I have a somewhat complex mathematical function for which I want to find the meeting point of two graphs. I went through some answers but couldn't get what I wanted. Code below
from scipy.special import gamma, factorial
mu=[0,1,2,3,4,5]
b_value=[]
N_0=8000
c=3.78
b=[9.922]*6 #straight line
for i in range (len(mu)):
b = ((3.6*(c**3)*3.14*1000*N_0*(10**-6)/12)*gamma(6.01+mu[i]))/(6*3.14*c*N_0*(10**- 4)*gamma(4.67+mu[i]))**((6.01+mu[i])/(4.67+mu[i]))
b_value.append(b)
plt.figure(570)
plt.scatter(mu,b_value,color='r',lw='2',label = 'b_theoretical',marker="o")
plt.plot(mu,b,':',color='b',lw='2',label = 'b')
Basically I want to find the value of their intersection.Scatter plot can be plotted as line as well.
Thanks in advance.
Related
I am trying to construct a grouped vertical bar chart in Bokeh from a pandas dataframe. I'm struggling with understanding the use of factor_cmap and how the color mapping works with this function. There's an example in the documentation (https://docs.bokeh.org/en/latest/docs/user_guide/categorical.html#pandas) that was helpful to follow, here:
from bokeh.io import output_file, show
from bokeh.palettes import Spectral5
from bokeh.plotting import figure
from bokeh.sampledata.autompg import autompg_clean as df
from bokeh.transform import factor_cmap
output_file("bar_pandas_groupby_nested.html")
df.cyl = df.cyl.astype(str)
df.yr = df.yr.astype(str)
group = df.groupby(by=['cyl', 'mfr'])
index_cmap = factor_cmap('cyl_mfr', palette=Spectral5, factors=sorted(df.cyl.unique()), end=1)
p = figure(plot_width=800, plot_height=300, title="Mean MPG by # Cylinders and Manufacturer",
x_range=group, toolbar_location=None, tooltips=[("MPG", "#mpg_mean"), ("Cyl, Mfr", "#cyl_mfr")])
p.vbar(x='cyl_mfr', top='mpg_mean', width=1, source=group,
line_color="white", fill_color=index_cmap, )
p.y_range.start = 0
p.x_range.range_padding = 0.05
p.xgrid.grid_line_color = None
p.xaxis.axis_label = "Manufacturer grouped by # Cylinders"
p.xaxis.major_label_orientation = 1.2
p.outline_line_color = None
show(p)
This yields the following (again, a screen shot from the documentation):
Grouped Vbar output
I understand how factor_cmap is working here, I think. The index for the dataframe has multiple factors and we're only taking the first by slicing (as seen with the end = 1). But when I try to instead set coloring based on the second index level, mfr, (setting start = 1 , end = 2) , the index mapping breaks and I get this. I based this change on my assumption that the factors were hierarchical and I needed to slice them to get the second level.
I think I must be thinking about the indexing with these categorical factors wrong, but I'm not sure what I'm doing wrong. How do I get a categorical mapper to color by the second level of the factor? I assumed the format of the factors was ('cyl', 'mfr') but maybe that assumption is wrong?
Here's the documentation for factor_cmap, although it wasn't very helpful: https://docs.bokeh.org/en/latest/docs/reference/transform.html#bokeh.transform.factor_cmap .
If you mean you are trying this:
index_cmap = factor_cmap('cyl_mfr',
palette=Spectral5,
factors=sorted(df.cyl.unique()),
start=1, end=2)
Then there are at least two issues:
2 is out of bounds for the length of the list of sub-factors ('cyl', 'mfr'). You would just want start=1 and leave end with its default value of None (which means to the end of the list, as usual for any Python slice).
In this specific case, with start=1 that means "colormap based on mfr sub-factors of the values", but you are still configuring the cololormapper with the cylinders as the factors for the map:
factors=sorted(df.cyl.unique())
When the colormapper goes to look up a value with mfr="mazda" in the mapping, it does not find anything (because you only put cylinder values in the mapping) so it gets shaded the default color grey (as expected).
So you could do something like this:
index_cmap = factor_cmap('cyl_mfr',
palette=Spectral5,
factors=sorted(df.mfr.unique()),
start=1)
Which "works" modulo the fact that there are way more manufacturer values than there are colors in the Spectral5 palette:
In the real situation you'll need to make sure you use a palette as least as big as the number of (sub-)factors that you configure.
I have two points to ask about:
1)
I would like to understand what is precisely returned from the np.random.randn from NumPy and torch.randn from PyTorch. They both return a tensor with random numbers from a normal distribution with mean 0 and std 1, hence, a standard normal distribution. However, it is not the same thing as puting x values in the standard normal distribution function here and getting its respective image values y. The values returned by PyTorch and NumPy does not seem like this.
For me, it seems that both np.random.randn and torch.randn from these libraries returns the x values from the functions, not the image y as I calculated below. Is that correct?
normal = np.array([(1/np.sqrt(2*np.pi))*np.exp(-(1/2)*(i**2)) for i in range(-38,39)])
Printing the normal variable shows me something like this.
array([1.10e-314, 2.12e-298, 1.51e-282, 3.94e-267, 3.79e-252, 1.34e-237,
1.75e-223, 8.36e-210, 1.47e-196, 9.55e-184, 2.28e-171, 2.00e-159,
6.45e-148, 7.65e-137, 3.34e-126, 5.37e-116, 3.17e-106, 6.90e-097,
5.52e-088, 1.62e-079, 1.76e-071, 7.00e-064, 1.03e-056, 5.53e-050,
1.10e-043, 8.00e-038, 2.15e-032, 2.12e-027, 7.69e-023, 1.03e-018,
5.05e-015, 9.13e-012, 6.08e-009, 1.49e-006, 1.34e-004, 4.43e-003,
5.40e-002, 2.42e-001, 3.99e-001, 2.42e-001, 5.40e-002, 4.43e-003,
1.34e-004, 1.49e-006, 6.08e-009, 9.13e-012, 5.05e-015, 1.03e-018,
7.69e-023, 2.12e-027, 2.15e-032, 8.00e-038, 1.10e-043, 5.53e-050,
1.03e-056, 7.00e-064, 1.76e-071, 1.62e-079, 5.52e-088, 6.90e-097,
3.17e-106, 5.37e-116, 3.34e-126, 7.65e-137, 6.45e-148, 2.00e-159,
2.28e-171, 9.55e-184, 1.47e-196, 8.36e-210, 1.75e-223, 1.34e-237,
3.79e-252, 3.94e-267, 1.51e-282, 2.12e-298, 1.10e-314])
2) Also, if we ask these libraries that I want a matrix of values from a standard normal distribution, it means that all rows and columns are draw from the same standard distribution? If I want i.i.d distributions in every row, I would need to call np.random.randn over a for loop for each row and then vstack them?
1) Yes, they give you x and not phi(x) since the formula for phi(x) gives the probability density of sampling a value x. If you want to know the probability of getting values in an interval [a,b] you need to integrate phi(x) between a and b. Intuitively, if you look at the function phi(x) you'll see that you're more likely to get values near zero than, say, values near 1.
An easy way to see it, is look at the histogram of the sampled values.
import numpy as np
import matplotlib.pyplot as plt
samples = np.random.normal(size=[1000])
plt.hist(samples)
2) they're iid. Just use a 2d size like so:
samples = np.random.normal(size=[10, 10])
If I have a table with three columns where the first column represents the name of each point, the second column represent numerical data (mean) and the last column represent (second column + fixed number). The following an example how is the data looks like:
I want to plot this table so I have the following figure
If it is possible how I can plot it using either Microsoft Excel or python or R (Bokeh).
Alright, I only know how to do it in ggplot2, I will answer regarding R here.
These method only works if the data-frame is in the format you provided above.
I rename your column to Name.of.Method, Mean, Mean.2.2
Preparation
Loading csv data into R
df <- read.csv('yourdata.csv', sep = ',')
Change column name (Do this if you don't want to change the code below or else you will need to go through each parameter to match your column names.
names(df) <- c("Name.of.Method", "Mean", "Mean.2.2")
Method 1 - Using geom_segment()
ggplot() +
geom_segment(data=df,aes(x = Mean,
y = Name.of.Method,
xend = Mean.2.2,
yend = Name.of.Method))
So as you can see, geom_segment allows us to specify the end position of the line (Hence, xend and yend)
However, it does not look similar to the image you have above.
The line shape seems to represent error bar. Therefore, ggplot provides us with an error bar function.
Method 2 - Using geom_errorbarh()
ggplot(df, aes(y = Name.of.Method, x = Mean)) +
geom_errorbarh(aes(xmin = Mean, xmax = Mean.2.2), linetype = 1, height = .2)
Usually we don't use this method just to draw a line. However, its functionality fits your requirement. You can see that we use xmin and ymin to specify the head and the tail of the line.
The height input is to adjust the height of the bar at the end of the line in both ends.
I would use hbar for this:
from bokeh.io import show, output_file
from bokeh.plotting import figure
output_file("intervals.html")
names = ["SMB", "DB", "SB", "TB"]
p = figure(y_range=names, plot_height=350)
p.hbar(y=names, left=[4,3,2,1], right=[6.2, 5.2, 4.2, 3.2], height=0.3)
show(p)
However Whisker would also be an option if you really want whiskers instead of interval bars.
%Sampling Frequency
f=8000;
%Sampling Time
t=5;
%Data imported from microsoft Excel
matrix=Book2S1;
%Size Matrix
s=size(matrix);
h=s(1,1);
w=s(1,2);
%Set Up Rows and Columns
rows=(0:(f/2)/(h-1):f/2);
columns=(0:t/(w-1):t);
%plot
mesh(columns,rows,matrix);
xlabel('Time, s')
ylabel('Frequency, Hz')
zlabel('Power Spectral Density, V^2/Hz')`enter code here
This is the code that I type in to attempt to get a 3D plot. The goal is for me to obtain a plot that looks like the image listed below, but I continue getting a mesh error
Error using mesh (line 139)
Data inputs must be numeric, datetime, duration, categorical arrays or objects which can be converted to
double.
Error in Lab_3_1 (line 21)
mesh(columns,rows,matrix);
What my plot is supposed to look like.
The picture didn't want to get saved after being cropped, sorry people.
The following is a link to half of the data being used for this plot.
https://docs.google.com/spreadsheets/d/e/2PACX-1vRMWfmFYDnwMSPzahD8k-aWAXHstbNRdlY4gmOHJoXkLaBb4PY7zF5-41yFkQHR4g0w3LrMFiz3ZqWJ/pubhtml
Try substituting your 4049x50 matrix replacing my random matrix f:
% t=5;
% fs = 8000;
lower = -60;
upper = 20;
f = (upper-lower).*rand(4049,50) + lower;
% s=size(f);
% h=s(1,1);
% w=s(1,2);
% rows=(0:(fs/2)/(h-1):fs/2);
% columns=(0:t/(w-1):t);
mesh(f);
colormap('jet');
colorbar;
xlabel('Time, s')
ylabel('Frequency, Hz')
zlabel('Power Spectral Density V^2/Hz')
ylim([0 4000])
zlim([-100 40])
Using the random data matrix f, I get this:
I figured out that the values that were imported into MATLAB were converted into string values. I stopped using the import button and used the xlsread function instead, and that allowed me to import the numerical values without them being converted into strings.
Finished Code
Resulting 3D Plot
Thank you guys for the help and looking over the problem.
I have a dataset with 80 variables. I am interested in creating a function that will automate the creation of a 20 X 4 GridSpec in Matplotlib. Each subplot would either contain a histogram or a barplot for each of the 80 variables in the data. As a first step, I successfully created two functions (I call them 'counts' and 'histogram') that contain the layout of the plot that I want. Both of them work when tested on individual variables. As a next step, I attempted to create a function that would take the column names, loop through a conditional to test whether the data type is an object or otherwise and call the right function based on the datatype as a new subplot. Here is the code that I have so far:
Creates list of coordinates we will need for subplot specification:
A = np.arange(21)
B = np.arange(4)
coords = []
for i in A:
for j in B:
coords.append([A[i], B[j]])
#Create the gridspec and layout the figure
import matplotlib.gridspec as gridspec
fig = plt.figure(figsize=(12,6))
gs = gridspec.GridSpec(2,4)
#Function that relies on what we've done above:
def grid(cols=['MSZoning', 'LotFrontage', 'LotArea', 'Street', 'Alley']):
for i in cols:
for vals in coords:
if str(train[i].dtype) == 'object':
plt.subplot('gs'+str(vals))
counts(cols)
else:
plt.subplot('gs'+str(vals))
histogram(cols)
When attempted, this code returns an error:
ValueError: Single argument to subplot must be a 3-digit integer
For purposes of helping you visualize, what I am hoping to achieve, I attach the screen shot below, which was produced by the line by line coding (with my created helper functions) I am trying to avoid:
Can anyone help me figure out where I am going wrong? I would appreciate any advice. Thank you!
The line plt.subplot('gs'+str(vals)) cannot work; which is also what the error tells you.
As can be seen from the matplotlib GridSpec tutorial, it needs to be
ax = plt.subplot(gs[0, 0])
So in your case you may use the values from the list as
ax = plt.subplot(gs[vals[0], vals[1]])
Mind that you also need to make sure that the coords list must have the n*m elements, if the gridspec is defined as gs = gridspec.GridSpec(n,m).