如何打印(加入)2个循环的结果

时间:2018-03-04 07:29:22

标签: python python-3.x for-loop

我有2个循环:

第一个循环:

number = 1234567
number_string = str(number)
for ch in number_string:
    print(ch)

这将打印:

1 
2 
3 
4 
5 
6 
7

我有另一个循环:

totalNum = len(str(abs(number)))
for i in range(totalNum, 0, -1):
  print("* 10^ ",int(i-1))

我会得到

* 10^  6
* 10^  5
* 10^  4
* 10^  3
* 10^  2
* 10^  1
* 10^  0

但是我如何加入这两个循环才能成为这样的结果:

1 * 10^  6
2 * 10^  5
3 * 10^  4
4 * 10^  3
5 * 10^  2
6 * 10^  1
7 * 10^  0

我是Python的新手,所以我无法弄清楚如何做到这一点。

4 个答案:

答案 0 :(得分:4)

这对您有何帮助:

number = 1234567
totalNum = len(str(abs(number)))
for i in range(totalNum, 0, -1):
  print(str(number)[::-1][i-1] + " * 10^ ",int(i-1))

输出:

1 * 10^  6
2 * 10^  5
3 * 10^  4
4 * 10^  3
5 * 10^  2
6 * 10^  1
7 * 10^  0

答案 1 :(得分:4)

有几种方法,我使用enumerate()索引字符串:

number = 1234567
number_string = str(number)
totalNum = len(str(abs(number)))

for j,i in enumerate(range(totalNum, 0, -1)):
  print(number_string[j], "* 10^ ", int(i-1))

给出:

1 * 10^  6
2 * 10^  5
3 * 10^  4
4 * 10^  3
5 * 10^  2
6 * 10^  1
7 * 10^  0

enumerate()遍历一个序列,并给出索引号(j)和项目(i)的元组。

答案 2 :(得分:2)

一种方法是先将字符串存储在列表中,然后打印出来:

>>> number = 1234567
>>> totalNum = len(str(abs(number)))
>>> string1 = [x for x in str(number)]
>>> string2 = ['* 10^ {}'.format(i-1) for i in range(totalNum, 0, -1)]
>>> for s1, s2 in zip(string1, string2):
...     print(s1, s2)
...
1 * 10^ 6
2 * 10^ 5
3 * 10^ 4
4 * 10^ 3
5 * 10^ 2
6 * 10^ 1
7 * 10^ 0

答案 3 :(得分:1)

这可能会有所帮助。您可以使用my_named_pipe="$HOME/foobar" mkfifo "$my_named_pipe" while read line; do on_fifo_msg "$line"; done < ${my_named_pipe} &

zip

<强>输出:

number = 1234567
totalNum = len(str(abs(number)))
for v, i in zip(str(number), range(totalNum, 0, -1)):
  print(v, "* 10^ ",int(i-1))