Playstore Download count is not updating even after 2 weeks

My game brick zen ( ) passed 500 downloads on April 1st 2019. However in playstore it still shows as 100+ downloads where as it should be 500+ downloads. Usually it takes a couple of days to reflect the changes in playstore but even after 2 weeks there is no effect.

I tried contacting the support team for developer console but they are not responding to my emails.

I did some research and found that there are games in the trending section of playstore with more reviews than the number of downloads (downloads as per playstore). perhaps they are having the same issue.

Is there anyone having the same issue ? Is there any possible fix ? Or any suggestion to get google’s attention to fix this problem because as far I checked they haven’t addressed this issue, which they should have being a responsible company.

Does the “Changeling” property count as an enhancement bonus?

The changeling property in the Magic Item Compendium p.31 names a flat price of 2000 gp. It does not say that it counts or costs as much as a +1 enhancement.

Does that mean that the property doesn’t count as an enhancement (and thus does not increase the price of the next enhancement and doesn’t count towards the limit)?

Or does it count as a +1 enhancement but always cost a flat 2000 gp?

Count the number of words in a list of titles, and working with data-types

I have a list of article titles, where i wish to count the number of occurrences for each word. (and remove some words and characters) The input is in a .csv file where the titles are in column ‘Titles’

I already have a code that does the job(pasted below), but maybe someone could help me do it more elegantly.

import numpy as np import pandas as pd #imports Counter, as we will need it later: from collections import Counter  df = pd.read_csv("Article_titles.csv") print (df.head(10))  #Selecting the titles into variable titles = [] titles = df.Title  remove_words_list = ["at","of","a","and","in","for","the","to","with","on","using","an","after","from","by","use","review","upper","new","system"] remove_characters_list = ".:,-%()[]?'"  huge_title_list = [] #create a list of all article titles: for i in range(len(titles)):   clean_title = titles[i].lower().translate({ord(i): None for i in remove_characters_list})   huge_title_list.append(clean_title)  total_words_string = " ".join(huge_title_list) #join all article titles into one huge string  querywords = total_words_string.split() #split the string into a series of words  resultwords = [word for word in querywords if word not in remove_words_list] #From stackoverflow  resultwords_as_list = list( Counter(resultwords).items())  #Convert resultwords_list to dataframe, then convert count to numbers and finally sorting. resultframe = pd.DataFrame(np.array(resultwords_as_list).reshape(-1,2), columns = ("Keyword","Count")) resultframe.Count = pd.to_numeric(resultframe.Count) sortedframe = resultframe.sort_values(by='Count',ascending=False).reset_index(drop=True) print(sortedframe[0:50]) 

example of Input:

Titles | other_field | other_field2  "Current status of prognostic factors in patients with metastatic renal cell carcinoma." |"asdf"|12 "Sentinel lymph node biopsy in clinically node-negative Merkel cell carcinoma: the Westmead Hospital experience." |"asdf"|15 

desired output:

Word | Count  carcinoma | 2  cell | 2  biopsy | 1  clinically | 1  ....  ... 

Spark SQL lazy count

I need to use a dataframe count as divisor for calculating percentages.

This is what I’m doing:

scala> val df = Seq(1,1,1,2,2,3).toDF("value") scala> val overallCount = df.count scala> df.groupBy("value")          .agg( count(lit(1)) / overallCount ) 

But I would like to avoid the action df.count as it will be evaluated immediately.

Accumulators won’t help as they will be evaluated in advance.

Is there a way to perform a lazy count over a dataframe?

Forcing a label with a 0 count in a barplot

I am plotting a barplot for a summary data.frame consisting of ten records. Each record lists a size and a frequency for that size. One size class has a zero frequency.

However, when run, the zero-count class vanishes and instead of the full ten classes, only nine appear in the plot.

I’ve tried an as.numeric, but that renders as a decimal and doesn’t reflect the factor value.

The ggpot2 code I am using is:

plt1 <- ggplot(szt, aes(x = as.factor(size, Freq))) +       geom_bar(stat = "identity")   plt1 + xlab("Debitage Size, 5 mm class") + ylab("Frequency") +        ggtitle("Debitage Size Distribution by 5 mm class") 

The data is:

size Freq 1 1 196 2 2 261 3 3 77 4 4 26 5 5 14 6 6 9 7 7 4 8 8 0 9 9 1 10 10 2

> str(szt) 'data.frame':   10 obs. of  2 variables:  $   size: Factor w/ 10 levels "1","2","3","4",..: 1 2 3 4 5 6 7 10 8 9  $   Freq: num  196 261 77 26 14 9 4 0 1 2 

As noted above, I wish to see all ten factors along the X-axis. However ggplot drops the zero-count element.

Count unique tags

I’m trying to list and count all tags in my table. For some reason, anything after comma, is getting a white space before the word. So if the tag is found on the beginning of a record, it is seen as different tag. example: “Service” and “ Service” are seen as 2 separate values, when they should be one, and not have that white space.

I’ve fiddled with trim, text join, etc. But I always seem to end up with this problem.

————————————— record1|tag1, tag2, tag3 record2|tag1, tag2, tag3 record3|tag3 

Here is the spreadsheet:

Here is the formula I was playing with:

The List:

=unique(transpose(split(CONCATENATE(arrayformula(B2:B9&", ")),", ", TRUE, TRUE))) 

The count:

=ARRAYFORMULA(if(countif(transpose(split(CONCATENATE((B2:B9&", ")),", ")),F2:F) = 0, "",countif(transpose(split(CONCATENATE((B2:B9&", ")),", ")),F2:F))) 

Does duplicating a spell with wish count as casting that spell?


Inspired by the line of reasoning in this answer. There are a number of spells that have wording “when you cast” or “as you cast” included. Specifically, some effects have predicates like “when you cast” in order to come into effect. If wish does not count as casting the a duplicated spell, it could have ramifications of the resulting effects.


Call lightning requires choosing a location

When you cast the spell, choose a point you can see under the cloud. A bolt of lightning flashes down from the cloud to that point. Each creature within 5 feet of that point…

Teleportation circle requires drawing a circle as you cast the spell for the portal to show up in.

As you cast the spell, you draw a 10-foot-diameter circle on the ground …

Does a wizard duplicating a spell with wish count as casting the duplicated spell?

Given matrix, count paths visiting each number exactly once

We are given matrix of size at most $ 21$ by $ 21$ , each number of the matrix is either $ -1$ , which means empty element, or integer between $ 1$ and $ 21$ .

We want to count paths that start in some cell, then moving in one of the four directions (up, down, left, right) visit all $ 21$ number exactly once.

It is impossible to move on cells marked as $ -1$ .

For example:

1,  2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, -1, 21 -1,-1,-1,-1,-1,-1,-1,-1,-1, -1, -1, -1, -1, -1, -1, -1, 18, 19, 20 We can start in the upper-left cell, move right until 17 then down, right again, and up on the end. The second path is the same as the first path but reversed (starting from 21). 

My idea is to use dynamic programming with three states $ i, j$ , coordinates of the current point, and bitmask of the visited cells. However this is pretty slow for numbers up to 21. Is there any way to speed up this calculation.