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The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
My question concerns the following two edits: [1] [2]
When I saw what looked like a normal vandalism revert on my watchlist, I looked into it. To my surprise the Internet Assigned Numbers Authority (IANA) has macon-tcp and macon-udp assigned to port 456. News to me. I would have reverted the apparent vandalism myself.
The obvious next step is to figure out what macon-tcp and macon-udp are and see what reliable sources say. I am having a heck of a time finding anything at all other than the bare fact that the IANA) assigned macon-tcp and macon-udp to port 456. Can anyone find anything on those services?
In the back of my mind I keep thinking that I may have used those names as non-public UDP/TCP placeholder values on a long-ago engineering project. I am really hoping that my temporary names didn't somehow leak out and get assigned... -- Guy Macon ( talk) 03:27, 7 November 2014 (UTC)
I am trying to re-plot: /info/en/?search=File:Poisson_pmf.svg I am using Python 2.7, and have a problem at the "bx = plt.legend(A, " part. Trying to run it like that produces the error message: "/usr/lib/pymodules/python2.7/matplotlib/legend.py:613: UserWarning: Legend does not support [<matplotlib.lines.Line2D object at 0x7fc82d6e0ad0>] Use proxy artist instead.
http://matplotlib.sourceforge.net/users/legend_guide.html#using-proxy-artist
(str(orig_handle),))"
Trying to solve the problem, I went to the docs:
"Using Proxy Artist When you want to display legend for an artist not supported by matplotlib, you may use another artist as a proxy. For example, you may create a proxy artist without adding it to the axes (so the proxy artist will not be drawn in the main axes) and feed it to the legend function.: p = Rectangle((0, 0), 1, 1, fc="r") legend([p], ["Red Rectangle"])"
So, I adapted the script above, removing A and introducing: c = [plt.Circle(1, fc=col[1]), plt.Circle(1, fc=col[4]), plt.Circle(0, fc=col[10])]
However, this does not depicts the legend icons as circles, but as rectangless, although the color is right. How can I solve this problem? Whatever I do, the colored element in the legend are depicted as rectangles, and not circles. -- Senteni ( talk) 17:46, 7 November 2014 (UTC)
#!/usr/bin/python
import numpy as np
import matplotlib.pyplot as plt
import scipy.special as sp
X = np.arange(0,21)
col = {1: 'orange', 4: 'purple', 10: 'lightblue'}
##
## PMF
##
plt.clf()
plt.figure(figsize=(4,3.2))
plt.axes([0.17,0.13,0.79,0.8])
plt.hold(True)
A = []
for L in 1,4,10:
P = -L + X*np.log(L) - sp.gammaln(X+1)
P = np.exp(P)
plt.plot(X, P, '-', color='grey')
a, = plt.plot(X, P, 'o', color=colL])
A.append(a)
plt.xlabel("k")
plt.ylabel(r"P(X=k)")
bx = plt.legend(A, (r"$\lambda=1$", r"$\lambda=4$", r"$\lambda=10$"),\
numpoints=1, handletextpad=0, loc="upper right")
bx.draw_frame(False)
plt.xlim(-1,21)
plt.savefig("poisson_pmf.pdf")
plt.savefig("poisson_pmf.svg")
#!/usr/bin/python
import numpy as np
import matplotlib.pyplot as plt
import scipy.special as sp
X = np.arange(0,21)
col = {1: 'orange', 4: 'purple', 10: 'lightblue'}
##
## PMF
##
plt.clf()
plt.figure(figsize=(4,3.2))
plt.axes([0.17,0.13,0.79,0.8])
plt.hold(True)
A = []
for L in 1,4,10:
P = -L + X*np.log(L) - sp.gammaln(X+1)
P = np.exp(P)
plt.plot(X, P, '-', color='grey')
a = plt.plot(X, P, 'o', color=colL])
A.append(a0])
plt.xlabel("k")
plt.ylabel(r"P(X=k)")
bx = plt.legend(A, (r"$\lambda=1$", r"$\lambda=4$", r"$\lambda=10$"),\
numpoints=1, handletextpad=0, loc="upper right")
bx.draw_frame(False)
plt.xlim(-1,21)
plt.savefig("poisson_pmf.pdf")
plt.savefig("poisson_pmf.svg")
Referring to this topic, I described what I was wanting to know to the people who answer ProBoards questions, and it took some effort from them to finally get the idea of what I expected. Here are the two answers I got:
It is really just that. There is a code to check where you have scrolled to on the page. Once you reach a scroll point where you have gone past where the pagination page initially sits on the page, an absolute positioning attribute gets added to stick it to the top of the page. It's not really software so to speak, just, as you mentioned before, CSS and Javascript [ jQuery ].
Fixed positioning, not absolute positioning. But the rest is right. :P
— Vchimpanzee • talk • contributions • 19:37, 7 November 2014 (UTC)
In Microsoft Excel (2013), you can perform operations on rows and also on columns. For example, I can total up (add) all of the numbers in a specific row (or in a specific column). Is it possible to perform a function (let's say, addition) for a range that includes both rows and columns? For example, let's say that I want a total of all of the values in the range from Columns A, B, C, D, and E and Rows 1, 2, 3, 4, and 5. So, that would be a "rectangle" of 25 values in 25 different cells (A1, A2, B1, B2, ... D4, D5, E4, E5). How can I total up all of the values in those 25 cells? Thanks. I am asking for a "one-step" formula. I already know that I can total up Column A, then total up Column B, and so forth with Columns C, D, and E ... and then add up all those five totals. But, I'd like to know if there is a single one-step function (as opposed to a combination of several functions with several steps) that will give the result I want? Thanks. Joseph A. Spadaro ( talk) 20:04, 7 November 2014 (UTC)
=SUM(A1:E5)
Computing desk | ||
---|---|---|
< November 6 | << Oct | November | Dec >> | November 8 > |
Welcome to the Wikipedia Computing Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
My question concerns the following two edits: [1] [2]
When I saw what looked like a normal vandalism revert on my watchlist, I looked into it. To my surprise the Internet Assigned Numbers Authority (IANA) has macon-tcp and macon-udp assigned to port 456. News to me. I would have reverted the apparent vandalism myself.
The obvious next step is to figure out what macon-tcp and macon-udp are and see what reliable sources say. I am having a heck of a time finding anything at all other than the bare fact that the IANA) assigned macon-tcp and macon-udp to port 456. Can anyone find anything on those services?
In the back of my mind I keep thinking that I may have used those names as non-public UDP/TCP placeholder values on a long-ago engineering project. I am really hoping that my temporary names didn't somehow leak out and get assigned... -- Guy Macon ( talk) 03:27, 7 November 2014 (UTC)
I am trying to re-plot: /info/en/?search=File:Poisson_pmf.svg I am using Python 2.7, and have a problem at the "bx = plt.legend(A, " part. Trying to run it like that produces the error message: "/usr/lib/pymodules/python2.7/matplotlib/legend.py:613: UserWarning: Legend does not support [<matplotlib.lines.Line2D object at 0x7fc82d6e0ad0>] Use proxy artist instead.
http://matplotlib.sourceforge.net/users/legend_guide.html#using-proxy-artist
(str(orig_handle),))"
Trying to solve the problem, I went to the docs:
"Using Proxy Artist When you want to display legend for an artist not supported by matplotlib, you may use another artist as a proxy. For example, you may create a proxy artist without adding it to the axes (so the proxy artist will not be drawn in the main axes) and feed it to the legend function.: p = Rectangle((0, 0), 1, 1, fc="r") legend([p], ["Red Rectangle"])"
So, I adapted the script above, removing A and introducing: c = [plt.Circle(1, fc=col[1]), plt.Circle(1, fc=col[4]), plt.Circle(0, fc=col[10])]
However, this does not depicts the legend icons as circles, but as rectangless, although the color is right. How can I solve this problem? Whatever I do, the colored element in the legend are depicted as rectangles, and not circles. -- Senteni ( talk) 17:46, 7 November 2014 (UTC)
#!/usr/bin/python
import numpy as np
import matplotlib.pyplot as plt
import scipy.special as sp
X = np.arange(0,21)
col = {1: 'orange', 4: 'purple', 10: 'lightblue'}
##
## PMF
##
plt.clf()
plt.figure(figsize=(4,3.2))
plt.axes([0.17,0.13,0.79,0.8])
plt.hold(True)
A = []
for L in 1,4,10:
P = -L + X*np.log(L) - sp.gammaln(X+1)
P = np.exp(P)
plt.plot(X, P, '-', color='grey')
a, = plt.plot(X, P, 'o', color=colL])
A.append(a)
plt.xlabel("k")
plt.ylabel(r"P(X=k)")
bx = plt.legend(A, (r"$\lambda=1$", r"$\lambda=4$", r"$\lambda=10$"),\
numpoints=1, handletextpad=0, loc="upper right")
bx.draw_frame(False)
plt.xlim(-1,21)
plt.savefig("poisson_pmf.pdf")
plt.savefig("poisson_pmf.svg")
#!/usr/bin/python
import numpy as np
import matplotlib.pyplot as plt
import scipy.special as sp
X = np.arange(0,21)
col = {1: 'orange', 4: 'purple', 10: 'lightblue'}
##
## PMF
##
plt.clf()
plt.figure(figsize=(4,3.2))
plt.axes([0.17,0.13,0.79,0.8])
plt.hold(True)
A = []
for L in 1,4,10:
P = -L + X*np.log(L) - sp.gammaln(X+1)
P = np.exp(P)
plt.plot(X, P, '-', color='grey')
a = plt.plot(X, P, 'o', color=colL])
A.append(a0])
plt.xlabel("k")
plt.ylabel(r"P(X=k)")
bx = plt.legend(A, (r"$\lambda=1$", r"$\lambda=4$", r"$\lambda=10$"),\
numpoints=1, handletextpad=0, loc="upper right")
bx.draw_frame(False)
plt.xlim(-1,21)
plt.savefig("poisson_pmf.pdf")
plt.savefig("poisson_pmf.svg")
Referring to this topic, I described what I was wanting to know to the people who answer ProBoards questions, and it took some effort from them to finally get the idea of what I expected. Here are the two answers I got:
It is really just that. There is a code to check where you have scrolled to on the page. Once you reach a scroll point where you have gone past where the pagination page initially sits on the page, an absolute positioning attribute gets added to stick it to the top of the page. It's not really software so to speak, just, as you mentioned before, CSS and Javascript [ jQuery ].
Fixed positioning, not absolute positioning. But the rest is right. :P
— Vchimpanzee • talk • contributions • 19:37, 7 November 2014 (UTC)
In Microsoft Excel (2013), you can perform operations on rows and also on columns. For example, I can total up (add) all of the numbers in a specific row (or in a specific column). Is it possible to perform a function (let's say, addition) for a range that includes both rows and columns? For example, let's say that I want a total of all of the values in the range from Columns A, B, C, D, and E and Rows 1, 2, 3, 4, and 5. So, that would be a "rectangle" of 25 values in 25 different cells (A1, A2, B1, B2, ... D4, D5, E4, E5). How can I total up all of the values in those 25 cells? Thanks. I am asking for a "one-step" formula. I already know that I can total up Column A, then total up Column B, and so forth with Columns C, D, and E ... and then add up all those five totals. But, I'd like to know if there is a single one-step function (as opposed to a combination of several functions with several steps) that will give the result I want? Thanks. Joseph A. Spadaro ( talk) 20:04, 7 November 2014 (UTC)
=SUM(A1:E5)