Feb 25, 2015Β Β· A list of lists would essentially represent a tree structure, where each branch would constitute the same type as its parent, and its leaf nodes would represent values. Given a dataframe, I want to groupby the first column and get second column as lists in rows, so that a dataframe like: a b A 1 A 2 B 5 B 5 B 4 C 6 becomes A [1,2] B [5,5,4] C [6] How do I do. Oct 5, 2017Β Β· @Sandy Chapman: List.of does return some ImmutableList type, its actual name is just a non-public implementation detail. If it was public and someone cast it to List again,.

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I have a piece of code here that is supposed to return the least common element in a list of elements, ordered by commonality: def getSingle(arr): from collections import Counter c =. Don't use quotes on the command line 1 Don't use type=list, as it will return a list of lists This happens because under the hood argparse uses the value of type to coerce each individual. If your list of lists comes from a nested list comprehension, the problem can be solved more simply/directly by fixing the comprehension; please see How can I get a flat result from a list. The first, [:], is creating a slice (normally often used for getting just part of a list), which happens to contain the entire list, and thus is effectively a copy of the list. The second, list(), is using the. The first way works for a list or a string; the second way only works for a list, because slice assignment isn't allowed for strings. Other than that I think the only difference is speed: it looks. C 7/1/2014 22000 18000 N C 8/1/2014 30000 28960 N C 9/1/2014 53000 51200 N I want to be able to return the contents of column 1 df['cluster'] as a list, so I can run a for-loop over it, and.

The first way works for a list or a string; the second way only works for a list, because slice assignment isn't allowed for strings. Other than that I think the only difference is speed: it looks. C 7/1/2014 22000 18000 N C 8/1/2014 30000 28960 N C 9/1/2014 53000 51200 N I want to be able to return the contents of column 1 df['cluster'] as a list, so I can run a for-loop over it, and.

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