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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]]

Previous Solution: As stated by Martijn Pieters, the correct, and fastest, format is:

if 1 in {x, y, z}:

The one major issue that does not seem to be addressed is that you want your output list to include each letter after a true if statement.

Using only Martijn Pieters' advice you would now have:

if 0 in {x, y, z}:

    Mylist.append("c")

elif 1 in {x, y, z}:

    Mylist.append("d")

...

Problem: The first if statement would return true, and you would never get to the following elif statement. So your list would simply return:

["c"]

What you want is to 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:

numToLetters = {0:"c", 1:"d", 2:"e", 3:"f"}

for number in numToLetters:

    if number in {x, y, z}:

        Mylist.append(numToLetters[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 ifs 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

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])

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))

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

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).

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)

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.

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]

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;

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)]

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']

It can be done easily as

for value in [var1,var2,var3]:

     li.append("targetValue")

The most mnemonic 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")

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}:

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

Maybe you need direct formula for output bits set.

x=0 or y=0 or z=0 is equivalent to xyz = 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

lets map to bits: 'c':1 'd':0xb10 'e':0xb100 'f':0xb1000

relation of isc:

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:

logic and is sum of squares of equations

logic or is product of equations

and you'll have 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 element would correspond to ready string.

array[sum] gives you the string.