Given two lists (a and b), I'd like to replace three elements of list 'a' with three elements of list 'b'. Currently I am using an expression like this:
a[0], a[5], a[7] = b[11], b[99], b[2]
As I need to do such operations very frequently with lots of different arrays I am wondering if there is a more compact solution for this problem (the number of elements I need to replace is always 3 though). I was thinking about something like:
a[0,5,7] = b[11,99,2]
Which obviously does not work.
If you've a python list you can do something like this :
toReplace = [0,5,7]
targetIndices = [11, 99, 2]
for i,j in zip(toReplace, targetIndices): a[i] = b[j]
If you've a numpy array, it's even simpler :
a[toReplace] = b[targetIndices]
#i.e, a[[0,5,7]] = b[[11, 99, 2]]
There might be some better solutions but this does the trick:
ind1 = [0,5,7]
ind2 = [11,99,2]
for i in range(len(ind1)):
a[ind1[i]]=b[ind2[i]]
Related
I need to create a list of named lists, so that I can iterate over them. This works, but I'm wondering if there isn't a better way.
terms = []; usedfors = []; broaders = []; narrowers = []
termlist = [terms, usedfors, broaders, narrowers]
The reason for doing it this way is that I have a function do_some_list_operation(l) and I want to do something like
for i in termlist:
do_some_list_operation(i)
rather than
do_some_list_operation(terms)
do_some_list_operation(usedfors)
do_some_list_operation(broaders)
do_some_list_operation(narrowers)
I've been searching for 'how to build list of named lists' to no avail.
The way you are doing it is fine but be aware that you are creating a list of lists when you defined termlist. If you want to have the name of the lists then termlist should be:
termlist = ["terms","usedfors", "broaders", "narrowers"]
Now if you want to use these strings as list names you can use globals() or locals() e.g.:
terms = [1]; usedfors = [2]; broaders = [3,4,8]; narrowers = [5]
termlist = ["terms","usedfors", "broaders", "narrowers"]
for i in termlist:
print(sum(locals()[i]))
output:
1
2
15
5
While I like the answer by #jayvee and upvoted it, I think a more conventional answer would be one based on using a dictionary as first suggested by #msaw328.
Starting with our new data structure:
term_data = {
"terms": [1],
"usedfors": [2],
"broaders": [3, 4, 5],
"narrowers": [6, 7]
}
Perhaps like:
for term in term_data:
print(f"{term} sums to {sum(term_data[term])}")
or even
for term, term_value in term_data.items():
print(f"{term} sums to {sum(term_value)}")
Both should give you:
terms sums to 1
usedfors sums to 2
broaders sums to 12
narrowers sums to 13
Provided with a list of lists. Here's an example myList =[[70,83,90],[19,25,30]], return a list of lists which contains the difference between the elements. An example of the result would be[[13,7],[6,5]]. The absolute value of (70-83), (83-90), (19-25), and (25-30) is what is returned. I'm not sure how to iterate through the list to subtract adjacent elements without already knowing the length of the list. So far I have just separated the list of lists into two separate lists.
list_one = myList[0]
list_two = myList[1]
Please let me know what you would recommend, thank you!
A custom generator can return two adjacent items at a time from a sequence without knowing the length:
def two(sequence):
i = iter(sequence)
a = next(i)
for b in i:
yield a,b
a = b
original = [[70,83,90],[19,25,30]]
result = [[abs(a-b) for a,b in two(sequence)]
for sequence in original]
print(result)
[[13, 7], [6, 5]]
Well, for each list, you can simply get its number of elements like this:
res = []
for my_list in list_of_lists:
res.append([])
for i in range(len(my_list) - 1):
# Do some stuff
You can then add the results you want to res[-1].
The task is:
User enters a number, you take 1 number from the left, one from the right and sum it. Then you take the rest of this number and sum every digit in it. then you get two answers. You have to sort them from biggest to lowest and make them into a one solid number. I solved it, but i don't like how it looks like. i mean the task is pretty simple but my code looks like trash. Maybe i should use some more built-in functions and libraries. If so, could you please advise me some? Thank you
a = int(input())
b = [int(i) for i in str(a)]
closesum = 0
d = []
e = ""
farsum = b[0] + b[-1]
print(farsum)
b.pop(0)
b.pop(-1)
print(b)
for i in b:
closesum += i
print(closesum)
d.append(int(closesum))
d.append(int(farsum))
print(d)
for i in sorted(d, reverse = True):
e += str(i)
print(int(e))
input()
You can use reduce
from functools import reduce
a = [0,1,2,3,4,5,6,7,8,9]
print(reduce(lambda x, y: x + y, a))
# 45
and you can just pass in a shortened list instead of poping elements: b[1:-1]
The first two lines:
str_input = input() # input will always read strings
num_list = [int(i) for i in str_input]
the for loop at the end is useless and there is no need to sort only 2 elements. You can just use a simple if..else condition to print what you want.
You don't need a loop to sum a slice of a list. You can also use join to concatenate a list of strings without looping. This implementation converts to string before sorting (the result would be the same). You could convert to string after sorting using map(str,...)
farsum = b[0] + b[-1]
closesum = sum(b[1:-2])
"".join(sorted((str(farsum),str(closesum)),reverse=True))
Say I have a list of strings, like so:
strings = ["abc", "def", "ghij"]
Note that the length of a string in the list can vary.
The way you generate a new string is to take one letter from each element of the list, in order. Examples: "adg" and "bfi", but not "dch" because the letters are not in the same order in which they appear in the list. So in this case where I know that there are only three elements in the list, I could fairly easily generate all possible combinations with a nested for loop structure, something like this:
for i in strings[0].length:
for ii in strings[1].length:
for iii in strings[2].length:
print(i+ii+iii)
The issue arises for me when I don't know how long the list of strings is going to be beforehand. If the list is n elements long, then my solution requires n for loops to succeed.
Can any one point me towards a relatively simple solution? I was thinking of a DFS based solution where I turn each letter into a node and creating a connection between all letters in adjacent strings, but this seems like too much effort.
In python, you would use itertools.product
eg.:
>>> for comb in itertools.product("abc", "def", "ghij"):
>>> print(''.join(comb))
adg
adh
adi
adj
aeg
aeh
...
Or, using an unpack:
>>> words = ["abc", "def", "ghij"]
>>> print('\n'.join(''.join(comb) for comb in itertools.product(*words)))
(same output)
The algorithm used by product is quite simple, as can be seen in its source code (Look particularly at function product_next). It basically enumerates all possible numbers in a mixed base system (where the multiplier for each digit position is the length of the corresponding word). A simple implementation which only works with strings and which does not implement the repeat keyword argument might be:
def product(words):
if words and all(len(w) for w in words):
indices = [0] * len(words)
while True:
# Change ''.join to tuple for a more accurate implementation
yield ''.join(w[indices[i]] for i, w in enumerate(words))
for i in range(len(indices), 0, -1):
if indices[i - 1] == len(words[i - 1]) - 1:
indices[i - 1] = 0
else:
indices[i - 1] += 1
break
else:
break
From your solution it seems that you need to have as many for loops as there are strings. For each character you generate in the final string, you need a for loop go through the list of possible characters. To do that you can make recursive solution. Every time you go one level deep in the recursion, you just run one for loop. You have as many level of recursion as there are strings.
Here is an example in python:
strings = ["abc", "def", "ghij"]
def rec(generated, k):
if k==len(strings):
print(generated)
return
for c in strings[k]:
rec(generated + c, k+1)
rec("", 0)
Here's how I would do it in Javascript (I assume that every string contains no duplicate characters):
function getPermutations(arr)
{
return getPermutationsHelper(arr, 0, "");
}
function getPermutationsHelper(arr, idx, prefix)
{
var foundInCurrent = [];
for(var i = 0; i < arr[idx].length; i++)
{
var str = prefix + arr[idx].charAt(i);
if(idx < arr.length - 1)
{
foundInCurrent = foundInCurrent.concat(getPermutationsHelper(arr, idx + 1, str));
}
else
{
foundInCurrent.push(str);
}
}
return foundInCurrent;
}
Basically, I'm using a recursive approach. My base case is when I have no more words left in my array, in which case I simply add prefix + c to my array for every c (character) in my last word.
Otherwise, I try each letter in the current word, and pass the prefix I've constructed on to the next word recursively.
For your example array, I got:
adg adh adi adj aeg aeh aei aej afg afh afi afj bdg bdh bdi
bdj beg beh bei bej bfg bfh bfi bfj cdg cdh cdi cdj ceg ceh
cei cej cfg cfh cfi cfj
I have another question that I'd like input on, of course no direct answers just something to point me in the right direction!
I have a string of numbers ex. 1234567890 and I want 1 & 0 to change places (0 and 9) and for '2345' & '6789' to change places. For a final result of '0678923451'.
First things I did was convert the string into a list with:
ex. original = '1234567890'
original = list(original)
original = ['0', '1', '2' etc....]
Now, I get you need to pull the first and last out, so I assigned
x = original[0]
and
y = original[9]
So: x, y = y, x (which gets me the result I'm looking for)
But how do I input that back into the original list?
Thanks!
The fact that you 'pulled' the data from the list in variables x and y doesn't help at all, since those variables have no connection anymore with the items from the list. But why don't you swap them directly:
original[0], original[9] = original[9], original[0]
You can use the slicing operator in a similar manner to swap the inner parts of the list.
But, there is no need to create a list from the original string. Instead, you can use the slicing operator to achieve the result you want. Note that you cannot swap the string elements as you did with lists, since in Python strings are immutable. However, you can do the following:
>>> a = "1234567890"
>>> a[9] + a[5:9] + a[1:5] + a[0]
'0678923451'
>>>