reshape

问题 I have a dataframe which I want to melt the data into multiple target columns. The below code I used grp2 = pd.lreshape(grp1, cols.groupby(cols.str.split('_').str[1])).sort_values('ACCT_NAME') The above line I lose the column names grp2 = pd.melt(grp1 , id_vars = ['Client' , 'Industry'] , var_name = "H Year" , value_name = 'Count') The above line I dont get multiple target columns From DF Client INDUSTRY 1H2016_6MO 2H2016_6MO 1H2017_6MO 2H2017_6MO 1H2016_12MO 2H2016_12MO 1H2017_12MO 2H2017

问题 I have a table as below (different row with same ID will have same gender and age value but different category and sub category value): ID product.category sub.category gender age 1 1 food chicken M young 2 1 kitchen napkin M young 3 1 food steak M young 4 2 electronic phone F mid 5 3 cloth shirt M old 6 3 kitchen bowl M old 7 4 alch beer F young And by combine different rows with same ID, I want to reform the table as below: ID product.category1 sub.category1 product.category2 sub.category2

问题 Hi I have a list flat which is length 2800, it contains 100 results for each of 28 variables: Below is an example of 4 results for 2 variables [0, 0, 1, 1, 2, 2, 3, 3] I would like to reshape the list to an array (2,4) so that the results for each variable are in a single element. [[0,1,2,3], [0,1,2,3]] 回答1: You can think of reshaping that the new shape is filled row by row (last dimension varies fastest) from the flattened original list/array. An easy solution is to shape the list into a

问题 I have a need to do what is really what tidyr::spread() does, but for multiple value columns. If I have a data set like this: te <- structure(list(Syllable = c("[pa]", "[ta]", "[ka]", "[pa]", "[ta]", "[ka]", "[pa]", "[ta]", "[ka]", "[pa]"), PA = c(15.9252335141423, 2.17504491982172, 5.26727958979289, 4.48590068583509, 2.1316282072803e-13, 14.1415335887116, 3.51720477328246, 0.839953301362556, 5.74712643678048, 7.01396701583887), transient_mean = c(4.43699436235785, 4.8733556527069, 5

问题 This question already has answers here : Split a column of concatenated comma-delimited data and recode output as factors (2 answers) Closed 2 years ago . As shown in the above pic, I've a column, genres, with a list of genres the corresponding movie belongs to. There are in total 19 unique genres. I'd like to know if I can manipulate this data into appending 19 columns to the data set each corresponding to each of the genres identifiers and label the corresponding cells as 0 or 1 indicating

问题 I have a data set which is very large, thousands of rows and hundreds of column. I try to alternately reshape the data for every nth row, and all the nth row column data. I tried like this: in=rand(71760,320); m=240; n=320; [R,C]=size(in); out=[]; R_out=R/m; for k=1:m %from row 1 to mth row for i=1:C %reshape every column of mth row out=[out;reshape(in(k:m:end,i),R_out,1)']; end end If you try out the code, it took very long time and not efficient at all, you won't even bother to let it

问题 I have a dataset dfs that i would like to reshape dfs # country.name indicator.name x1990 x1991 x1992 # 507 andorra GDP at market prices (current US$) 1.028989e+09 1.106891e+09 1.209993e+09 # 510 andorra GDP growth (annual %) 3.781393e+00 2.546001e+00 9.292154e-01 # 1347 albania GDP at market prices (current US$) 2.101625e+09 1.139167e+09 7.094526e+08 # 1350 albania GDP growth (annual %) -9.575640e+00 -2.958900e+01 -7.200000e+00 # 3587 austria GDP at market prices (current US$) 1.660624e+11 1

问题 I have picked up an awful public data set that needs a lot of work to make it useful. Here is a simplification: Molten<-data.frame(ID=round(runif(100, 0, 50),0), Element=c(rep("Au", 20), rep("Fe", 10), rep("Al", 30),rep("Cu", 20),rep("Au", 20)), Measure=rnorm(100), Units=c(rep("ppm",10), rep("pct",10), rep("ppb", 80))) Molten$UnitElement<-paste(Molten$Element, Molten$Units, sep="_") Molten<-Molten[!duplicated(Molten[,c("ID", "Element")]),] I have arrived at a data frame with the IDs and a

问题 I have a binary 3D array of the size 1024 by 1024 by 1024. I want to use a function ( convhull ), which has the following input: X is of size mpts-by-ndim, where mpts is the number of points and ndim is the dimension of the space where the points reside, 2 ≦ ndim ≦ 3 How can I reshape my array into the array X which is required by this function? Maybe "reshape" isn't the best word, because using the "reshape" function isn't enough. 回答1: What convhull is looking for is a list of subscripts of

问题 trainX.size == 43120000 trainX = trainX.reshape([-1, 28, 28, 1]) (1)Does reshape accept a list as an argment instead of a tuple? (2)Are the following two statements equivalent? trainX = trainX.reshape([-1, 28, 28, 1]) trainX = trainX.reshape((55000, 28, 28, 1)) 回答1: Try the variations: In [1]: np.arange(12).reshape(3,4) Out[1]: array([[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11]]) In [2]: np.arange(12).reshape([3,4]) Out[2]: array([[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11]]) In [3]: np