cfl | a Compileable statically typed Functional programming | Compiler library
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kandi X-RAY | cfl Summary
a Compileable statically typed Functional programming Language
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QUESTION
I am trying to execute a code using the gfortran compiler. In order to compile, I use :
...ANSWER
Answered 2021-May-26 at 09:50Your structure basically is
QUESTION
I am looking to scrape data for teams over a period of years across countries from https://www.transfermarkt.com/premier-league/startseite/wettbewerb/GB1/plus/?saison_id=2019
This site is an example of what I am looking to scrape including the table with squad size, etc. in the middle of the page as well as the table with the match data on the right side of the page. I am using Beautiful Soup in python.
Here is the code I have so far:
...ANSWER
Answered 2021-May-25 at 16:48It'd be helpful if you'd specify what part of your code throws the error. I'm assuming its the part where you initialize df_soccer1.
Your problem is that try: executes until it doesn't, which means if there are only 5 in a , text is appended to team, squad and age, then an error is thrown because you are iterating over more than there are and nothing is appended to foreigners and the other two data points. This means your arrays are of uneven length.
Following code seperates the steps, it first extracs the text from all and only if all of them were returned, the information is appended, else '' is appended.
QUESTION
I've never programed anything before and have no experience :) I'm using a program which I get from github and now I have to make a json settings file for it.
basicly, I'll pick fix(ASMAP or BARPE) and runway(35L or 36) and the program should give me the right "dep"/"nap" and "cfl" for these settings. But there are more than one possible cfl and runway for a fix which is causing me duplicate key warning. can anybody help me about how to overcome this problem? I'll share just a little part of the code for you to see the structure:
...ANSWER
Answered 2021-May-13 at 12:23It looks like you are using the wrong data structure.
A hashmap's (what you are using) keys must be unique. You have multiple keys that are not unique:
QUESTION
Considering the following Leapfrog scheme used to discretize a vectorial wave equation with given initial conditions and periodic boundary conditions. I have implemented the scheme and now I want to make numerical convergence tests to show that the scheme is of second order in space and time.
I'm mainly struggling with two points here:
- I'm not 100% sure if I implemented the scheme correctly. I really wanted to use slicing because it is so much faster than using loops.
- I don't really know how to get the right error plot, because I'm not sure which norm to use. In the examples I have found (they were in 1D) we've always used the L2-Norm.
ANSWER
Answered 2020-Nov-05 at 13:11Apart from the initialization, I see no errors in your code.
As to the initialization, consider the first step. There you should compute, per the method description, approximations for p(dt,j*dx) from the values of p(0,j*dx) and u(0.5*dt, (j+0.5)*dx). This means that you need to initialize at time==0
QUESTION
I have a netty server and client in the project and want to exchange message between them.
The netty server code:
...ANSWER
Answered 2020-Jun-10 at 10:02You should check the ChannelFuture that is returned by writeAndFlush to be understand if the write failed.
For doing so add a ChannelFutureListener to it:
QUESTION
%matplotlib notebook
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.animation as animation
from math import pi
# wave speed
c = 1
# spatial domain
xmin = 0
xmax = 1
#time domain
m=500; # num of time steps
tmin=0
T = tmin + np.arange(m+1);
tmax=500
n = 50 # num of grid points
# x grid of n points
X, dx = np.linspace(xmin, xmax, n+1, retstep=True);
X = X[:-1] # remove last point, as u(x=1,t)=u(x=0,t)
# for CFL of 0.1
CFL = 0.3
dt = CFL*dx/c
# initial conditions
def initial_u(x):
return np.sin(2*pi*x)
# each value of the U array contains the solution for all x values at each timestep
U = np.zeros((m+1,n),dtype=float)
U[0] = u = initial_u(X);
def derivatives(t,u,c,dx):
uvals = [] # u values for this time step
for j in range(len(X)):
if j == 0: # left boundary
uvals.append((-c/(2*dx))*(u[j+1]-u[n-1]))
elif j == n-1: # right boundary
uvals.append((-c/(2*dx))*(u[0]-u[j-1]))
else:
uvals.append((-c/(2*dx))*(u[j+1]-u[j-1]))
return np.asarray(uvals)
# solve for 500 time steps
for k in range(m):
t = T[k];
k1 = derivatives(t,u,c,dx)*dt;
k2 = derivatives(t+0.5*dt,u+0.5*k1,c,dx)*dt;
k3 = derivatives(t+0.5*dt,u+0.5*k2,c,dx)*dt;
k4 = derivatives(t+dt,u+k3,c,dx)*dt;
U[k+1] = u = u + (k1+2*k2+2*k3+k4)/6;
# plot solution
plt.style.use('dark_background')
fig = plt.figure()
ax1 = fig.add_subplot(1,1,1)
line, = ax1.plot(X,U[0],color='cyan')
ax1.grid(True)
ax1.set_ylim([-2,2])
ax1.set_xlim([0,1])
def animate(i):
line.set_ydata(U[i])
return line,
ANSWER
Answered 2020-Jan-10 at 11:49While superficially your computation steps are related to the RK4 method, they deviate from the RK4 method and the correct space discretization too much to mention it all.
The traditional way to apply ODE integration methods is to have a function derivatives(t, state, params) and then apply that to compute the Euler step or the RK4 step. In your case it would be
QUESTION
I'd like to calculate % of stocks above rolling mean, therefore, I need to group the data by 'Date' and want to keep the 'Date' column. Percentages are calculated correctly, however, instead of actual dates I'm getting 'NaN' values. The 'Date' column is not the data frame index.
...ANSWER
Answered 2020-Jan-09 at 16:02You have to remove the 'Date' from the [ ], you are already grouping by it. And don't drop the index, Date is your new index in your returning dataframe and you want to keep it
QUESTION
I'm trying to create a cfg generating following language:
Is This language context-free and could be generated by a cfg? if yes , how could the grammar generating this language be created?
I'm not experienced so much in creating cfg's for cfl's. I would be glad if any help or solution is given
...ANSWER
Answered 2020-Jan-03 at 13:57To start you off, do you know how to create a CFG for the language {a^n d^t | n = t}? This will be your starting point.
QUESTION
I have a simple stores program developed using Codeigniter. Following are the main tables.
store_item ...ANSWER
Answered 2019-Oct-27 at 19:32You have set $where = NULL 2 times, so its removing 1st condition from the where clause,
just remove the 2nd $where= NULL, it will be fine.
QUESTION
Let L(G) be the language generated by a context free grammar G. Is the following decision problem decidable ?
Whether L(G) is deterministic context free language ?
I understood why the above problem is undecidable from this link, but I had a doubt.
We know that CFL's and PDA's are equivalent (reference), i.e. for every CFL, G, there is a PDA M such that L(G) = L(M) and vice versa.
A context free language is deterministic if it can be accepted by a DPDA.
A deterministic PDA is one in which there is at most one possible transition from any state based on the current input.
Since we can create a PDA for every CFL and distinguish between PDA's being deterministic or not, could we say that the problem of whether L(G) is deterministic context free language is decidable ? Or am I missing something ?
ANSWER
Answered 2019-Sep-26 at 15:56You are missing something. You say:
A context free language is deterministic if it can be accepted by a DPDA.
and
we can create a PDA for every CFL
and
[we can] distinguish between PDA's being deterministic or not
The problem is that the PDA you get for the CFL might be nondeterministic even if the language is deterministic. While it's true that every deterministic CFL has a DPDA that accepts it, is is NOT true that every deterministic CFL is accepted ONLY by DPDAs. Indeed, every deterministic CFL is accepted by many nondeterministic PDAs... it's not hard to see that any DPDA can be transformed into an equivalent nondeterministic PDA by adding new states and branches that don't lead to accepting anything.
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