Python's equivalent of && (logical-and) in an

Question

Here's my code:

def front_back(a, b):
  # +++your code here+++
  if len(a) % 2 == 0 && len(b) % 2 == 0:
    return a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):] 
  else:
    #todo! Not yet done. :P
  return

I'm getting an error in the IF conditional. What am I doing wrong?

Answer 1

Two comments:

• Use and and or for logical operations in Python.
• Use 4 spaces to indent instead of 2. You will thank yourself later because your code will look pretty much the same as everyone else's code. See PEP 8 for more details.

Answer 2

You would want and instead of &&.

Answer 3

Use of "and" in conditional. I often use this when importing in Jupyter Notebook:

def find_local_py_scripts():
    import os # does not cost if already imported
    for entry in os.scandir('.'):
        # find files ending with .py
        if entry.is_file() and entry.name.endswith(".py") :
            print("- ", entry.name)
find_local_py_scripts()

-  googlenet_custom_layers.py
-  GoogLeNet_Inception_v1.py

Answer 4

Probably this is not best code for this task, but is working -

def front_back(a, b):

 if len(a) % 2 == 0 and len(b) % 2 == 0:
    print a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):]

 elif len(a) % 2 == 1 and len(b) % 2 == 0:
    print a[:(len(a)/2)+1] + b[:(len(b)/2)] + a[(len(a)/2)+1:] + b[(len(b)/2):] 

 elif len(a) % 2 == 0 and len(b) % 2 == 1:
     print a[:(len(a)/2)] + b[:(len(b)/2)+1] + a[(len(a)/2):] + b[(len(b)/2)+1:] 

 else :
     print a[:(len(a)/2)+1] + b[:(len(b)/2)+1] + a[(len(a)/2)+1:] + b[(len(b)/2)+1:]

Answer 5

A single & (not double &&) is enough or as the top answer suggests you can use 'and'. I also found this in pandas

cities['Is wide and has saint name'] = (cities['Population'] > 1000000) 
& cities['City name'].apply(lambda name: name.startswith('San'))

if we replace the "&" with "and", it won't work.

Answer 6

I went with a purlely mathematical solution:

def front_back(a, b):
  return a[:(len(a)+1)//2]+b[:(len(b)+1)//2]+a[(len(a)+1)//2:]+b[(len(b)+1)//2:]