user3570022user3570022
164
asked Mar 6 at 15:22
user3570022user3570022
164
asked Mar 6 at 15:22
user3570022user3570022
164
asked Mar 6 at 15:22
user3570022user3570022
164
asked Mar 6 at 15:22
user3570022user3570022
164
164
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
I think this does what you're looking for; using np.where is a bit overkill for this purpose:
X = [1000, 2000, 3000, 4500]
y = [200000, 3000, 200000, 200000]
colors = list(map(lambda x: 'r' if x == 3000 else 'b', X))
plt.scatter(X, y, color=colors)
plt.xlabel('X')
plt.ylabel('y')
plt.show()
answered Mar 6 at 15:37
rgkrgk
38339
Awesome. Thanks rgk. worked like a charm
– user3570022
Mar 6 at 15:51
1
You can also add a list comprehension to avoid lambda and map:colors = ['r' if x == 3000 else 'b' for x in X]
– Bazingaa
Mar 6 at 16:08
add a comment |
You've set colors_z to include colors[4] but there are only 4 colors in the list colors. The index for colors_z should be out of range. I'd dump the np.where in favor of a simple if statement or ternary operator. Something like:
# ternary operator example
plt.scatter(x, y, color = [colors[3] if x == 3000 else colors[0] for i in x])
Note that this will only work when x is exactly == 3000, but it doesn't throw a syntactical error on my console, so it should work in your regression.
answered Mar 6 at 15:37
SamSam
917
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I like your approach because it's the cleanest and simplest to read. For the color argument, shouldn't it be a list like socolor = [colors[3] if x == 3000 else colors[0] for x in X]?
– Guimoute
Mar 6 at 15:47
Because as of now, you evaluate ifx == 3000which is always false ([1000, 2000, 3000, 4500] != 3000) so you are stuck with a single color. You do want a list.
– Guimoute
Mar 6 at 15:53
Yeah that's probably better Guimote.
– Sam
Mar 6 at 16:09
That ran much nicer. Thanks!
– Sam
Mar 6 at 16:13
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
I think this does what you're looking for; using np.where is a bit overkill for this purpose:
X = [1000, 2000, 3000, 4500]
y = [200000, 3000, 200000, 200000]
colors = list(map(lambda x: 'r' if x == 3000 else 'b', X))
plt.scatter(X, y, color=colors)
plt.xlabel('X')
plt.ylabel('y')
plt.show()
answered Mar 6 at 15:37
rgkrgk
38339
Awesome. Thanks rgk. worked like a charm
– user3570022
Mar 6 at 15:51
1
You can also add a list comprehension to avoid lambda and map:colors = ['r' if x == 3000 else 'b' for x in X]
– Bazingaa
Mar 6 at 16:08
add a comment |
I think this does what you're looking for; using np.where is a bit overkill for this purpose:
X = [1000, 2000, 3000, 4500]
y = [200000, 3000, 200000, 200000]
colors = list(map(lambda x: 'r' if x == 3000 else 'b', X))
plt.scatter(X, y, color=colors)
plt.xlabel('X')
plt.ylabel('y')
plt.show()
answered Mar 6 at 15:37
rgkrgk
38339
Awesome. Thanks rgk. worked like a charm
– user3570022
Mar 6 at 15:51
1
You can also add a list comprehension to avoid lambda and map:colors = ['r' if x == 3000 else 'b' for x in X]
– Bazingaa
Mar 6 at 16:08
add a comment |
I think this does what you're looking for; using np.where is a bit overkill for this purpose:
X = [1000, 2000, 3000, 4500]
y = [200000, 3000, 200000, 200000]
colors = list(map(lambda x: 'r' if x == 3000 else 'b', X))
plt.scatter(X, y, color=colors)
plt.xlabel('X')
plt.ylabel('y')
plt.show()
answered Mar 6 at 15:37
rgkrgk
38339
I think this does what you're looking for; using np.where is a bit overkill for this purpose:
X = [1000, 2000, 3000, 4500]
y = [200000, 3000, 200000, 200000]
colors = list(map(lambda x: 'r' if x == 3000 else 'b', X))
plt.scatter(X, y, color=colors)
plt.xlabel('X')
plt.ylabel('y')
plt.show()
answered Mar 6 at 15:37
rgkrgk
38339
answered Mar 6 at 15:37
rgkrgk
38339
answered Mar 6 at 15:37
rgkrgk
38339
answered Mar 6 at 15:37
rgkrgk
38339
38339
Awesome. Thanks rgk. worked like a charm
– user3570022
Mar 6 at 15:51
1
You can also add a list comprehension to avoid lambda and map:colors = ['r' if x == 3000 else 'b' for x in X]
– Bazingaa
Mar 6 at 16:08
add a comment |
Awesome. Thanks rgk. worked like a charm
– user3570022
Mar 6 at 15:51
1
You can also add a list comprehension to avoid lambda and map:colors = ['r' if x == 3000 else 'b' for x in X]
– Bazingaa
Mar 6 at 16:08
Awesome. Thanks rgk. worked like a charm
– user3570022
Mar 6 at 15:51
Awesome. Thanks rgk. worked like a charm
– user3570022
Mar 6 at 15:51
1
1
You can also add a list comprehension to avoid lambda and map:
colors = ['r' if x == 3000 else 'b' for x in X]– Bazingaa
Mar 6 at 16:08
You can also add a list comprehension to avoid lambda and map:
colors = ['r' if x == 3000 else 'b' for x in X]– Bazingaa
Mar 6 at 16:08
add a comment |
You've set colors_z to include colors[4] but there are only 4 colors in the list colors. The index for colors_z should be out of range. I'd dump the np.where in favor of a simple if statement or ternary operator. Something like:
# ternary operator example
plt.scatter(x, y, color = [colors[3] if x == 3000 else colors[0] for i in x])
Note that this will only work when x is exactly == 3000, but it doesn't throw a syntactical error on my console, so it should work in your regression.
answered Mar 6 at 15:37
SamSam
917
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I like your approach because it's the cleanest and simplest to read. For the color argument, shouldn't it be a list like socolor = [colors[3] if x == 3000 else colors[0] for x in X]?
– Guimoute
Mar 6 at 15:47
Because as of now, you evaluate ifx == 3000which is always false ([1000, 2000, 3000, 4500] != 3000) so you are stuck with a single color. You do want a list.
– Guimoute
Mar 6 at 15:53
Yeah that's probably better Guimote.
– Sam
Mar 6 at 16:09
That ran much nicer. Thanks!
– Sam
Mar 6 at 16:13
add a comment |
You've set colors_z to include colors[4] but there are only 4 colors in the list colors. The index for colors_z should be out of range. I'd dump the np.where in favor of a simple if statement or ternary operator. Something like:
# ternary operator example
plt.scatter(x, y, color = [colors[3] if x == 3000 else colors[0] for i in x])
Note that this will only work when x is exactly == 3000, but it doesn't throw a syntactical error on my console, so it should work in your regression.
answered Mar 6 at 15:37
SamSam
917
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I like your approach because it's the cleanest and simplest to read. For the color argument, shouldn't it be a list like socolor = [colors[3] if x == 3000 else colors[0] for x in X]?
– Guimoute
Mar 6 at 15:47
Because as of now, you evaluate ifx == 3000which is always false ([1000, 2000, 3000, 4500] != 3000) so you are stuck with a single color. You do want a list.
– Guimoute
Mar 6 at 15:53
Yeah that's probably better Guimote.
– Sam
Mar 6 at 16:09
That ran much nicer. Thanks!
– Sam
Mar 6 at 16:13
add a comment |
You've set colors_z to include colors[4] but there are only 4 colors in the list colors. The index for colors_z should be out of range. I'd dump the np.where in favor of a simple if statement or ternary operator. Something like:
# ternary operator example
plt.scatter(x, y, color = [colors[3] if x == 3000 else colors[0] for i in x])
Note that this will only work when x is exactly == 3000, but it doesn't throw a syntactical error on my console, so it should work in your regression.
answered Mar 6 at 15:37
SamSam
917
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You've set colors_z to include colors[4] but there are only 4 colors in the list colors. The index for colors_z should be out of range. I'd dump the np.where in favor of a simple if statement or ternary operator. Something like:
# ternary operator example
plt.scatter(x, y, color = [colors[3] if x == 3000 else colors[0] for i in x])
Note that this will only work when x is exactly == 3000, but it doesn't throw a syntactical error on my console, so it should work in your regression.
answered Mar 6 at 15:37
SamSam
917
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 6 at 16:13
answered Mar 6 at 15:37
SamSam
917
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 6 at 15:37
SamSam
917
answered Mar 6 at 15:37
SamSam
917
917
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I like your approach because it's the cleanest and simplest to read. For the color argument, shouldn't it be a list like socolor = [colors[3] if x == 3000 else colors[0] for x in X]?
– Guimoute
Mar 6 at 15:47
Because as of now, you evaluate ifx == 3000which is always false ([1000, 2000, 3000, 4500] != 3000) so you are stuck with a single color. You do want a list.
– Guimoute
Mar 6 at 15:53
Yeah that's probably better Guimote.
– Sam
Mar 6 at 16:09
That ran much nicer. Thanks!
– Sam
Mar 6 at 16:13
add a comment |
1
I like your approach because it's the cleanest and simplest to read. For the color argument, shouldn't it be a list like socolor = [colors[3] if x == 3000 else colors[0] for x in X]?
– Guimoute
Mar 6 at 15:47
Because as of now, you evaluate ifx == 3000which is always false ([1000, 2000, 3000, 4500] != 3000) so you are stuck with a single color. You do want a list.
– Guimoute
Mar 6 at 15:53
Yeah that's probably better Guimote.
– Sam
Mar 6 at 16:09
That ran much nicer. Thanks!
– Sam
Mar 6 at 16:13
1
1
I like your approach because it's the cleanest and simplest to read. For the color argument, shouldn't it be a list like so
color = [colors[3] if x == 3000 else colors[0] for x in X]?– Guimoute
Mar 6 at 15:47
I like your approach because it's the cleanest and simplest to read. For the color argument, shouldn't it be a list like so
color = [colors[3] if x == 3000 else colors[0] for x in X]?– Guimoute
Mar 6 at 15:47
Because as of now, you evaluate if
x == 3000 which is always false ([1000, 2000, 3000, 4500] != 3000) so you are stuck with a single color. You do want a list.– Guimoute
Mar 6 at 15:53
Because as of now, you evaluate if
x == 3000 which is always false ([1000, 2000, 3000, 4500] != 3000) so you are stuck with a single color. You do want a list.– Guimoute
Mar 6 at 15:53
Yeah that's probably better Guimote.
– Sam
Mar 6 at 16:09
Yeah that's probably better Guimote.
– Sam
Mar 6 at 16:09
That ran much nicer. Thanks!
– Sam
Mar 6 at 16:13
That ran much nicer. Thanks!
– Sam
Mar 6 at 16:13
add a comment |
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