Having this input:
a = (1,2,3)
b = 'something'
I want to create a list which will look like that:
['something', 1, 2, 3]
I tried to do:
[b, i for i in a]
But got a syntax error.
Note that, I'm looking for a one line solution out of curiosity.
If your variable 'a' will always be a tuple and 'b' will always be a string then you can try one thing...
a = (1,2,3)
b = 'something'
c = [b] + [i for i in a]
Related
I have a string from user. It must containt a comma to split by it and assign to variables. But what if user miss a comma? Surely I can check len of splitted string in if-else branches, but maybe there is another way, I mean assignment during list has a values. For example
a, b, c, d, e = list(range(3)) # 'a' and 'b' are None or not exists
You could do something like this:
>>> alist=list(range(3))
>>> alist
[0, 1, 2]
>>> a,b,c,*d=alist
>>> a,b,c
(0, 1, 2)
>>> d
[]
If there are no more elements, d is an empty list. It uses the unpacking operator *. Not the best possible solution for large lists, so I would still define a function for that. For small cases, it works well. (You could assume that is d==[], there are no more elements in alist)For example, you could add:
return False if not d else return True
You are free to extend the list with None values before extraction.
li = list(range(3))
expected_size = 5
missing_size = expected_size - len(li)
none_li = [None] * missing_size
new_li = li + none_li
a, b, c, d, e = new_li
I have a tuple like this:
('TRM',)
I would like to extract the string only from this tuple.
I did the following and it works but just thinking if there is a better way to do that.
>>> b = [x for x in a]
>>> b
['TRM']
>>> for i in b:
... print(i)
...
TRM
>>>
This will help you
b = [x for x in a if type(x) == type("Str")]
print(b)
If you need to access a specific element of the tuple you can do it by index:
>>> a[0]
TRM
Not exactly sure what you're asking for but you could do:
for i in b:
if isinstance(i, str):
print(i)
and this will print
TRM
Edit: Now it checks whether an element is a string.
In Python you can use a.intersection(b) to find the items common to both sets.
Is there a way to do the disjoint opposite version of this? Items that are not common to both a and b; the unique items in a unioned with the unique items in b?
You are looking for the symmetric difference; all elements that appear only in set a or in set b, but not both:
a.symmetric_difference(b)
From the set.symmetric_difference() method documentation:
Return a new set with elements in either the set or other but not both.
You can use the ^ operator too, if both a and b are sets:
a ^ b
while set.symmetric_difference() takes any iterable for the other argument.
The output is the equivalent of (a | b) - (a & b), the union of both sets minus the intersection of both sets.
a={1,2,4,5,6}
b={5,6,4,9}
c=(a^b)&b
print(c) # you got {9}
The best way is a list comprehension.
a = [ 1,2,3,4]
b = [ 8,7,9,2,1]
c = [ element for element in a if element not in b]
d = [ element for element in b if element not in a]
print(c)
# output is [ 3,4]
print(d)
# output is [8,7,9]
You can join both lists
Try this code for (set(a) - intersection(a&b))
a = [1,2,3,4,5,6]
b = [2,3]
for i in b:
if i in a:
a.remove(i)
print(a)
the output is [1,4,5,6]
I hope, it will work
e, f are two list you want to check disjoint
a = [1,2,3,4]
b = [8,7,9,2,1]
c = []
def loop_to_check(e,f):
for i in range(len(e)):
if e[i] not in f:
c.append(e[i])
loop_to_check(a,b)
loop_to_check(b,a)
print(c)
## output is [3,4,8,7,9]
This loops around to list and returns the disjoint list
After entering a command I am given data, that I then transform into a list. Once transformed into a list, how do I copy ALL of the data from that list [A], and save it - so when I enter a command and am given a second list of data [B], I can compare the two; and have data that is the same from the two lists cancel out - so what is not similar between [A] & [B] is output. For example...
List [A]
1
2
3
List [B]
1
2
3
4
Using Python, I now want to compare the two lists to each other, and then output the differences.
Output = 4
Hopefully this makes sense!
You can use set operations.
a = [1,2,3]
b = [1,2,3,4]
print set(b) - set(a)
to output the data in list format you can use the following print statement
print list(set(b) - set(a))
>>> b=[1,2,3,4]
>>> a=[1,2,3]
>>> [x for x in b if x not in a]
[4]
for element in b:
if element in a:
a.remove(element)
This answer will return a list not a set, and should take duplicates into account. That way [1,2,1] - [1,2] returns [1] not [].
Try itertools.izip_longest
import itertools
a = [1,2,3]
b = [1,2,3,4]
[y for x, y in itertools.izip_longest(a, b) if x != y]
# [4]
You could easily modify this further to return a duple for each difference, where the first item in the duple is the position in b and the second item is the value.
[(i, pair[1]) for i, pair in enumerate(itertools.izip_longest(a, b)) if pair[0] != pair[1]]
# [(3, 4)]
For entering the data use a loop:
def enterList():
result = []
while True:
value = raw_input()
if value:
result.append(value)
else:
return result
A = enterList()
B = enterList()
For comparing you can use zip to build pairs and compare each of them:
for a, b in zip(A, B):
if a != b:
print a, "!=", b
This will truncate the comparison at the length of the shorter list; use the solution in another answer given here using itertools.izip_longest() to handle that.
I want to do a functional like pattern match to get the first two elements, and then the rest of an array return value.
For example, assume that perms(x) returns a list of values, and I want to do this:
seq=perms(x)
a = seq[0]
b = seq[1]
rest = seq[2:]
Of course I can shorten to:
[a,b] = seq[0:2]
rest = seq[2:]
Can I use some notation to do this?
[a,b,more] = perms(x)
or conceptually:
[a,b,more..] = perms(x)
PROLOG & functional languages do list decomposition so nicely like this!
You can do it in Python 3 like this:
(a, b, *rest) = seq
See the extended iterable unpacking PEP for more details.
In python 2, your question is very close to an answer already:
a, b, more = (seq[0], seq[1], seq[2:])
or:
(a, b), more = (seq[0:2], seq[2:])
For Python 2, I know you can do it with a function:
>>> def getValues(a, b, *more):
return a, b, more
>>> seq = [1,2,3,4,5]
>>> a, b, more = getValues(*seq)
>>> a
1
>>> b
2
>>> more
(3, 4, 5)
But not sure if there's any way of doing it like Ayman's Python 3 suggestion
Very nice, thanks.
The suggestions where one dissects the array on the fight-hand side don't work so well for me, as I actually wanted to pattern match on the returns from a generator expression.
for (a, b, more) in perms(seq): ...
I like the P3 solution, but have to wait for Komodo to support it!