## Matplotlib Graph Data

Matplotlib Graph Data

## How to get the cartesian product of a series of lists

Use itertools.product, which has been available since Python 2.6. import itertools somelists = [ [1, 2, 3], [‘a’, ‘b’], [4, 5] ] for element in itertools.product(*somelists): print(element) This is the same as: for element in itertools.product([1, 2, 3], [‘a’, ‘b’], [4, 5]): print(element)

## How to access the elements of a 2D array?

If you have a=[[1,1],[2,1],[3,1]] b=[[1,2],[2,2],[3,2]] Then a[1][1] Will work fine. It points to the second column, second row just like you wanted. I’m not sure what you did wrong. To multiply the cells in the third column you can just do c = [a[2][i] * b[2][i] for i in range(len(a[2]))] Which will work for any … Read more

## Attribute Error: ‘list’ object has no attribute ‘split’

I think you’ve actually got a wider confusion here. The initial error is that you’re trying to call split on the whole list of lines, and you can’t split a list of strings, only a string. So, you need to split each line, not the whole thing. And then you’re doing for points in Type, … Read more

## Check if list items contains substrings from another list

If you just want a test, join the target list into a string and test each element of bad like so: >>> my_list = [‘abc-123’, ‘def-456’, ‘ghi-789’, ‘abc-456’, ‘def-111’, ‘qwe-111’] >>> bad = [‘abc’, ‘def’] >>> [e for e in bad if e in ‘\n’.join(my_list)] [‘abc’, ‘def’] From your question, you can test each element … Read more

## SwiftUI List color background

iOS 16 Since Xcode 14 beta 3, You can change the background of all lists and scrollable contents using this modifier: .scrollContentBackground(.hidden) You can pass in .hidden to make it transparent. So you can see the color or image underneath. iOS 15 and below All SwiftUI’s Lists are backed by a UITableView (until iOS 16). … Read more

## Python split for lists

itertools.groupby is one approach (as it often is): >>> l = [“data”,”more data”,””,”data 2″,”more data 2″,”danger”,””,”date3″,”lll”] >>> from itertools import groupby >>> groupby(l, lambda x: x == “”) <itertools.groupby object at 0x9ce06bc> >>> [list(group) for k, group in groupby(l, lambda x: x == “”) if not k] [[‘data’, ‘more data’], [‘data 2’, ‘more data 2’, … Read more

## python: most elegant way to intersperse a list with an element

I would have written a generator myself, but like this: def joinit(iterable, delimiter): it = iter(iterable) yield next(it) for x in it: yield delimiter yield x

## Change a string of integers separated by spaces to a list of int

The most simple solution is to use .split()to create a list of strings: x = x.split() Alternatively, you can use a list comprehension in combination with the .split() method: x = [int(i) for i in x.split()] You could even use map map as a third option: x = list(map(int, x.split())) This will create a list … Read more

## How can I format a list to print each element on a separate line in python? [duplicate]

Embrace the future! Just to be complete, you can also do this the Python 3k way by using the print function: from __future__ import print_function # Py 2.6+; In Py 3k not needed mylist = [’10’, 12, ’14’] # Note that 12 is an int print(*mylist,sep=’\n’) Prints: 10 12 14 Eventually, print as Python statement … Read more