I'm trying to make a function that will compare multiple variables to an integer and output a string of three letters. I was wondering if there was a way to translate this into Python. So say:
x = 0
y = 1
z = 3
mylist = []
if x or y or z == 0 :
mylist.append("c")
if x or y or z == 1 :
mylist.append("d")
if x or y or z == 2 :
mylist.append("e")
if x or y or z == 3 :
mylist.append("f")
which would return a list of:
["c", "d", "f"]
Is something like this possible?
This question is related to
python
if-statement
comparison
match
boolean-logic
It can be done easily as
for value in [var1,var2,var3]:
li.append("targetValue")
If you want something very interesting solution, refer below:
x = 0
y = 1
z = 3
mylist = []
if any(i in [0] for i in[x,y,z]):
mylist.append("c")
if any(i in [1] for i in[x,y,z]):
mylist.append("d")
if any(i in [2] for i in[x,y,z]):
mylist.append("e")
if any(i in [3] for i in[x,y,z]):
mylist.append("f")
It is a mix of list comprehension and any keyword.
Set is the good approach here, because it orders the variables, what seems to be your goal here. {z,y,x}
is {0,1,3}
whatever the order of the parameters.
>>> ["cdef"[i] for i in {z,x,y}]
['c', 'd', 'f']
This way, the whole solution is O(n).
Maybe you need direct formula for output bits set.
x=0 or y=0 or z=0 is equivalent to x*y*z = 0
x=1 or y=1 or z=1 is equivalent to (x-1)*(y-1)*(z-1)=0
x=2 or y=2 or z=2 is equivalent to (x-2)*(y-2)*(z-2)=0
Let's map to bits: 'c':1 'd':0xb10 'e':0xb100 'f':0xb1000
Relation of isc (is 'c'):
if xyz=0 then isc=1 else isc=0
Use math if formula https://youtu.be/KAdKCgBGK0k?list=PLnI9xbPdZUAmUL8htSl6vToPQRRN3hhFp&t=315
[c]: (xyz=0 and isc=1) or (((xyz=0 and isc=1) or (isc=0)) and (isc=0))
[d]: ((x-1)(y-1)(z-1)=0 and isc=2) or (((xyz=0 and isd=2) or (isc=0)) and (isc=0))
...
Connect these formulas by following logic:
and
is the sum of squares of equationsor
is the product of equationsand you'll have a total equation express sum and you have total formula of sum
then sum&1 is c, sum&2 is d, sum&4 is e, sum&5 is f
After this you may form predefined array where index of string elements would correspond to ready string.
array[sum]
gives you the string.
d = {0:'c', 1:'d', 2:'e', 3: 'f'}
x, y, z = (0, 1, 3)
print [v for (k,v) in d.items() if x==k or y==k or z==k]
You can use dictionary :
x = 0
y = 1
z = 3
list=[]
dict = {0: 'c', 1: 'd', 2: 'e', 3: 'f'}
if x in dict:
list.append(dict[x])
else:
pass
if y in dict:
list.append(dict[y])
else:
pass
if z in dict:
list.append(dict[z])
else:
pass
print list
Looks like you're building some kind of Caesar cipher.
A much more generalized approach is this:
input_values = (0, 1, 3)
origo = ord('c')
[chr(val + origo) for val in inputs]
outputs
['c', 'd', 'f']
Not sure if it's a desired side effect of your code, but the order of your output will always be sorted.
If this is what you want, the final line can be changed to:
sorted([chr(val + origo) for val in inputs])
To test multiple variables with one single value: if 1 in {a,b,c}:
To test multiple values with one variable: if a in {1, 2, 3}:
Without dict, try this solution:
x, y, z = 0, 1, 3
offset = ord('c')
[chr(i + offset) for i in (x,y,z)]
and gives:
['c', 'd', 'f']
You can unite this
x = 0
y = 1
z = 3
in one variable.
In [1]: xyz = (0,1,3,)
In [2]: mylist = []
Change our conditions as:
In [3]: if 0 in xyz:
...: mylist.append("c")
...: if 1 in xyz:
...: mylist.append("d")
...: if 2 in xyz:
...: mylist.append("e")
...: if 3 in xyz:
...: mylist.append("f")
Output:
In [21]: mylist
Out[21]: ['c', 'd', 'f']
This code may be helpful
L ={x, y, z}
T= ((0,"c"),(1,"d"),(2,"e"),(3,"f"),)
List2=[]
for t in T :
if t[0] in L :
List2.append(t[1])
break;
I think this will handle it better:
my_dict = {0: "c", 1: "d", 2: "e", 3: "f"}
def validate(x, y, z):
for ele in [x, y, z]:
if ele in my_dict.keys():
return my_dict[ele]
Output:
print validate(0, 8, 9)
c
print validate(9, 8, 9)
None
print validate(9, 8, 2)
e
The most pythonic way of representing your pseudo-code in Python would be:
x = 0
y = 1
z = 3
mylist = []
if any(v == 0 for v in (x, y, z)):
mylist.append("c")
if any(v == 1 for v in (x, y, z)):
mylist.append("d")
if any(v == 2 for v in (x, y, z)):
mylist.append("e")
if any(v == 3 for v in (x, y, z)):
mylist.append("f")
As stated by Martijn Pieters, the correct, and fastest, format is:
if 1 in {x, y, z}:
Using his advice you would now have separate if-statements so that Python will read each statement whether the former were True
or False
. Such as:
if 0 in {x, y, z}:
mylist.append("c")
if 1 in {x, y, z}:
mylist.append("d")
if 2 in {x, y, z}:
mylist.append("e")
...
This will work, but if you are comfortable using dictionaries (see what I did there), you can clean this up by making an initial dictionary mapping the numbers to the letters you want, then just using a for-loop:
num_to_letters = {0: "c", 1: "d", 2: "e", 3: "f"}
for number in num_to_letters:
if number in {x, y, z}:
mylist.append(num_to_letters[number])
The direct way to write x or y or z == 0
is
if any(map((lambda value: value == 0), (x,y,z))):
pass # write your logic.
But I dont think, you like it. :) And this way is ugly.
The other way (a better) is:
0 in (x, y, z)
BTW lots of if
s could be written as something like this
my_cases = {
0: Mylist.append("c"),
1: Mylist.append("d")
# ..
}
for key in my_cases:
if key in (x,y,z):
my_cases[key]()
break
While the pattern for testing multiple values
>>> 2 in {1, 2, 3}
True
>>> 5 in {1, 2, 3}
False
is very readable and is working in many situation, there is one pitfall:
>>> 0 in {True, False}
True
But we want to have
>>> (0 is True) or (0 is False)
False
One generalization of the previous expression is based on the answer from ytpillai:
>>> any([0 is True, 0 is False])
False
which can be written as
>>> any(0 is item for item in (True, False))
False
While this expression returns the right result it is not as readable as the first expression :-(
All of the excellent answers provided here concentrate on the specific requirement of the original poster and concentrate on the if 1 in {x,y,z}
solution put forward by Martijn Pieters.
What they ignore is the broader implication of the question:
How do I test one variable against multiple values?
The solution provided will not work for partial hits if using strings for example:
Test if the string "Wild" is in multiple values
>>> x = "Wild things"
>>> y = "throttle it back"
>>> z = "in the beginning"
>>> if "Wild" in {x, y, z}: print (True)
...
or
>>> x = "Wild things"
>>> y = "throttle it back"
>>> z = "in the beginning"
>>> if "Wild" in [x, y, z]: print (True)
...
for this scenario it's easiest to convert to a string
>>> [x, y, z]
['Wild things', 'throttle it back', 'in the beginning']
>>> {x, y, z}
{'in the beginning', 'throttle it back', 'Wild things'}
>>>
>>> if "Wild" in str([x, y, z]): print (True)
...
True
>>> if "Wild" in str({x, y, z}): print (True)
...
True
It should be noted however, as mentioned by @codeforester
, that word boundries are lost with this method, as in:
>>> x=['Wild things', 'throttle it back', 'in the beginning']
>>> if "rot" in str(x): print(True)
...
True
the 3 letters rot
do exist in combination in the list but not as an individual word. Testing for " rot " would fail but if one of the list items were "rot in hell", that would fail as well.
The upshot being, be careful with your search criteria if using this method and be aware that it does have this limitation.
If you want to use if, else statements following is another solution:
myList = []
aList = [0, 1, 3]
for l in aList:
if l==0: myList.append('c')
elif l==1: myList.append('d')
elif l==2: myList.append('e')
elif l==3: myList.append('f')
print(myList)
you can develop it through two ways
def compareVariables(x,y,z):
mylist = []
if x==0 or y==0 or z==0:
mylist.append('c')
if x==1 or y==1 or z==1:
mylist.append('d')
if x==2 or y==2 or z==2:
mylist.append('e')
if x==3 or y==3 or z==3:
mylist.append('f')
else:
print("wrong input value!")
print('first:',mylist)
compareVariables(1, 3, 2)
Or
def compareVariables(x,y,z):
mylist = []
if 0 in (x,y,z):
mylist.append('c')
if 1 in (x,y,z):
mylist.append('d')
if 2 in (x,y,z):
mylist.append('e')
if 3 in (x,y,z):
mylist.append('f')
else:
print("wrong input value!")
print('second:',mylist)
compareVariables(1, 3, 2)
One line solution:
mylist = [{0: 'c', 1: 'd', 2: 'e', 3: 'f'}[i] for i in [0, 1, 2, 3] if i in (x, y, z)]
Or:
mylist = ['cdef'[i] for i in range(4) if i in (x, y, z)]
To check if a value is contained within a set of variables you can use the inbuilt modules itertools
and operator
.
For example:
Imports:
from itertools import repeat
from operator import contains
Declare variables:
x = 0
y = 1
z = 3
Create mapping of values (in the order you want to check):
check_values = (0, 1, 3)
Use itertools
to allow repetition of the variables:
check_vars = repeat((x, y, z))
Finally, use the map
function to create an iterator:
checker = map(contains, check_vars, check_values)
Then, when checking for the values (in the original order), use next()
:
if next(checker) # Checks for 0
# Do something
pass
elif next(checker) # Checks for 1
# Do something
pass
etc...
This has an advantage over the lambda x: x in (variables)
because operator
is an inbuilt module and is faster and more efficient than using lambda
which has to create a custom in-place function.
Another option for checking if there is a non-zero (or False) value in a list:
not (x and y and z)
Equivalent:
not all((x, y, z))
This will help you.
def test_fun(val):
x = 0
y = 1
z = 2
myList = []
if val in (x, y, z) and val == 0:
myList.append("C")
if val in (x, y, z) and val == 1:
myList.append("D")
if val in (x, y, z) and val == 2:
myList.append("E")
test_fun(2);
If you ARE very very lazy, you can put the values inside an array. Such as
list = []
list.append(x)
list.append(y)
list.append(z)
nums = [add numbers here]
letters = [add corresponding letters here]
for index in range(len(nums)):
for obj in list:
if obj == num[index]:
MyList.append(letters[index])
break
You can also put the numbers and letters in a dictionary and do it, but this is probably a LOT more complicated than simply if statements. That's what you get for trying to be extra lazy :)
One more thing, your
if x or y or z == 0:
will compile, but not in the way you want it to. When you simply put a variable in an if statement (example)
if b
the program will check if the variable is not null. Another way to write the above statement (which makes more sense) is
if bool(b)
Bool is an inbuilt function in python which basically does the command of verifying a boolean statement (If you don't know what that is, it is what you are trying to make in your if statement right now :))
Another lazy way I found is :
if any([x==0, y==0, z==0])
Your problem is more easily addressed with a dictionary structure like:
x = 0
y = 1
z = 3
d = {0: 'c', 1:'d', 2:'e', 3:'f'}
mylist = [d[k] for k in [x, y, z]]
You can try the method shown below. In this method, you will have the freedom to specify/input the number of variables that you wish to enter.
mydict = {0:"c", 1:"d", 2:"e", 3:"f"}
mylist= []
num_var = int(raw_input("How many variables? ")) #Enter 3 when asked for input.
for i in range(num_var):
''' Enter 0 as first input, 1 as second input and 3 as third input.'''
globals()['var'+str('i').zfill(3)] = int(raw_input("Enter an integer between 0 and 3 "))
mylist += mydict[globals()['var'+str('i').zfill(3)]]
print mylist
>>> ['c', 'd', 'f']
Source: Stackoverflow.com