WebTo get the surrounding ones: mask = pd.Index (base).union (pd.Index (base - 1)).union (pd.Index (base + 1)) I used Indexes and unions to remove duplicates. You may want to keep them, in which case you can use np.concatenate Be careful with matches on the very first or last rows :) Share Improve this answer Follow edited Jul 29, 2016 at 12:23 jk. WebSep 27, 2024 · You can see that the index of this second row, here b, is kept as the name of the series. Hence we can use it to find the position in the index: >>> ser = df.iloc [1] >>> df.index.get_indexer ( [ser.name]) [0] 1 Note that Index.get_indexer only works with arrays, hence the need to get the first element of the answer.
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WebJan 20, 2016 · To reference a row in a data frame use df [row,] To get the first position in a vector of something use match (item,vector), where the vector could be one of the columns of your data frame, eg df$cname if the column name is cname. Edit: To combine these you would write: df [match (item,df$cname),] Web3 hours ago · Thanks for the help and sorry if there is anything wrong with my question. This function: shifted_df.index = pd.Index (range (2, len (shifted_df) + 2)) is the first one which as actually changing the index of my dataframe but it just overwrites the given index with the numbers 2 to len (shifted_df) pandas. dataframe. gail osborne
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WebApr 6, 2024 · Get Indexes of a Pandas DataFrames in array format. We can get the indexes of a DataFrame or dataset in the array format using “ index.values “. Here, the below … WebApr 11, 2024 · You can first rank and then use pd.Series.last_valid_index to get the last valid values. df.rank(pct=True).mul(100).apply(lambda x: x[x.last_valid_index()], axis=1) ... Deleting DataFrame row in Pandas based on column value. 1321. Get a list from Pandas DataFrame column headers. 506. Python Pandas: Get index of rows where column … WebMar 25, 2015 · The following gives you the last index value: df.index [-1] Example: In [37]: df.index [-1] Out [37]: Timestamp ('2015-03-25 00:00:00') Or you could access the index attribute of the tail: In [40]: df.tail (1).index [0] Out [40]: Timestamp ('2015-03-25 00:00:00') Share Improve this answer Follow answered Sep 1, 2015 at 22:24 EdChum black and white whisky aldi