我想做几个语句,给出标准输出,而不看到换行之间的语句。
具体来说,假设我有:
for item in range(1,100):
print item
结果是:
1
2
3
4
.
.
.
如何让它看起来像:
1 2 3 4 5 ...
更好的是,是否可以将单个数字打印在最后一个数字之上,这样一次只有一个数字出现在屏幕上?
我想做几个语句,给出标准输出,而不看到换行之间的语句。
具体来说,假设我有:
for item in range(1,100):
print item
结果是:
1
2
3
4
.
.
.
如何让它看起来像:
1 2 3 4 5 ...
更好的是,是否可以将单个数字打印在最后一个数字之上,这样一次只有一个数字出现在屏幕上?
当前回答
在Python 3中,你可以这样做:
for item in range(1,10):
print(item, end =" ")
输出:
1 2 3 4 5 6 7 8 9
Tuple:你可以对Tuple做同样的事情:
tup = (1,2,3,4,5)
for n in tup:
print(n, end = " - ")
输出:
1 - 2 - 3 - 4 - 5 -
另一个例子:
list_of_tuples = [(1,2),('A','B'), (3,4), ('Cat', 'Dog')]
for item in list_of_tuples:
print(item)
输出:
(1, 2)
('A', 'B')
(3, 4)
('Cat', 'Dog')
你甚至可以像这样解包你的元组:
list_of_tuples = [(1,2),('A','B'), (3,4), ('Cat', 'Dog')]
# Tuple unpacking so that you can deal with elements inside of the tuple individually
for (item1, item2) in list_of_tuples:
print(item1, item2)
输出:
1 2
A B
3 4
Cat Dog
另一个变化:
list_of_tuples = [(1,2),('A','B'), (3,4), ('Cat', 'Dog')]
for (item1, item2) in list_of_tuples:
print(item1)
print(item2)
print('\n')
输出:
1
2
A
B
3
4
Cat
Dog
其他回答
我认为一个简单的连接应该工作:
nl = []
for x in range(1,10):nl.append(str(x))
print ' '.join(nl)
将打印项更改为:
在Python 2.7中,打印项 print(item, end=" "
如果你想动态打印数据,请使用以下语法:
打印(item, sep=' ', end= ", flush=True
In [9]: print?
Type: builtin_function_or_method
Base Class: <type 'builtin_function_or_method'>
String Form: <built-in function print>
Namespace: Python builtin
Docstring:
print(value, ..., sep=' ', end='\n', file=sys.stdout)
Prints the values to a stream, or to sys.stdout by default.
Optional keyword arguments:
file: a file-like object (stream); defaults to the current sys.stdout.
sep: string inserted between values, default a space.
end: string appended after the last value, default a newline.
对于那些像我一样挣扎的人,我提出了以下似乎在python 3.7.4和3.5.2中都可以工作的方法。
I expanded the range from 100 to 1,000,000 because it runs very fast and you may not see the output. This is because one side effect of setting end='\r' is that the final loop iteration clears all of the output. A longer number was needed to demonstrate that it works. This result may not be desirable in all cases, but was fine in mine, and OP didn't specify one way or another. You could potentially circumvent this with an if statement that evaluates the length of the array being iterated over, etc. The key to get it working in my case was to couple the brackets "{}" with .format(). Otherwise, it didn't work.
以下应按原样工作:
#!/usr/bin/env python3
for item in range(1,1000000):
print("{}".format(item), end='\r', flush=True)
注意:我之所以指出这个解决方案,是因为如果下一次打印的长度小于前一次打印的长度,我所见过的大多数其他解决方案都不起作用。
如果您知道要删除什么,并且可以使用全局变量,那么只需用空格覆盖最后一行。
在打印之前,将字符串的长度存储为' n '。 打印它,但以' \r '结尾(它返回行首)。 下次,在打印信息之前,在该行上打印“n”个空格。
_last_print_len = 0
def reprint(msg, finish=False):
global _last_print_len
# Ovewrites line with spaces.
print(' '*_last_print_len, end='\r')
if finish:
end = '\n'
# If we're finishing the line, we won't need to overwrite it in the next print.
_last_print_len = 0
else:
end = '\r'
# Store len for the next print.
_last_print_len = len(msg)
# Printing message.
print(msg, end=end)
例子:
for i in range(10):
reprint('Loading.')
time.sleep(1)
reprint('Loading..')
time.sleep(1)
reprint('Loading...')
time.sleep(1)
for i in range(10):
reprint('Loading.')
time.sleep(1)
reprint('Loading..')
time.sleep(1)
reprint('Loading...', finish=True)
time.sleep(1)