如何从字符串中删除空格

Question

我希望我的代码要做的是"-"最终会与"0"发生冲突。但是现在它所做的就是继续追逐"0"并且永远不要碰它。

import time
global gunshot, gun
gun = 0
gunshot = "-           0"

while True:
    global gunshot
    gunshot = " " + gunshot
    gunshot.replace(' 0', '0')
    print ('\r {0}'.format(gunshot)),
    if gunshot.find("-0") == 1:
        gunshot = "-"
    time.sleep(.1)

这就是我想要的：

     -     0
      -    0
       -   0

这就是它正在做的事情

     -     0
     -     0
     -     0

Answer 1

replace返回一个新字符串，它不会改变该位置的变量。

gunshot = gunshot.replace(' 0', '0')

这将解决您的直接问题，但您应该考虑使用@MSeiferts代码，因为它要好得多。

Answer 2

您可以在此处使用str.format和str.rjust功能：

bullet = '-'
target = 'O'

distance = 3
field = 10

while distance >= 0:
    print('{}{}{}'.format(bullet, ' '*distance, target).rjust(field))
    distance -= 1

打印：

     -   O
      -  O
       - O
        -O

Answer 3

你也可以使用双端队列并只旋转值：

from collections import deque


def gunshot_path(distance):
    return deque(['-'] + ([''] * (distance - 1)))


def print_gunshot_path(distance, target="0"):
    path = gunshot_path(distance)
    for i in range(distance):
        print(" ".join(path) + target)
        path.rotate()


print_gunshot_path(5)
print_gunshot_path(10, target='X')

打印哪些：

-    0
 -   0
  -  0
   - 0
    -0
-         X
 -        X
  -       X
   -      X
    -     X
     -    X
      -   X
       -  X
        - X
         -X