通过引用另一个列表中存在的次数来显示列表中的项目。

Question

I am trying to display each item in string_list. But the number of times to display each item is in display_times list. The code should display a 2 times, b 3 times and c once. This is what I have got which is not correct. What should I do next?

string_list = ['a', 'b', 'c']
display_times = ['2', '3', '1']

for item in string_list:
    for times in display_times:
        print("It is " + item)

Answer 1

Here's a solution

In [12]: for s, t in zip(string_list, display_times):
    for i in range(int(t)):
        print "It is %s" % s
   ....:         
It is a
It is a
It is b
It is b
It is b
It is c

Answer 2

I think this is a good understandable way to do it:

zipped = zip(string_list, display_times) #equal to [('a', '2'), ('b', '3'), ('c', '1')]
for value, time in zipped:
    for i in range(int(time)):
        print value

result:

a
a
b
b
b
c

Answer 3

The first problem you have that is the display_times list is storing strings and not numbers.

There are lots of shortcut ways to do this, but lets start with the easy way:

>>> for position,item in enumerate(string_list):
...     how_many = int(display_times[position])
...     print('It is {}'.format(item*how_many))
...
It is aa
It is bbb
It is c

You can use zip to combine the two lists:

>>> zip(string_list, display_times)
[('a', '2'), ('b', '3'), ('c', '1')]

Now you can do:

>>> for item,how_many in zip(string_list, display_times):
...     print('It is {}'.format(item*int(how_many)))
...
It is aa
It is bbb
It is c

There are lot of other ways but they will all involve some variation of the above two.

Answer 4

for i, v in enumerate(display_times):
    for j in range(int(v)):
        print (string_list[i])

Or using a LC:

r = [[string_list[i] for j in range(int(v))] for i,v in enumerate(display_times)]
for i in r:
   print (i)