WebAssuming you wanted to create a new column c2, equivalent to c1 except where c1 is Value, in which case, you would like to assign it to 10: First, you could create a new column c2, and set it to equivalent as c1, using one of the following two lines (they essentially do the same thing): df = df.assign(c2 = df['c1']) # OR: df['c2'] = df['c1'] WebApr 22, 2015 · How to add value to a pandas dataframe column by row depending a key value in a dictionary? 0 How to create multiple new dataframe columns from a nested dict, based on the dict key
Adding a Column to a Pandas DataFrame Based on an If-Else ... - Medium
WebApr 8, 2024 · I want to add a new column based on the below condition and distinct values. ... How do I select rows from a DataFrame based on column values? 960. Deleting DataFrame row in Pandas based on column value. 17. In pandas, how to concatenate horizontally and then remove the redundant columns. 2. WebDec 12, 2024 · A Computer Science portal for geeks. It contains well written, well thought and well explained computer science and programming articles, quizzes and practice/competitive programming/company interview Questions. dialysis experiment biochemistry
pandas - add new column to dataframe from dictionary
WebNov 20, 2024 · The following is slower than the approaches timed here, but we can compute the extra column based on the contents of more than one column, and more than two values can be computed for the extra column.. Simple example using just the "Set" column: def set_color(row): if row["Set"] == "Z": return "red" else: return "green" df = … WebFeb 15, 2024 · 1. I have a dataframe as shown below. ID Price Duration 1 100 60 2 200 2 3 1 366 4 1 365. I would like to create a flag column based on condition in Price column and Duration column. Steps: If Price is less than 20 flag it as False else flag it as True. If Duration is less than 30 flag it as False else flag it as True. Expected Output: WebOct 7, 2015 · 3 Answers. In [231]: df ['lunch'] = (df ['hour']<=11) & (df ['hour']<=1) In [232]: df ['lunch'] Out [232]: 0 True 1 True 2 True 3 False 4 False Name: lunch, dtype: bool In [233]: df ['lunch'].astype (int) Out [233]: 0 1 1 1 2 1 3 0 4 0 Name: lunch, dtype: int32. You can have a vectorized approach (the minus operator is here to negate the ... cipher\u0027s zx