Merging cells in a spreadsheet means taking two or more cells and constructing a single cell out of them. When you merge two or more adjacent horizontal or vertical cells, the cells become one larger cell that is displayed across multiple columns or rows. When you merge multiple cells, the contents of only one cell … Read more
If you read the documentation, you will learn it’s: key, old value, and new value. h1.merge(h2) { |key, old, new| new – old } means the final result will have value h2[key] – h1[key]. Since you are merging h2 over h1, values from h2 will be new values and those from h1 will be old … Read more
def intersect(a, b): a, b = set(a), set(b) return list(a & b) def merge_dicts(d1, d2): return {k: intersect(v1, v2) for (k, v1), v2 in zip(d1.items(), d2.values())} dictionary_1 = {‘A’ :[‘B’,’C’,’D’],’B’ :[‘A’]} dictionary_2 = {‘A’ :[‘A’,’B’], ‘B’ :[‘A’,’F’]} merge = merge_dicts(dictionary_1, dictionary_2) print(merge) # {‘A’: [‘B’], ‘B’: [‘A’]} 4 solved How to merge two dictionaries with … Read more
It’s easy: Read you first file line by line (you can use a Scanner for that). For each line, write it to the output file (you can use a PrintWriter for that). Also store the line in a HashSet. Read your second file line by line. For each line, check if the line is in … Read more
From the documentation of merge: If the columns in the data frames not used in merging have any common names, these have suffixes (“.x” and “.y” by default) appended to try to make the names of the result unique. If this is not possible, an error is thrown. Based on your names results before and … Read more
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
The easiest solution would be to iterate through the intersection of the keys and add the corresponding balance and price as an entry in a new dictionary. The intersection isn’t necessary if you can guarantee that if a key exists in either of balance of price, it will exist in the other. As I cannot … Read more
See this. I had a datframe and it showed me like this, but the datframe was one intact data frame… solved Merges and joins in pandas
This operation is just doing the “merge” step of a Merge Sort and can be done in O(l1+l2) time. The general premise is to iterate over both (already sorted) lists at the same time, but only advance the list with the lowest head value, while using the advanced value in the resulting output. The operation … Read more
If there are maximum of two rows, this will work select Srn, SrnPrs,sDayDate,max(sTime) as sTime1, min(sTime) as sTime2 from table group by Srn, SrnPrs,sDayDate solved merge same rows and show different fields in new column in sql
It’s more or less the same as your other question. The only change is that you’ll have to set “company,date” as the key columns to perform the join on (note that the order is important – it’ll first sort by company and then by date). require(data.table) ## 1.9.2 setDT(df) setDT(event) setkey(df, company1, date1) setkey(event, company2, … Read more
One of the shortest way would be to prepare a list/set of all the keys from all the dictionaries and call that key on all the dictionary in the list. list_of_dict = [{‘a’: 1, ‘b’: 2, ‘c’: 3}, {‘a’: 3, ‘b’: 5}, {‘k’: 5, ‘j’: 5}, {‘a’: 3, ‘k’: 5, ‘d’: 4}, {‘a’: 3}] # … Read more
I think your problem is that these lines: if (que1.empty()){ for (int m = j; m < counterq2; m++){ que_merge.push_back(que2.at(m)); } } else { for (int l = i; l < counterq1; ++l) { que_merge.push_back(que1.at(l)); } } doesn’t do what you expect. As far as I can see, your idea is to merge the remaining … Read more
Commented for explanation of approach. Have found two addresses where ID from sheet2 comes back onto sheet1 import io sheeta = pd.read_csv(io.StringIO(“”” house_number street suburb 0 43 Smith Street Frewville 1 45 Smith Street Frewville 2 47 Smith Street Frewville 3 49 Smith Street Frewville 4 51 Smith Street Frewville 5 53 Smith Street Frewville … Read more
Using dplyr from the tidyverse you should be able to do this: library(tidyverse) AB <- B %>% rename(ID = `Cust No`, Name = `Customer Name`) %>% full_join(A, ., by = “ID”) rename will let you change the names in B manually so that they match A in the form new_name = old_name. full_join will keep … Read more