blocks|key|622314|text|更整洁一些：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|622315|>>>+s+=+"CREDITCARDNUMBER"
>>>+s[-4:].rjust(len(s),+"*")
'************MBER'|code-block|syntax|javascript|622316|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|$]]

Neater still:

<pre><code>&gt;&gt;&gt; s = "CREDITCARDNUMBER"
&gt;&gt;&gt; s[-4:].rjust(len(s), "*")
'************MBER'
</code></pre>

blocks|key|622203|text|也许更优雅一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|622204|card+=+"CREDITCARDNUMBER"
print+"*"+*+(len(card)+-+4)+%2B+card[-4:]|code-block|syntax|javascript|622205|请注意，我避免使用名称str，因为它已经是内置字符串类型的名称。使用隐藏在内置名称后面的名称通常不是一个好主意。|offset|length|style|CODE|622206|entityMap^0|0|0|B|3|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Perhaps slightly more elegant:

<pre><code>card = "CREDITCARDNUMBER"
print "*" * (len(card) - 4) + card[-4:]
</code></pre>

Note that I've avoided using the name <code>str</code> because that is already the name of the built-in string type. It's usually not a good idea to use names that shadow the built-in names.

blocks|key|406737|text|对于那些想要保留Issuer+Identification+Number+(以前称为“银行识别码”)(通常是前6位数字)的人来说，这应该可以做到这一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|406738|print+card[:6]+%2B+'X'+*+6+%2B+card[-4:]
'455694XXXXXX6539'|code-block|syntax|javascript|406739|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Bank_card_number#Issuer_identification_number_.28IIN.29^0|8|S|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

For those who want to keep <a href="http://en.wikipedia.org/wiki/Bank_card_number#Issuer_identification_number_.28IIN.29" rel="noreferrer">Issuer Identification Number</a> (IIN) (previously called the "Bank Identification Number" (BIN)), which is usually the first 6 digits, that should do the job:

<pre><code>print card[:6] + 'X' * 6 + card[-4:]
'455694XXXXXX6539'
</code></pre>

blocks|key|622321|text|使用Format+String和Slicings|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|622322|'{:*>16}'.format(card[-4:])|code-block|syntax|javascript|622323|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/string.html#formatstrings|1|http://docs.python.org/reference/expressions.html#index-34^0|2|D|0|G|8|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]]]

With <a href="http://docs.python.org/library/string.html#formatstrings" rel="nofollow">Format String</a> and <a href="http://docs.python.org/reference/expressions.html#index-34" rel="nofollow">Slicings</a>:

<pre><code>'{:*&gt;16}'.format(card[-4:])
</code></pre>

blocks|key|622250|text|你可以把它缩短一点，就像这样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|622251|str+=+"CREDITCARDNUMBER";
print+"*"+*+(len(str)+-+4)+%2B+str[-4:];|code-block|syntax|javascript|622252|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|$]]

You could make it a little shorter like so:

<pre><code>str = "CREDITCARDNUMBER";
print "*" * (len(str) - 4) + str[-4:];
</code></pre>

blocks|key|406810|text|我想要一种灵活的方式来遮盖东西，所以我想我应该分享...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|406811|def+mask(string,+start=0,+end=0,+chunk=4,+char="*"):

++++if+isinstance(string,+str):

++++++++string_len+=+len(string)

++++++++if+string_len+<+3:
++++++++++++return+"input+len+=+{},+what's+the+point?".format(string_len)

++++++++if+start+<=+0+or+end+<=+0:
++++++++++++start+=+string_len+//+chunk
++++++++++++end+=+start

++++++++if+string_len+<=+4:
++++++++++++mask_len+=+string_len+//+2
++++++++++++start+=+1+if+mask_len+-+1+==+0+else+mask_len+-+1
++++++++++++end+=+start
++++++++else:
++++++++++++mask_len+=+string_len+-+abs(start)+-+abs(end)

++++++++return+string[:abs(start)]+%2B+(char+*+mask_len)+%2B+string[-end:]



if+__name__+==+"__main__":

++++s+=+"abcdefghijklmnopqrstuvwxyz01234567890a0b0c0d0e0f1a1b1c1d1e1f2a2b"
++++l+=+[i+for+i+in+s]
++++message+=+"ip:+{}\nop:+{}\n"
++++section+=+"-"*79

++++print("\nSTRINGS")
++++print(section)
++++print(message.format(s[:4],+mask(s[:4])))
++++print(message.format(s[:3],+mask(s[:3])))
++++print(message.format(s[:2],+mask(s[:2])))
++++print(message.format(s,+mask(s)))
++++print(message.format(s,+mask(s,start=2,end=4,char="#")))
++++print(message.format(s,+mask(s,start=3,end=3,char="x")))|code-block|syntax|javascript|406812|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|$]]

I wanted a flexible way to mask stuff, so I thought I'd share ...

<pre><code>def mask(string, start=0, end=0, chunk=4, char="*"):

 if isinstance(string, str):

 string_len = len(string)

 if string_len &lt; 3:
 return "input len = {}, what's the point?".format(string_len)

 if start &lt;= 0 or end &lt;= 0:
 start = string_len // chunk
 end = start

 if string_len &lt;= 4:
 mask_len = string_len // 2
 start = 1 if mask_len - 1 == 0 else mask_len - 1
 end = start
 else:
 mask_len = string_len - abs(start) - abs(end)

 return string[:abs(start)] + (char * mask_len) + string[-end:]



if __name__ == "__main__":

 s = "abcdefghijklmnopqrstuvwxyz01234567890a0b0c0d0e0f1a1b1c1d1e1f2a2b"
 l = [i for i in s]
 message = "ip: {}\nop: {}\n"
 section = "-"*79

 print("\nSTRINGS")
 print(section)
 print(message.format(s[:4], mask(s[:4])))
 print(message.format(s[:3], mask(s[:3])))
 print(message.format(s[:2], mask(s[:2])))
 print(message.format(s, mask(s)))
 print(message.format(s, mask(s,start=2,end=4,char="#")))
 print(message.format(s, mask(s,start=3,end=3,char="x")))
</code></pre>

blocks|key|622432|text|我认为使用一个简单的正则表达式替换和两个字符串片就可以得到一个很好的可读性解决方案，而不用担心str的len：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|622433|import+re
str+=+'CREDITCARDNUMBER'
print(re.sub('.',+'*',+str[:-4])+%2B+str[-4:])|code-block|syntax|javascript|622434|我不确定这是不是比原来的更“优雅”，但它就是这样。|622435|entityMap^0|1B|3|1F|3|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

I think you can get a decently readable solution without worrying about the <code>len</code> of the <code>str</code> using a simple regex replace and two string slices:

<pre><code>import re
str = 'CREDITCARDNUMBER'
print(re.sub('.', '*', str[:-4]) + str[-4:])
</code></pre>

I am not sure this is more "elegant" than the original, but there it is.

blocks|key|622462|text|我们也可以使用这个函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|622463|def+maskify(cc):
++++if+len(cc)+<+4:
++++++++return+cc
++++return+"#"+*+(len(cc)-4)+%2B+cc[-4:]|code-block|syntax|javascript|622464|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|$]]

We also can use this function:

<pre><code>def maskify(cc):
 if len(cc) &lt; 4:
 return cc
 return "#" * (len(cc)-4) + cc[-4:]
</code></pre>

blocks|key|622466|text|#Simple+code+to+mask+16-digit+credit+card+number

def+cardNumberHider():
++++str+=+input("Enter+the+card+number:+")
++++masked+=+len(str)+-+4
++++slimstr+=+str[masked:]
++++print("Your+masked+card+number+is:+",+end="")
++++print("*"+*+masked+%2B+slimstr)
cardNumberHider()|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|622467|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre class="lang-py prettyprint-override"><code>#Simple code to mask 16-digit credit card number

def cardNumberHider():
 str = input(&quot;Enter the card number: &quot;)
 masked = len(str) - 4
 slimstr = str[masked:]
 print(&quot;Your masked card number is: &quot;, end=&quot;&quot;)
 print(&quot;*&quot; * masked + slimstr)
cardNumberHider()
</code></pre>

So I have a csv file that contains full credit-card numbers.. We dont need the full number, and so I am writing a quick script to parse through the csv and replace the cc number with a masked representation. (all *'s except the last four). I am pretty new to python and hacked this up, and it works, but in order to learn I want to know if it could be done easier.

Assume that "str" will be a full creditcard number. But for the sake of my example I am just using the string "CREDITCARDNUMBER".

<pre><code>str = "CREDITCARDNUMBER";
strlength = len(str)
masked = strlength - 4
slimstr = str[masked:]
print "*" * masked + slimstr
</code></pre>

The output is exactly what I want

<pre><code>************MBER
</code></pre>

But I am sure there is a more elegant solution. :) Thanks!

Is there a better way to mask a Credit Card number in python?

所以我有一个包含完整信用卡号码的csv文件..我们不需要完整的数字，所以我编写了一个快速脚本来解析csv，并用掩码表示替换cc数字。(除了最后四个之外都是*)。我是python的新手，我把它破解出来了，而且它还能用，但是为了学习，我想知道它是否可以更容易地完成。假设"str“将是一个完整的信用卡号。但是对于我的例子，我只是使用字符串"CREDITCARDNUMBER“。str = "CREDITC

问在python中有没有更好的方法来屏蔽信用卡号码？
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问在python中有没有更好的方法来屏蔽信用卡号码？
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