How can ı replace the number to white space in python

Question

I have a encrypt decrypt work in phyton, first ı am changed the whitespaces to random numbers. In order to do this job, I converted the string into a list with this code a = [x for x in text] . ı replaced the whitespaces using this code :

    for i in range(lenghtOfText):
        if ( a[i]==" "):
            a[i]=str((random.randint(0, 9)))

now ı need one more time change the numbers with whitespaces. ı tried this but itsn't work

    for i in range(lenghtOfText):
        if (a[i] in range(0,9)):
            a[i]=" "

I tried one more way for the do this. but it's still not work. the other code is

    for i in range(lenghtOfText):
        if (a[i] == 0 or a[i] == 1 or a[i] == 2 or a[i] == 3 or a[i] == 4  or a[i] == 5 or a[i] == 6 or a[i] == 7 or a[i] == 8 or a[i] == 9 ):
            a[i]=" "

Answer 1

That's because the character '1' (ascii 49) is not the same as the number 1.

Try this:

for i in range(lenghtOfText):
    try:
        if (int(a[i]) in range(0,10)):
            a[i]=" "
    except:
        pass

Note: Remember range's end is EXCLUSIVE.

Slightly better way:

digits = [str(i) for i in range(10)]
for i in range(lenghtOfText):
    if a[i] in digits:
        a[i] = " "

Even better:

for i in range(lenghtOfText):
    if a[i].isdigit():
        a[i] = " "

One more, this time using map:

a = list(map(lambda e: " " if e.isdigit() else e, a))

The only caveat is that this will create a copy of the array, leaving the original one unmodified.

Answer 2

You seem to save the whitespaces as strings, but query them as integer. Try a[i] = "0" and such.

Answer 3

It doesn't work because you try to compare string with int values.

a[i] is string type range(0,9) is list with int values.

Try this:

    for i in range(lenghtOfText):
       if (int(a[i]) in range(0,9)):
           a[i]=" "