[ad_1] Given the explanation of your problem, the following may work: # get all overlapping names bindNames <- intersect(names(ww), names(dd03)) # get a list of rbinded data.frames, keeping unique observations newList <- lapply(bindNames, function(i) unique(rbind(ww[[i]], dd03[[i]]))) If at this point, you want to append all of your data.frames into a single data.frame, you can once … Read more
[ad_1] You may use map and Convert you data as you required: try below: import pandas as pd df = pd.DataFrame({‘name’:[‘geeks’, ‘gor’, ‘geeks’, ‘is’,’portal’, ‘for’,’geeks’]}) df[‘name’]=df[‘name’].map(lambda x: x[:2].upper()+x[2:]) print (df) output: name 0 GEeks 1 GOr 2 GEeks 3 IS 4 POrtal 5 FOr 6 GEeks demo 1 [ad_2] solved Convert first 2 letters of … Read more
[ad_1] Option 1 You can nest numpy.where statements: org[‘LT’] = np.where(org[‘ID’].isin(ltlist_set), 1, np.where(org[‘ID2’].isin(ltlist_set), 2, 0)) Option 2 Alternatively, you can use pd.DataFrame.loc sequentially: org[‘LT’] = 0 # default value org.loc[org[‘ID2’].isin(ltlist_set), ‘LT’] = 2 org.loc[org[‘ID’].isin(ltlist_set), ‘LT’] = 1 Option 3 A third option is to use numpy.select: conditions = [org[‘ID’].isin(ltlist_set), org[‘ID2’].isin(ltlist_set)] values = [1, 2] org[‘LT’] … Read more
[ad_1] You can use .isin() directly in pandas filtering – recent_indicators_filtered = recent_indicators[recent_indicators[‘CountryCode’].isin(developed_countries)] Also, you can come up with a boolean column that says True if developed – recent_indicators[‘Developed’] = recent_indicators[‘CountryCode’].isin(developed_countries) [ad_2] solved How to check whether data of a row is in list, inside of np.where()?
[ad_1] The problem is that R is trying to be “helpful”, and simplifying your data for you. The solution is to do the following (note two commas, not one): df[-1,, drop = FALSE] This will remove the specified row, and leave your data.frame otherwise untouched. [ad_2] solved How do I delete rows from a data … Read more
[ad_1] The table can be pretty formatted in Pandas by assembling the two missing formatting conditions into a single df. I made the following two changes to the original code. Hide index numbers with hide_index() df[[“Unit”, “Abbreviation”, “Storage”]].style.hide_index() To apply to a subset of columns, you can use the subset parameter. Left align the first … Read more
[ad_1] Basically np.random.randn returns random float values of normal distributions with mean = 0 and variance = 1. Now np.random.randn takes shape you would like to return of those distributions. For example: np.random.randn(1,2) returns an array of one row and two columns. Similarly, you can give np.random.randn(1,.,.,.,9) which gives you out a complicated array. Since … Read more
[ad_1] Hopefully I got it this time: subdf = df.iloc[3:, 1:4] df[‘flag’] = 1 if subdf.values.sum()/subdf.size >= 0.1 else 0 output: unit A B C row_num flag 0 ABC 1 1 1 7 1 1 DEF 1 1 1 6 1 2 GEH 1 1 1 5 1 3 IJK 0 1 0 4 1 … Read more
[ad_1] Assuming your split criteria is by fixed number of characters (e.g. 5 here), you can use: df[‘dfnewcolumn1’] = df[‘dfcolumn’].str[:5] df[‘dfnewcolumn2’] = df[‘dfcolumn’].str[5:] Result: dfcolumn dfnewcolumn1 dfnewcolumn2 0 PUEF2CarmenXFc034DpEd PUEF2 CarmenXFc034DpEd 1 PUEF2BalulanFc034CamH PUEF2 BalulanFc034CamH 2 CARF1BalulanFc013Baca CARF1 BalulanFc013Baca If your split criteria is by the first digit in the string, you can use: df[[‘dfnewcolumn1’, … Read more
[ad_1] You can use pandas groupby method with list comprehension which will do the JOb like Below: >>> df X Y 0 Yes 1 1 No 2 2 Yes 3 3 Yes 4 4 No 2 5 No 1 6 Yes 0 7 No 4 8 No 4 9 No 5 >>> {k: v[“Y”].tolist() for … Read more
[ad_1] We could use shift from data.table library(data.table) m1 <- na.omit(do.call(cbind, shift(df1$col1, 0:4, type=”lead”))) rowSums(m1*(1:5)[col(m1)]/5) #[1] 13.60 12.20 31.24 25.58 30.48 32.58 44.88 Or another option m1 <- embed(df1$col1,5) rowSums(m1*(5:1)[col(m1)]/5) #[1] 13.60 12.20 31.24 25.58 30.48 32.58 44.88 [ad_2] solved in R, How to sum by flowing row in a data frame
[ad_1] How to perform self join with same row of previous group(month) to bring in additional columns with different expressions in Pyspark [ad_2] solved How to perform self join with same row of previous group(month) to bring in additional columns with different expressions in Pyspark
[ad_1] If I understand the logic correctly . . . # imports import pandas as pd from io import StringIO # sample data s1 = “””id Name score 111 Jack 2.17 112 Nick 1.11 113 Zoe 4.12″”” s2 = “””id Name score 111 Jack 2.17 112 Sick 1.10 113 Zoe 4.12 114 Jay 12.3″”” df1 … Read more
[ad_1] Just use the internal bits from janitor::clean_names(): # #’ ‘Clean’ a character/factor vector like `janitor::clean_names()` does for data frame columns # #’ # #’ Most of the internals are from `janitor::clean_names()` # #’ # #’ @param x a vector of strings or factors # #’ @param refactor if `x` is a factor, return a … Read more
[ad_1] maybe not the best way to do it, but will get the job done. vars_df = unique(df$x) for (i in 1:length(vars_df)) { assign(paste0(vars_df[i]), df %>% filter(x == vars_df[i]), envir = .GlobalEnv) } [ad_2] solved Automatically subset data frame by factor