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I'm currently trying to complete Computer Science Circles online but I am stuck on part 14: Methods. Here is the question.
Using index and other list methods, write a function replace(list, X, Y) which replaces all occurrences of X in list with Y. For example, if L = [3, 1, 4, 1, 5, 9] then replace(L, 1, 7) would change the contents of L to [3, 7, 4, 7, 5, 9]. To make this exercise a challenge, you are not allowed to use [].
Note: you don't need to use return.
I would probably be able to do this if we were allowed to use square brackets.
Here is what I have so far.
def replace(L, X, Y):
while X in L:
var = L.index(X)
var = Y
return(L)
I'll give some tips since this is an exercise.
1) You already found out the index where you're supposed to replace one element with another. What other way is there to replace a value in a given index? Check all the methods of list.
2) A list comprehension also allows an elegant solution:
[...???... for value in list]
You'll need to figure out what the expression should be, and how to make the comprehension modify your original list, not just create a new one.
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For a homework assignment I am given this task:
Read a list of numbers and generate a new list which squares the members of the original list.
generate a third list of ints that ranks the second list of ints.
What does it mean to "rank the list"? What is this asking me to do?
I am NOT asking for a solution to the programming assignment, please do not provide one.
To create a list of integers between 0 and N, a solution is to use the range(N) function.
Then you can use the append() method which adds an item to the end of the list while getting the square of each element.
Finally use the sort() method which sorts the list ascending by default. You can also make a function to decide the sorting criteria(s)
def printSortedList():
l = list()
for i in range(1,21):
l.append(i**2)
l.sort()
print(l)
printSortedList()
Ranking means to number the items in a list from lowest to highest, or vice-versa.
So, for example, “ranking” the list [639, 810, 150, 162, 461, 853, 648] produces the list [4, 6, 1, 2, 3, 7, 5], where 1 corresponds to the lowest value (150), 2 to the next-lowest value (162).
Or maybe you're supposed to rank them the other way, giving [4, 2, 7, 6, 5, 1, 3].
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How can I add a constant (ie 10) to a list without creating a new list?
For example,
list = [2, 4, 6, 8]
How can I overwrite the elements in this list without creating another list (append method)? I understand that there is a map lambda method. Any way to go about by using a for loop?
When x = 10, expected final output should be
list = [12, 14, 16, 18]
x = 10
for i in range(0, len(list)):
list[i] = list[i] + x
I kept the variable name as list, but I recommend you don't use built-in names as names for your variables.
You can for sure use the map function in Python as follows. I changed your list name to list1 to not confuse it with the datatype.
list1 = [2,4,6,8]
print(list(map(lambda x : x+10, list1)))
#Jax Teller 's solution is the standard.
Here is a generator, not changing the original, but instead yielding mapped values
Also, using the word list isn't good, because it is a Python construct. Use something else instead like my_list
my_list = [2,4,6,8]
val = 10
gen = (val + e for e in my_list)
This is actually the same as map.
Just notice you can't use list(gen) because it would create a new list, not just iterate.
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I have 5 arrays with 6 questions in each one.
I need the script to pick 2 questions from each array and create an input() function. The part I can't seem to think of is how to make an output for a correct answer for the questions. I understand how a specified input would work but what about randomized.
I think you're looking for something like this:
randomNumber1=***some generated number (0 thru 6)
randomNumber2=***some generated number (0 thru 6)
array1=['what is the meaning of life', 'how far away is the sun',...]
array2=['what did is your favorite color', 'how many pennies are in 1 dollar'...]
q1=array1[randomNumber1]
q2=array2[randomNumber2]
input1=input(q1)
input2=input(q2)
#stores answers in a dictionary
answers={q1:input1, q2:input2}
I do not think the random module has the function that you want.
But it is easy to build one if you like. Python is easy.
Does this work?
import random
from typing import Iterable
def get_sub_random_list(sub_length: int, iterable: Iterable) -> list:
iterable_copy = list(iterable)
result = []
for __ in range(sub_length):
length = len(iterable_copy)
if length == 0:
raise ValueError(f"the iterable should longer than {sub_length}")
index = random.choice(range(length))
result.append(iterable_copy[index])
del iterable_copy[index]
return result
example:
>>> get_sub_random_list(1, [1, 2, 3, 4, 5, 6])
[5]
>>> get_sub_random_list(6, [1, 2, 3, 4, 5, 6])
[4, 1, 5, 2, 6, 3]
The complexity is O(n+m): n is the length of iterable, and the m is the the times of the loop.
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Hi I need a single list of nested lists in python without any inbuilt function. I did a lot of research but not able to find without flatten or ravel function().
for ex:
input = [1,2,[2,3,[4,5]],[6,7],[8,[9]],10]
output = [1,2,2,3,4,5,6,7,8,9,10]
You can use recursion to implement your own flatten function. Here's a nice tutorial explaining the concepts of recursion.
The idea here is that you loop through each element in your list and if you encounter a sublist you recursively call the same method with the current list. If it's a valid element you add it to to your output list.
Example:
def flatten(input_list, output_list=[]):
for i in input_list:
if isinstance(i, list):
flatten(i, output_list)
else:
output_list.append(i)
return output_list
input_list = [1,2,[2,3,[4,5]],[6,7],[8,[9]],10]
print(flatten(input_list))
Outputs:
[1, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10]
I found a solution on my own. Below is the solution:
c = []
def test(a):
for val in a:
if type(val) == list:
test(val)
else:
c.append(val)
return c
a=[1,2,[2,3,[4,5]],[6,7],[8,[9]],10]
print test(a)
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I am going through the exercises here: https://www.machinelearningplus.com/python/101-pandas-exercises-python/
Problem #16 has a solution (#1) using np.where() that I am having trouble understanding.
import pandas as pd
import numpy as np
print('pandas: {}'.format(pd.__version__))
print('NumPy: {}'.format(np.__version__))
print('-----')
ser1 = pd.Series([10, 9, 6, 5, 3, 1, 12, 8, 13])
ser2 = pd.Series([1, 3, 10, 13])
# Get the positions of items of 'ser2' in 'ser1' as a list.
# Solution 1
list1 = [np.where(i == ser1)[0].tolist()[0] for i in ser2]
print(list1)
print()
# Solution 2
list2 = [pd.Index(ser1).get_loc(i) for i in ser2]
print(list2)
I have looked up np.where() here:
# https://stackoverflow.com/questions/34667282/numpy-where-detailed-step-by-step-explanation-examples
# https://thispointer.com/numpy-where-tutorial-examples-python/
# https://www.geeksforgeeks.org/numpy-where-in-python/
To be precise, I am not understanding the function and placement of both
bracketed zero's ( [0] ).
np.where outputs a tuple (output of numpy.where(condition) is not an array, but a tuple of arrays: why?), so you'd have to index it (hence the first [0]), then, the output is a numpy array of elements. There is only one in this case, so the second [0] works. the tolist() is completely redundant though
It'd be better to extend list1 with the found indexes, because this code fails when an element occurs more than once:
list1 = []
[list1.extend(np.where(i == ser1)[0]) for i in ser2]
print(list1)
print()
Not the best code imo.
tip, just check the output of stuff yourself, and you would have figured this out. just run np.where(i==ser1) and you'd have seen it returns a tuple, and you need to index it. etc.