I've got a question about arrays in Python. So I have to replace chars which are neighbors to each other INSIDE of array. For example (let's say input will be Mike ):
stack = []
word = input("Type Your word here: ")
wordChars = list(word)
for i in range(len(wordChars)):
stack.append(wordChars[i])
print(stack)
result: ['M', 'i', 'k', 'e']
So, when I have even number of chars - I need to replace neighbors, so:
'M' with 'i' and so on: 'iMek'.
The same thing with odd number of chars but last number is staying without replacing, so for 'Lover': 'L' with 'o' and so on; then 'r' is staying at the end: 'oLevr'.
I know how to find if word is even or odd with modulo but still can't figure out how to replace neighbors.
a = ['L', 'o', 'v', 'e', 'r']
for i in range(0,len(a)-1,2):
a[i], a[i+1] = a[i+1], a[i]
print(a)
You can write a function that replaces 2 indexes in an array:
def replace(arr, i, j):
tmp = arr[i]
arr[i] = arr[j]
arr[j] = tmp
Now you can go over the list and replace every 2 neighbors:
for i in range(0, len(wordChars) - 1, 2):
replace(wordChars, i, i + 1)
As mentioned by DarrylG, in python you can swap 2 elements by doing the following:
arr[i], arr[j] = arr[j], arr[i]
Thus, making the code simpler:
for i in range(0, len(wordChars) - 1, 2):
arr[i], arr[j] = arr[j], arr[i]
Alternative approach with list.pop(0)
def flip(a):
#convert input string to list
lst = list(a)
#initialise output string
out = ''
# if more than 1 element is left, remove the first two elements and invert
while len(lst)>1:
a = lst.pop(0)
b = lst.pop(0)
out = out + b + a
# if list has at least one element, add this one to the output string
if lst:
out = out + lst.pop()
return out
Without giving it too much thought, loop through (based on modulo as you say) and do something like this:
for x in range(0, len(wordChars) % 2, 2):
char_a = stack[i]
char_b = stack[i + 1]
stack[i] = char_b
stack[i + 1] = char_a
This will basically advance through the list in increments of 2 where each increment therefore swaps items 0 and 1, 2 and 3, 4 and 5 etc etc.
You can take two letters at a time and iterate like the following:
length = len(stack)
for i in range(0,length,2):
if(i+1 >= length ):
break
temp = stack[i]
stack[i] = stack[i+1]
stack[i+1] = temp
Use this Code to do it -
def is_even(stack):
for i in range(0, len(stack), 2):
temp = stack[i]
stack[i] = stack[i+1]
stack[i+1] = temp
return stack
def is_odd(stack):
for i in range(0, len(stack - 1), 2):
temp = stack[i]
stack[i] = stack[i+1]
stack[i+1] = temp
return stack
stack = []
word = input("Type Your word here: ")
wordChars = list(word)
for i in range(len(wordChars)):
stack.append(wordChars[i])
print(stack)
if len(stack) % 2 == 0:
new_stack = is_even(stack)
else:
new_stack = is_odd(stack)
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