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Here is my code to merge all tuple in to dictionary,
x = (1,2,3)
y = ('car',"truck","plane")
z=("merc","scania","boeing")
products={}
for i in x,y,z:
products[x[i]]= {y[i]:z[i]}
output:
error:
6 for i in x,y,z:
----> 7 products[x[i]]= {y[i]:z[i]}
8
9 print(products)
TypeError: tuple indices must be integers or slices, not a tuple
Now if i use indexing method inside loop for identifying positions like below code,
for i in x,y,z:
products[x[0]]= {y[0]:z[0]}
print(products)
out:
{1: {'car': 'merc'}}
here, I could only create what I need but only for a specified index how do create a complete dictionary using multiple lists/tuples??
is it also possible to use Zip & map functions?
Use zip to iterate over your separate iterables/tuples in parallel
list(zip(x, y, z)) # [(1, 'car', 'merc'), (2, 'truck', 'scania'), (3, 'plane', 'boeing')]
x = (1, 2, 3)
y = ("car", "truck", "plane")
z = ("merc", "scania", "boeing")
products = {i: {k: v} for i, k, v in zip(x, y, z)}
print(products) # {1: {'car': 'merc'}, 2: {'truck': 'scania'}, 3: {'plane': 'boeing'}}
You should use integer as indices.
x = (1,2,3)
y = ('car',"truck","plane")
z=("merc","scania","boeing")
products={}
for i in range(len(x)):
products[x[i]]= {y[i]:z[i]}
This should solve your problem
To add for above answer, I'm posting a solution using map,
x = (1,2,3)
y = ('car',"truck","plane")
z=("merc","scania","boeing")
products=dict(map(lambda x,y,z:(x,{y:z}),x,y,z))
print(products)
I sometimes need to iterate a list in Python looking at the "current" element and the "next" element. I have, till now, done so with code like:
for current, next in zip(the_list, the_list[1:]):
# Do something
This works and does what I expect, but is there's a more idiomatic or efficient way to do the same thing?
Some answers to this problem can simplify by addressing the specific case of taking only two elements at a time. For the general case of N elements at a time, see Rolling or sliding window iterator?.
The documentation for 3.8 provides this recipe:
import itertools
def pairwise(iterable):
"s -> (s0, s1), (s1, s2), (s2, s3), ..."
a, b = itertools.tee(iterable)
next(b, None)
return zip(a, b)
For Python 2, use itertools.izip instead of zip to get the same kind of lazy iterator (zip will instead create a list):
import itertools
def pairwise(iterable):
"s -> (s0, s1), (s1, s2), (s2, s3), ..."
a, b = itertools.tee(iterable)
next(b, None)
return itertools.izip(a, b)
How this works:
First, two parallel iterators, a and b are created (the tee() call), both pointing to the first element of the original iterable. The second iterator, b is moved 1 step forward (the next(b, None)) call). At this point a points to s0 and b points to s1. Both a and b can traverse the original iterator independently - the izip function takes the two iterators and makes pairs of the returned elements, advancing both iterators at the same pace.
Since tee() can take an n parameter (the number of iterators to produce), the same technique can be adapted to produce a larger "window". For example:
def threes(iterator):
"s -> (s0, s1, s2), (s1, s2, s3), (s2, s3, 4), ..."
a, b, c = itertools.tee(iterator, 3)
next(b, None)
next(c, None)
next(c, None)
return zip(a, b, c)
Caveat: If one of the iterators produced by tee advances further than the others, then the implementation needs to keep the consumed elements in memory until every iterator has consumed them (it cannot 'rewind' the original iterator). Here it doesn't matter because one iterator is only 1 step ahead of the other, but in general it's easy to use a lot of memory this way.
Roll your own!
def pairwise(iterable):
it = iter(iterable)
a = next(it, None)
for b in it:
yield (a, b)
a = b
Starting in Python 3.10, this is the exact role of the pairwise function:
from itertools import pairwise
list(pairwise([1, 2, 3, 4, 5]))
# [(1, 2), (2, 3), (3, 4), (4, 5)]
or simply pairwise([1, 2, 3, 4, 5]) if you don't need the result as a list.
I’m just putting this out, I’m very surprised no one has thought of enumerate().
for (index, thing) in enumerate(the_list):
if index < len(the_list):
current, next_ = thing, the_list[index + 1]
#do something
Since the_list[1:] actually creates a copy of the whole list (excluding its first element), and zip() creates a list of tuples immediately when called, in total three copies of your list are created. If your list is very large, you might prefer
from itertools import izip, islice
for current_item, next_item in izip(the_list, islice(the_list, 1, None)):
print(current_item, next_item)
which does not copy the list at all.
Iterating by index can do the same thing:
#!/usr/bin/python
the_list = [1, 2, 3, 4]
for i in xrange(len(the_list) - 1):
current_item, next_item = the_list[i], the_list[i + 1]
print(current_item, next_item)
Output:
(1, 2)
(2, 3)
(3, 4)
I am really surprised nobody has mentioned the shorter, simpler and most importantly general solution:
Python 3:
from itertools import islice
def n_wise(iterable, n):
return zip(*(islice(iterable, i, None) for i in range(n)))
Python 2:
from itertools import izip, islice
def n_wise(iterable, n):
return izip(*(islice(iterable, i, None) for i in xrange(n)))
It works for pairwise iteration by passing n=2, but can handle any higher number:
>>> for a, b in n_wise('Hello!', 2):
>>> print(a, b)
H e
e l
l l
l o
o !
>>> for a, b, c, d in n_wise('Hello World!', 4):
>>> print(a, b, c, d)
H e l l
e l l o
l l o
l o W
o W o
W o r
W o r l
o r l d
r l d !
This is now a simple Import As of 16th May 2020
from more_itertools import pairwise
for current, next in pairwise(your_iterable):
print(f'Current = {current}, next = {nxt}')
Docs for more-itertools
Under the hood this code is the same as that in the other answers, but I much prefer imports when available.
If you don't already have it installed then:
pip install more-itertools
Example
For instance if you had the fibbonnacci sequence, you could calculate the ratios of subsequent pairs as:
from more_itertools import pairwise
fib= [1,1,2,3,5,8,13]
for current, nxt in pairwise(fib):
ratio=current/nxt
print(f'Curent = {current}, next = {nxt}, ratio = {ratio} ')
As others have pointed out, itertools.pairwise() is the way to go on recent versions of Python. However, for 3.8+, a fun and somewhat more concise (compared to the other solutions that have been posted) option that does not require an extra import comes via the walrus operator:
def pairwise(iterable):
a = next(iterable)
yield from ((a, a := b) for b in iterable)
A basic solution:
def neighbors( list ):
i = 0
while i + 1 < len( list ):
yield ( list[ i ], list[ i + 1 ] )
i += 1
for ( x, y ) in neighbors( list ):
print( x, y )
Pairs from a list using a list comprehension
the_list = [1, 2, 3, 4]
pairs = [[the_list[i], the_list[i + 1]] for i in range(len(the_list) - 1)]
for [current_item, next_item] in pairs:
print(current_item, next_item)
Output:
(1, 2)
(2, 3)
(3, 4)
code = '0016364ee0942aa7cc04a8189ef3'
# Getting the current and next item
print [code[idx]+code[idx+1] for idx in range(len(code)-1)]
# Getting the pair
print [code[idx*2]+code[idx*2+1] for idx in range(len(code)/2)]
Please help me understand how to code the following task in Python using input
Programming challenge description:
Write a short Python program that takes two arrays a and b of length n
storing int values, and returns the dot product of a and b. That is, it returns
an array c of length n such that c[i] = a[i] · b[i], for i = 0,...,n−1.
Test Input:
List1's input ==> 1 2 3
List2's input ==> 2 3 4
Expected Output: 2 6 12
Note that the dot product is defined in mathematics to be the sum of the elements of the vector c you want to build.
That said, here is a possibiliy using zip:
c = [x * y for x, y in zip(a, b)]
And the mathematical dot product would be:
sum(x * y for x, y in zip(a, b))
If the lists are read from the keyboard, they will be read as string, you have to convert them before applying the code above.
For instance:
a = [int(s) for s in input().split(",")]
b = [int(s) for s in input().split(",")]
c = [x * y for x, y in zip(a, b)]
Using for loops and appending
list_c = []
for a, b in zip(list_a, list_b):
list_c.append(a*b)
And now the same, but in the more compact list comprehension syntax
list_c = [a*b for a, b in zip(list_a, list_b)]
From iPython
>>> list_a = [1, 2, 3]
>>> list_b = [2, 3, 4]
>>> list_c = [a*b for a, b in zip(list_a, list_b)]
>>> list_c
[2, 6, 12]
The zip function packs the lists together, element-by-element:
>>> list(zip(list_a, list_b))
[(1, 2), (2, 3), (3, 4)]
And we use tuple unpacking to access the elements of each tuple.
From fetching the input and using map & lambda functions to provide the result. If you may want to print the result with spaces between (not as list), use the last line
list1, list2 = [], []
list1 = list(map(int, input().rstrip().split()))
list2 = list(map(int, input().rstrip().split()))
result_list = list(map(lambda x,y : x*y, list1, list2))
print(*result_list)
I came out with two solutions. Both or them are the ones that are expected in a Python introductory course:
#OPTION 1: We use the concatenation operator between lists.
def dot_product_noappend(list_a, list_b):
list_c = []
for i in range(len(list_a)):
list_c = list_c + [list_a[i]*list_b[i]]
return list_c
print(dot_product_noappend([1,2,3],[4,5,6])) #FUNCTION CALL TO SEE RESULT ON SCREEN
#OPTION 2: we use the append method
def dot_product_append(list_a, list_b):
list_c = []
for i in range(len(list_a)):
list_c.append(list_a[i]*list_b[i])
return list_c
print(dot_product_append([1,2,3],[4,5,6])) #FUNCTION CALL TO SEE RESULT ON SCREEN
Just note that the first method requires that you cast the product of integers to be a list before you can concatenate it to list_c. You do that by using braces ([[list_a[i]*list_b[i]] instead of list_a[i]*list_b[i]). Also note that braces are not necessary in the last method, because the append method does not require to pass a list as parameter.
I have added the two function calls with the values you provided, for you to see that it returns the correct result. Choose whatever function you like the most.
I use Python 3
I want to check if all of my tested values in the nested dictionary are non 0.
So here is the simplified example dict:
d = {'a': {'1990': 10, '1991': 0, '1992': 30},
'b': {'1990': 15, '1991': 40, '1992': 0}}
and I want to test if for both dicts 'a' and 'b' the values of the keys '1990' and '1991' are not zero
for i in d:
for k in range(2):
year = 1990
year = year + k
if all((d[i][str(year)]) != 0):
print(d[i])
so it should only return b, because a['1991']=0
but this is the first time I work with the all() function and I get the error core: TypeError: 'bool' object is not iterable
the error is in the if all() line
thank you very much!
This can done a bit more generally with a list comprehension where you iterate over the items in dict d. A simple comprehension to iterate over the keys and values in our dictionary looks like this:
>>> [k for k, v in d.items()]
['a', 'b']
In the above k will contain the keys and v the values. The comprehension also has an if clause. With that you can filter out the items you don't want. So we define years = ('1990', '1991'). Now we can do another comprehension to test our year values.
To iterate over only 'a', we could do this:
>>> [d['a'][y] for y in years]
[10, 0]
>>> all([d['a'][y] for y in years])
False
Gluing the whole thing together:
>>> d={'a' :{ '1990': 10, '1991':0, '1992':30},'b':{ '1990':15, '1991':40, '1992':0}}
>>> years = ('1990', '1991')
>>> [k for k, v in d.items() if all([v[y] for y in years])]
['b']
See the python docs for more information on list comprehensions.
I can get this loop to work properly:
for x in range(0,len(l)):
for k in d:
if l[x] in d[k]:
l[x] = k
This looks through a list and checks if the value is in any of the dictionary items and then calculates it equal to the dictionary key it is found within (the dictionary contains lists.)
However, I want to convert to a list comprehension or other single line statement for use in a pandas dataframe - to populate a field based on whether or not another field's value is in the labeled dictionary keys and assign it the dictionary key value.
Here is my best attempt, but it does not work:
l = [ k for x in range(0,len(l)) if l[x] in d[k] for k in d ]
Thanks
Assuming I understand what you're after (example data that can be copied and pasted is always appreciated), I'd do something like this:
>>> l = ["a", "b", "c", "d"]
>>> d = {1: ["a"], 3: ["d", "c"]}
>>> l2 = [next((k for k,v in d.items() if lx in v), lx) for lx in l]
>>> l2
[1, 'b', 3, 3]
Don't forget to think about what behaviour you want if an entry in l is found in multiple lists in d, of course, although that may not be an issue with your data.
You can't do it with a list comprehension, because you have an assignment:
l[x] = k
which is an statement, and a list comprehension can't have them.