I am trying to write a code where it outputs the number of consecutive letters that come after one another as it is running through list_. So all the outputs will start with 1 consecutive number followed by a second and a third if the following value is equivalent. So for example the order ['b','b','b',r] would produce the output of 1,2,3,1 as the consecutive list is disrupt by the upcoming r in the loop so it goes back to 1 consecutive int from 3.
Code for the consecutive bs and rs
list_ = ['b', 'r', 'b', 'r', 'b', 'r', 'b', 'r', 'b', 'b', 'b', 'r', 'b', 'b', 'r', 'r', 'b', 'r', 'r', 'r', 'r', 'r', 'b', 'b', 'b', 'b', 'b', 'r']
for k, gp in groupby(list_):
print(k,list(gp))
This is rle:
What you are looking for is:
from itertools import groupby
rle = lambda x : [k for i,j in groupby(x) for k in range(1,len(list(j)) + 1)]
print(rle(['b','b','b','r']))
[1, 2, 3, 1]
Simple one-liner:
alist = ['b', 'b', 'b', 'r']
print([idx for k, group in itertools.groupby(alist)
for idx, v in enumerate(group, start=1)])
Output:
[1, 2, 3, 1]
Related
I want to generate different combinations of 3 elements a, b, and c. The length of these combinations needs to be 4. I want to have a maximum of 4 times from 'a' and a maximum 1 time from each 'b' and 'c' element. So, for example, we can have ['a',' a',' a','a'] or ['a','a','b','c'] but not ['a','b','b','b'].
There is a similar question in 1, but, as far as I know, using the last 'gen' function, the length of a generation is controlled by the multiplication of a maximum number of repetitions (4 in my case). Also, cases were limited to tuples with exactly 1 'b' and 1 'c' and the rest are 'a'. For the last issue, I replaced 'combinations' with 'combinations_with_replacement', but it still produces tuples with 4 elements and there is no ['a',' a',' a','a'].
How can I tackle this problem?
Here is the code:
from itertools import combinations_with_replacement
def gen(ns, elems=None, C=None, out=None):
if elems is None:
elems = list(range(len(ns)))
else:
assert len(elems) == len(ns)
if out is None:
N = 1
for n in ns:
N *= n
out = [elems[0]]*N
C = range(N)
if len(ns) == 1:
yield out
else:
n = ns[-1]
e = elems[-1]
for c in combinations_with_replacement(C,n):
out_ = out.copy()
for i in c:
out_[i] = e
C_ = [i for i in C if i not in c]
yield from gen(ns[:-1], elems[:-1], C_, out_)
for tmp_list in gen([4,1,1], elems=['a', 'b', 'c'], C=None, out=None):
print(tmp_list)
output:
['c', 'b', 'a', 'a']
['c', 'a', 'b', 'a']
['c', 'a', 'a', 'b']
['b', 'c', 'a', 'a']
['a', 'c', 'b', 'a']
['a', 'c', 'a', 'b']
['b', 'a', 'c', 'a']
['a', 'b', 'c', 'a']
['a', 'a', 'c', 'b']
['b', 'a', 'a', 'c']
['a', 'b', 'a', 'c']
['a', 'a', 'b', 'c']
Please note that since I care about the execution time, I want to generate the tuples in a loop.
I am trying to run a function where it will recognize if 3 consecutive bs or rs are after one another within list_ if it is I want the code to print('divisible by 3').
list_ = ['b', 'r', 'b', 'r', 'b', 'r', 'b', 'r', 'b', 'b', 'b', 'r', 'b', 'b', 'r', 'r', 'b', 'r', 'r', 'r', 'r', 'r', 'b', 'b', 'b', 'b', 'b', 'r']
if (list_%3==0):
print('divisible by 3')
#firefireeyyy - do you know itertools.groupby() in Python? I've change your list_ variable to lst.
Then you can try this and figure out the answer next:
from itertools import groupby
for k, gp in groupby(lst):
print(k, list(gp))
I have the following 2D list:
test_list = [['A', 'B', 'C'], ['I', 'L', 'A', 'C', 'K', 'B'], ['J', 'I', 'A', 'B', 'C']]
I want to compare the 1st list elements of the 2D array test_list[0] with all other lists. If the elements ['A', 'B', 'C'] are present in all other lists then it should print any message such as "All elements are similar".
I have tried this piece of code but it is not working as I expected:
test_list = [['A', 'B', 'C'], ['I', 'L', 'A', 'C', 'K', 'B'], ['J', 'I', 'A', 'B', 'C']]
for idx,ele in enumerate(p):
result = set(test_list [0]).intersection(test_list [(idx + 1) % len(temp_d)])
print(result)
Expected Output:
The elements of the list ['A', 'B', 'C'] are present in all other lists.
You can use the all(...) function - or remove all elements from the bigger list from your smaller one converted to set. If the set.difference() is Falsy (i.e. all elements were removed) they were all contained in it:
test_list = [['A', 'B', 'C'], ['I', 'L', 'A', 'C', 'K', 'B'], ['J', 'I', 'A', 'B', 'C']]
s = test_list[0]
for e in test_list[1:]:
if all(v in e for v in s):
print(e, "contains all elements of ", s)
s = set(s)
for e in test_list[1:]:
# if all elements of s are in e the difference will be an empty set == Falsy
if not s.difference(e):
print(e, "contains all elements of ", s)
Output:
['I', 'L', 'A', 'C', 'K', 'B'] contains all elements of ['A', 'B', 'C']
['J', 'I', 'A', 'B', 'C'] contains all elements of ['A', 'B', 'C']
['I', 'L', 'A', 'C', 'K', 'B'] contains all elements of {'A', 'B', 'C'}
['J', 'I', 'A', 'B', 'C'] contains all elements of {'A', 'B', 'C'}
For each letter in the first list see if they are in the second and third list and return the boolean.
Then see if the set of the new list equals True
test_list = [['A', 'B', 'C'], ['I', 'L', 'A', 'C', 'K', 'B'], ['J', 'I', 'A', 'B', 'C']]
bool = ([x in test_list[1]+test_list[2] for x in test_list[0]])
if list(set(bool))[0] == True:
print('All elements are similar')
>>> All elements are similar
I have a large number of identical lists 'old' which I want to transform in the same way into a list 'new'. The way I want to do it, is to make an example of the desired list 'new'. Then I turn the difference between the two lists 'old' and 'new' into a rule, and then use that rule to turn my other lists 'old_2' into 'new_2'.
I cannot figure out how to do the first step and the second step does not give me the expected result. Is there an elegant way to do this?
import numpy
# 0 1 2 3 4 5
old_1 = ['A', 'B', 'C', 'D', 'E', 'F']
new = ['B', 'C', 'D', 'E', 'A']
# 01 Get the difference new - /- old_1 based on index positions of
# the list elements, to get something like this:
order = [1,2,3,4,0]
# 02 Then use this order to transform a second identical list, old_2.
# For this I wanted to use the following:
old_2 = ['A', 'B', 'C', 'D', 'E', 'F']
old_2 = numpy.array(old_2)
order = numpy.array(order)
inds = order.argsort()
print('inds =', inds) # As a check, this gives the wrong order: [4 1 0 2 3]
new_2 = old_2[inds]
# I expected this to result in what I want, which is:
print(new_2)
['C', 'B', 'D', 'E', 'A']
# But what I get in reality is this:
inds = [4 1 0 2 3]
['E' 'B' 'A' 'C' 'D']
Any suggestions to get the desired result?
new_2 = ['B', 'C', 'D', 'E', 'A']
From what I understand, I tried to edit your code. Hopefully it helps.
import numpy as np
def get_order(new, old):
order = []
for element in new:
order.append(old.index(element))
return order
def main():
old_1 = ['A', 'B', 'C', 'D', 'E', 'F']
new = ['B', 'C', 'D', 'E', 'A']
order = get_order(new, old_1)
print(order)
old_2 = ['A', 'B', 'C', 'D', 'E', 'F']
old_2 = np.array(old_2)
order = np.array(order)
#inds = order.argsort()
#print('inds =', inds) # As a check, this gives the wrong order: [4 1 0 2 3]
new_2 = old_2[order]
print(new_2)
if __name__ == '__main__':
main()
Output
[1, 2, 3, 4, 0]
['B' 'C' 'D' 'E' 'A']
I am very new to python and i was stuck with this. I need to create a list of lists that is
formed from this list c: ['asdf','bbnm','rtyu','qwer'].
I need to create something like this:
b: [['a','s','d','f'],['b','b','n','m'],['r','t','y','u'],['q','w','e','r']]
I tried using a for-loop, but it is not working out. I don't know what mistake I am making.
You can use a list comprehension with list():
>>> c = ['asdf','bbnm','rtyu','qwer']
>>>
>>> b = [list(s) for s in c]
>>> b
[['a', 's', 'd', 'f'], ['b', 'b', 'n', 'm'], ['r', 't', 'y', 'u'], ['q', 'w', 'e', 'r']]
Notice that calling list() with a string argument returns a list containing the characters of that string:
>>> list('abc')
['a', 'b', 'c']
What we're doing above is applying this to every element of the list via the comprehension.
Use map function.
>>> a= ['asdf','bbnm','rtyu','qwer']
>>> map(list ,a )
[['a', 's', 'd', 'f'], ['b', 'b', 'n', 'm'], ['r', 't', 'y', 'u'], ['q', 'w', 'e', 'r']]
>>>
You could do something that's easier to understand:
b = []
for x in c:
list(x).append(b)