gsf | Grid Solutions

Various locator strategies you can use (as per the DOM provided in the query):

All are xpaths

The design is character class. Should the formula be reversed? Based on the ?lm documentation

Models for lm are specified symbolically. A typical model has the form response ~ terms where response is the (numeric) response vector and terms is a series of terms which specifies a linear predictor for response

Can you confirm that

1. transect (in the expected output) is the same thing as sample_site (in the incoming dataset
2. The expected dataset (which has values for "BLC" species) wasn't produced from the incoming dataset (which doesn't).

If so, dplyr's group_by() and summarize() is all you need.

from the documentation on the function you are using "If size is a named vector, the function will check to see whether the length of the vector matches the number of groups and that the names match the group names" - it then goes on to provide an example of exactly what your question asks.

As someone upvoted my question, i am putting my solution here that i have found on my own. As i said on OCT 8th, the reason is setState() so the solution is simple. I used Navigator.push... or Navigator.pushReplacement... instead of all setState() functions in my app and the error fixed. But this was pretty hard in somewhere because i needed to define new final required parametres of the class to fresh the page rightly or define some other parametres in İnherited Widget Class to use them how i want after freshing the page. So the problem solved. I found out there are some reasons like provider but i dont have enough time to learn how to use it so choose this way that i have known. If somebody explain any other solution it will be better because i dont feel ok with this solution. There must be better or more stable fix.

Is there a problem if you install the application on your device directly from Android Studio via a USB Debugging mode? If not, then the problem is with generating a signed APK for your device.

Also you need to secure your API key first: Using API Keys. Take out the API KEY in string resources like that

You have authorized your application with

The problem comes from the matrix size (206 MiB). Indeed, the matrix is pretty big and (virtually) filled with zeros every time the function is called. Assuming the all values would be physically written to memory in 12.2 ms, the throughput would be 5206**2*8/12.2e-3/1024**3 = 16.5 GiB/s which is pretty good for a sequential workload (especially on an average personal computer which often barely reach more than 25 GiB/s in sequential). In practice, the memory is not set to zero and most of the computation time comes from the management of virtual memory.

np.zeros((leny, leny)) internally request your operating systems (OS) to reserve some space and initialize it with zero. Your OS will virtually allocate many pages (chunks of typically 4 KiB) and mark them as to be set to zero later (delayed initialization). As a result, this process is very cheap. However, when J[0,0] = -10 is executed later, it write data in given first-touched page that need to be initialized. This is called a page fault. This process is expensive (especially in sequential) as it stop the execution of your process, find some place in physical memory, fill a whole page, and then resume your process, for each page! In your case, many array cells to be initialized cause pages to be filled which is much more costly than just filling actually the cells.

CPU caches also matter a lot in this code as page filling will be much slower when the computation is done in RAM. This happens when the amount of data is to big to fit in the last level-cache (of typically 1-64 MiB).

The memory access pattern strongly impacts performance on quite big data structures (ie. not fitting in CPU caches). Indeed, a random access pattern or strided access with huge stride are really bad for performance since the processor cannot easily predict the location of the RAM chunks to load ahead of time (not to mention the RAM latency is quite big). Contiguous accesses are much faster.

All of this should be taken into account when you measure the time on simpler cases.

One simple way to see the impact of virtual memory is to use np.full((leny, leny), 0.0) instead of np.zeros(leny, leny). This is 3 times slower on my machine. All the pages needs to be filled in that case. You can also create measure the time to fille multiple time the J matrix. The first time should be significantly slower (2.2 times slower on my machine).

One solution is to allocate and initialize the array once and fill/recycle only some values but this is a bit tricky to do. In the worst case, most of the matrix need to be zeroised which will be too slow in your case since it will be bound by your memory throughput.

One simple solution is to use a space matrix. Space matrices can be created using the module scipy.sparse. They are often a slower to compute, but faster and much more compact if there are a lot of zeros (which is your case). Here is an example:

This splits it into 2 stages, first gather the data with a min and max years for each item. When setting the min and max's, it checks the previous value (or defaults it if not found) and applies the appropriate min() or max()...