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 again … Read more
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 solved Convert first 2 letters of all records … Read more
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
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) solved How to check whether data of a row is in list, inside of np.where()?
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. solved How do I delete rows from a data frame when … Read more
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 column … Read more
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 you … Read more
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 4 … Read more
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’, ‘dfnewcolumnX’]] … Read more
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 k,v … Read more
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 solved in R, How to sum by flowing row in a data frame
How to perform self join with same row of previous group(month) to bring in additional columns with different expressions in Pyspark solved How to perform self join with same row of previous group(month) to bring in additional columns with different expressions in Pyspark
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
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 ref-factored … Read more
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) } solved Automatically subset data frame by factor