blocks|key|4116393|text|只需再次追加语句即可。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4116394|array+=+[]
add_x+=+input('enter+no.+of+x\'s:+')
add_x+=+int(add_x)
for+i+in+range(0,add_x,1):
+++++array.append('X')
+++++array.append("+")


#+Multiply+the+space+by+the+number+of+spaces+you+want+to+append+each+time.
#+array.append("+"*1)|code-block|syntax|javascript|4116395|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Just you can append the statement once again.

<pre><code>array = []
add_x = input('enter no. of x\'s: ')
add_x = int(add_x)
for i in range(0,add_x,1):
 array.append('X')
 array.append(" ")


# Multiply the space by the number of spaces you want to append each time.
# array.append(" "*1)
</code></pre>

blocks|key|4116412|text|由于您特别需要带空格的奇数位置，因此可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4116413|def+two_numbers():

++++array+=+[]
++++add_x+=+input('enter+no.+of+x\'s:+')
++++add_x+=+int(add_x)
++++for+i+in+range(0,add_x,1):
++++++++#+Add+space+if+array+length+is+even.
++++++++if+len(array)+%25+2+==+1:
++++++++++++array.append('+')
++++++++array.append('X')

two_numbers()

#+['X',+'+',+'X',+'+',+'X',+'+',+'X',+'+',+'X',+'+',+'X',+'+',+'X',+'+',+'X',+'+',+'X',+'+',+'X']|code-block|syntax|javascript|4116414|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Since you specifically want odd numbered positions with a space, you could do this:

<pre><code>def two_numbers():

 array = []
 add_x = input('enter no. of x\'s: ')
 add_x = int(add_x)
 for i in range(0,add_x,1):
 # Add space if array length is even.
 if len(array) % 2 == 1:
 array.append(' ')
 array.append('X')

two_numbers()

# ['X', ' ', 'X', ' ', 'X', ' ', 'X', ' ', 'X', ' ', 'X', ' ', 'X', ' ', 'X', ' ', 'X', ' ', 'X']
</code></pre>

blocks|key|4116416|text|这里有一行代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4116417|num_x+=+4
output+=+list('+'.join('X'+*+num_x))

print(output)
>>>+['X',+'+',+'X',+'+',+'X',+'+',+'X']|code-block|syntax|javascript|4116418|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is a one liner:

<pre class="lang-py prettyprint-override"><code>num_x = 4
output = list(' '.join('X' * num_x))

print(output)
&gt;&gt;&gt; ['X', ' ', 'X', ' ', 'X', ' ', 'X']

</code></pre>

blocks|key|5584784|text|您可以使用str.join|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|5584785|add_x+=+input('enter+no.+of+x\'s:+')
#+input+3
array+=+list('+'.join('X'+*+add_x)
print(array)|code-block|syntax|javascript|5584786|输出：|5584787|['X',+'+',+'X',+'+',+'X',+'+',+'X']|5584788|单行解决方案：|5584789|array+=+list('+'.join('X'+*+int(input('enter+no.+of+x\'s:+'))))|5584790|或者：|5584791|array+=+(['X',+'']+*+int(input('enter+no.+of+x\'s:+')))[:-1]|5584792|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/stdtypes.html#str.join^0|5|8|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|Q|3|R|5|6|7|1B|8|@]|9|@]|C|$]]|$1|S|3|T|5|F|7|1C|8|@]|9|@]|C|$G|H]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]]]

you can use <a href="https://docs.python.org/3/library/stdtypes.html#str.join" rel="nofollow noreferrer">str.join</a>:

<pre><code>add_x = input('enter no. of x\'s: ')
# input 3
array = list(' '.join('X' * add_x)
print(array)
</code></pre>

output:

<pre><code>['X', ' ', 'X', ' ', 'X', ' ', 'X']
</code></pre>

one-line solution:

<pre><code>array = list(' '.join('X' * int(input('enter no. of x\'s: '))))
</code></pre>

or: 

<pre><code>array = (['X', ''] * int(input('enter no. of x\'s: ')))[:-1]
</code></pre>

blocks|key|1501535|text|只是为了在已有的解决方案基础上添加其他解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1501536|import+itertools

def+two_numbers(n):
++++tmp+=+zip(['X'+for+_+in+range(n)],+['+'+for+_+in+range(n)+])
++++return+list(itertools.chain(*tmp))|code-block|syntax|javascript|1501537|或|1501538|def+two_numbers(n):
++++array+=+[]
++++for+_+in+range(n):
++++++++array+%2B=+['X',+'+']
++++return+array|1501539|返回的列表将以'+'终止，如果您愿意，可以通过调用列表的pop方法或通过返回切片(例如return+array[:-1]+)来删除它。|offset|length|style|CODE|1501540|entityMap^0|0|0|0|0|7|3|S|3|17|H|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]|$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

just to add other solutions to the ones already here.

<pre><code>import itertools

def two_numbers(n):
 tmp = zip(['X' for _ in range(n)], [' ' for _ in range(n) ])
 return list(itertools.chain(*tmp))
</code></pre>

or

<pre><code>def two_numbers(n):
 array = []
 for _ in range(n):
 array += ['X', ' ']
 return array
</code></pre>

The returned lists will terminate with a <code>' '</code>, if you want to you can remove it by calling the <code>pop</code> method of the list or by returning a slice (e.g. <code>return array[:-1]</code> )

blocks|key|4116551|text|解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4116552|下面是您想要做的事情的单行解决方案。|offset|length|style|BOLD|4116553|n+=+8+#+number+of+X's
(['X',+'+']*n)[:-1]|code-block|syntax|javascript|4116554|输出：|4116555|['X',+'+',+'X',+'+',+'X',+'+',+'X',+'+',+'X',+'+',+'X',+'+',+'X',+'+',+'X']|4116556|等效的其他方法|4116557|#+Method-1
nrep+=+5
','.join(['X,+']*nrep)[:-2)].split(',')+

#+Method-1+wrapped+as+a+convenience+function
def+patterned_array(nrep:int+=+5,+sep+=+'+',+base_string+=+'X')+->+list:
++++spattern+=+''.join([base_string,+sep])
++++return+','.join(spattern*n)[:-len(spattern)].split(',')

#+generate+patterned+array+(a+list)
patterned_array(nrep+=+8)|4116558|在现有列表的元素之间插入空格字符|4116559|如果你已经有了一个项目列表，但它们不一定都是一样的，那该怎么办？您可以在列表中的这些元素之间插入所需的字符，如下所示。|4116560|array+=+['a',+'b',+'c',+'d',+'e']
bigarray+=+['+']*(len(array)-1)
for+i,+e+in+enumerate(array):+bigarray.insert(2*i,+e)
bigarray|4116561|4116562|['a',+'+',+'b',+'+',+'c',+'+',+'d',+'+',+'e']|4116563|entityMap^0|0|B|6|0|0|0|2|0|0|0|0|0|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|16|8|@$D|17|E|18|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|19|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1A|8|@$D|1B|E|1C|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|1D|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1E|8|@]|9|@]|A|$]]|$1|S|3|T|5|J|7|1F|8|@]|9|@]|A|$K|L]]|$1|U|3|V|5|6|7|1G|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1H|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|J|7|1I|8|@]|9|@]|A|$K|L]]|$1|10|3|N|5|6|7|1J|8|@$D|1K|E|1L|F|G]]|9|@]|A|$]]|$1|11|3|12|5|J|7|1M|8|@]|9|@]|A|$K|L]]|$1|13|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|14|$]]

<h1>Solution</h1>

Here is a one-line solution to what you want to do.

<pre class="lang-py prettyprint-override"><code>n = 8 # number of X's
(['X', ' ']*n)[:-1]
</code></pre>

Output: 

<pre><code>['X', ' ', 'X', ' ', 'X', ' ', 'X', ' ', 'X', ' ', 'X', ' ', 'X', ' ', 'X']
</code></pre>

<h2>Equivalent Other Method(s)</h2>

<pre class="lang-py prettyprint-override"><code># Method-1
nrep = 5
','.join(['X, ']*nrep)[:-2)].split(',') 

# Method-1 wrapped as a convenience function
def patterned_array(nrep:int = 5, sep = ' ', base_string = 'X') -&gt; list:
 spattern = ''.join([base_string, sep])
 return ','.join(spattern*n)[:-len(spattern)].split(',')

# generate patterned array (a list)
patterned_array(nrep = 8)
</code></pre>

<h1>Insert a space-character in between elements of an existing list</h1>

What if you already have a list of items and they are not necessarily all the same? You could insert a desired character in between those elements in the list as follows.

<pre class="lang-py prettyprint-override"><code>array = ['a', 'b', 'c', 'd', 'e']
bigarray = [' ']*(len(array)-1)
for i, e in enumerate(array): bigarray.insert(2*i, e)
bigarray
</code></pre>

Output: 

<pre><code>['a', ' ', 'b', ' ', 'c', ' ', 'd', ' ', 'e']
</code></pre>

Essentially, I am inputting a certain number of Xs in an array. However, after that, I want to add a space after each X so that it goes from ['X','X','X','X'] to [X,' ',X,' ',X,' ',X] subject to how many spaces a user enters. 

Any help would be appreciated, please.

<pre class="lang-py prettyprint-override"><code>def two_numbers():

 array = []
 add_x = input('enter no. of x\'s: ')
 add_x = int(add_x)
 for i in range(0,add_x,1):
 array.append('X')


two_numbers()
</code></pre>

Python-How do I add a space for each odd numbered position in an array?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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本质上，我是在一个数组中输入一定数量的X。然而，在那之后，我想在每个X后面添加一个空格，以便它从'X'，'X‘到X，’'，X取决于用户输入了多少空格。如果有任何帮助，我将不胜感激。def two_numbers():    array = []    add_x = input('enter no. of x\'s: ')    add_x = int(add_x)    for i in ran

问Python-如何为数组中每个奇数编号的位置添加空格？
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回答 6

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