How to find number of matches in the array or dictionary? for example if the dictionary or the array has the number 4 occurring 5 times, how can I find out about it?
List elements:
my_list = [1,4,7,4,5,7,1,3]
print my_list.count(4)
Dictionary values using generator expression:
my_dict = {0: 1, 2: 1, 4: 5, 6: 3, 8: 4, 10: 4, 12: 1}
print sum(1 for x in my_dict.values() if x == 4)
As pointed out by zondo, the last line can be more simply written as:
print sum(x == 4 for x in my_dict.values())
due to the fact that True == 1.
For a list:
my_list = [x for x in [1,4,7,4,5,7,1,3]
len([x for x in my_list if x == 4])
For a dictionary (assuming you want to count the values):
my_dict = {0: 1, 2: 1, 4: 5, 6: 3, 8: 4, 10: 4, 12: 1}
len([x for x in my_dict.values() if x == 4])
The two code segments are counting the number of elements that match the criteria after if (in your case, that they equal 4). The len function requires a list (not a generator), requiring the extra-looking [] in there.
Related
Hey I'm new to lists and modules can you show me how to do this. I actually tried this code but it seems to be giving me a wrong answer
l = [1, 2, 3, 4, 5]
fact = 1
for i in l:
for number in range(1,i+1):
fact=fact*number
print ("Factorial of", i, "is", fact)
def factorial(n):
if n < 2:
return 1
else:
return n * factorial(n-1)
l = [1, 2, 3, 4, 5]
for number in l:
fact = factorial(number)
print("Facotrial of ",number," is",fact)
Here is a working example.
You can use the balise code to better show your code, I have trouble to read it so I can't really comment on what you did wrong
e.g. [a,a,c,e] and [a,b,c,d,e]. This is what I mean by same order. It shouldn't matter if one of the lists is only partial. To be clear, I am uninterested in sorting anything.
Edit: Apparently I should mention that I mean lists with some different values. e.g. [a,a,c,e] and [a,b,c,d,e] are in the same order. Simply getting rid of duplicates and cutting the bigger list won't work.
We can use itertools.groupby to squash identical consecutive items into a single item of that value. We can then zip together the squashed values with the other list to see if they are the same
from itertools import izip, groupby
def squash(iterable):
return (k for k, _ in groupby(iterable))
def same_order(iterable, reference):
return all(x == y for x, y in zip(squash(iterable), reference))
print(same_order((1, 2, 2, 3, 4), (1, 2, 3, 4, 5)))
# True
print(same_order((1, 2, 2, 1, 3, 4), (1, 2, 3, 4, 5)))
# False
You can first take the unique set of b, then truncate to the size of a (or vice versa)
a = [1,2,3,4]
b = [1,1,2,2,3,4,4,5]
a == list(set(b))[:len(a)]
You can achieve this by first getting the unique elements from partial list a (in the order of a), then get those elements from full list b (in the order of b). Then you can compare the two and see if they are identical.
>>> a = [1, 1, 4, 3]
>>> a = sorted(set(a), key=a.index) # Remove duplicate elements, but keep the original order
>>> a
[1, 4, 3]
>>> b = [1, 2, 4, 3, 5]
>>> b = [x for x in sorted(set(b), key=b.index) if x in a] # Get unique elements of a from b in order of b, and filter them out to get only elements present in a
>>> a == b
True
>>> a = [4, 3, 2, 1]
>>> a = sorted(set(a), key=a.index) # Remove duplicate elements, but keep the original order
>>> b = [1, 2, 4, 3, 5]
>>> b = [x for x in sorted(set(b), key=b.index) if x in a] # Get unique elements of a from b in order of b, and filter them out to get only elements present in a
>>> a == b
False
If I have an input like this (1, 2, 3, 4, 5, 6)
The output has to be ... [[1, 2], [3, 4], [5, 6]].
I know how to deal with if it's one element but not two.
x=[]
for number in numbers:
x.append([number])
I'll appreciate your any help!
Something like this would work:
out = []
lst = (1,2,3,4,5,6,7,8,9,10)
for x in range(len(lst)):
if x % 2 == 0:
out.append([lst[x], lst[x+1]])
else:
continue
To use this, just set lst equal to whatever list of numbers you want. The final product is stored in out.
There is a shorter way of doing what you want:
result = []
L = (1,2,3,4,5,6,7,8,9,10)
result = [[L[i], L[i + 1]] for i in range(0, len(L) - 1, 2)]
print(result)
You can use something like this. This solution also works for list of odd length
def func(lst):
res = []
# Go through every 2nd value | 0, 2, 4, ...
for i in range(0, len(lst), 2):
# Append a slice of the list, + 2 to include the next value
res.append(lst[i : i + 2])
return res
# Output
>>> lst = [1, 2, 3, 4, 5, 6]
>>> func(lst)
[[1, 2], [3, 4], [5, 6]]
>>> lst2 = [1, 2, 3, 4, 5, 6, 7]
>>> func(lst2)
[[1, 2], [3, 4], [5, 6], [7]]
List comprehension solution
def func(lst):
return [lst[i:i+2] for i in range(0, len(lst), 2)]
Slicing is better in this case as you don't have to account for IndexError allowing it to work for odd length as well.
If you want you can also add another parameter to let you specify the desired number of inner elements.
def func(lst, size = 2): # default of 2 it none specified
return [lst[i:i+size] for i in range(0, len(lst), size)]
There's a few hurdles in this problem. You want to iterate through the list without going past the end of the list and you need to deal with the case that list has an odd length. Here's one solution that works:
def foo(lst):
result = [[x,y] for [x,y] in zip(lst[0::2], lst[1::2])]
return result
In case this seems convoluted, let's break the code down.
Index slicing:
lst[0::2] iterates through lst by starting at the 0th element and proceeds in increments of 2. Similarly lst[1::2] iterates through starting at the 1st element (colloquially the second element) and continues in increments of 2.
Example:
>>> lst = (1,2,3,4,5,6,7)
>>> print(lst[0::2])
(1,3,5,7)
>>> print(lst[1::2])
(2,4,6)
zip: zip() takes two lists (or any iterable object for that matter) and returns a list containing tuples. Example:
>>> lst1 = (10,20,30, 40)
>>> lst2 = (15,25,35)
>>> prit(zip(lst1, lst2))
[(10,15), (20,25), (30,35)]
Notice that zip(lst1, lst2) has the nice property that if one of it's arguments is longer than the other, zip() stops zipping whenever the shortest iterable is out of items.
List comprehension: python allows iteration quite generally. Consider the statement:
>>> [[x,y] for [x,y] in zip(lst1,lst2)]
The interior bit "for [x,y] in zip(lst1,lst2)" says "iterate through all pairs of values in zip, and give their values to x and y". In the rest of the statement
"[[x,y] for [x,y] ...]", it says "for each set of values x and y takes on, make a list [x,y] to be stored in a larger list". Once this statement executes, you have a list of lists, where the interior lists are all possible pairs for zip(lst1,lst2)
Very Clear solution:
l = (1, 2, 3, 4, 5, 6)
l = iter(l)
w = []
for i in l:
sub = []
sub.append(i)
sub.append(next(l))
w.append(sub)
print w
a = [1 5 2 4 6 3]
I have this list. Is there a way to make a second corresponding list with the same number but 6 from a list must be 1 in the b list and 1 in a list must be 6 in b list .I want exactly this:
b = [6 2 5 3 1 4]
Use zip of the sorted, and reversed-sorted list to create a reverse mapping from reversed-tuples:
>>> a = [1, 5, 2, 4, 6, 3]
>>> reverse_mapping = dict(zip(sorted(a), reversed(sorted(a))))
Creates the following mapping:
{1: 6, 2: 5, 3: 4, 4: 3, 5: 2, 6: 1}
Now use that to construct the new list:
>>> b = [reverse_mapping[item] for item in a]
>>> b
[6, 2, 5, 3, 1, 4]
This can be optimized if needed, I just kept it clear.
I have a nested sequence that I want to flatten into a single list of values.
Please try this general solution:
Write a recursive generator function involving a yield from
statement. For example:
from collections import Iterable
def flatten(items, ignore_types=(str, bytes)):
for x in items:
if isinstance(x, Iterable) and not isinstance(x, ignore_types):
yield from flatten(x, ignore_types)
else:
yield x
items = [1, 2, [3, 4, [5, 6], 7], 8]
# Produces 1 2 3 4 5 6 7 8
for x in flatten(items):
print(x)
I'd go with recursion but split in a balanced way.
def flatten(lst):
n = len(lst)
if n == 0:
return lst
elif n == 1:
item = lst[0]
try:
return flatten(list(item))
except TypeError:
return [item]
else:
k = n // 2
return flatten(lst[:k]) + flatten(lst[k:])
Demo
items = [1, 2, [3, 4, [5, 6], 7], 8]
flatten(items)
[1, 2, 3, 4, 5, 6, 7, 8]