ValueError：list.remove（x）：x不在列表中

Question

What is wrong in my code, but I can get my expected result.

I am trying to remove all "#" in the list.

funds_U is the list of data:

In [3]: funds_U
Out[3]: 
[u'#',
 u'#',
 u'MMFU_U',
 u'#',
 u'#',
 u'AAI_U',
 u'TGI_U',
 u'JAS_U',
 u'TAG_U',
 u'#',
 u'#',
 u'AAT_U',
 u'BGR_U',
 u'BNE_U',
 u'IGE_U',
 u'#',
 u'#',
 u'DGF_U',
 u'BHC_U',
 u'FCF_U',
 u'SHK_U',
 u'VCF_U',
 u'#',
 u'JEM_U',
 u'SBR_U',
 u'TEM_U',
 u'#',
 u'#',
 u'BAB_U',
 u'BGA_U',
 u'#']

The following is the code:

In [4]: for fund_U in funds_U[:]:
   ...:     funds_U.remove(u"#")
   ...:     

The following is the error:

ValueError                                Traceback (most recent call last)
<ipython-input-4-9aaa02e32e76> in <module>()
      1 for fund_U in funds_U[:]:
----> 2     funds_U.remove(u"#")
      3 

ValueError: list.remove(x): x not in list

Answer 1

As per the documentation , if the item doesn't exist in the list, remove() will throw an error. Right now your code iterates through every item in the list and tries to remove that many # s. Since not every item is a # , remove() will throw an error as the list runs out of # s.

Try a list comprehension like this:

funds_U = [x for x in funds_U if x != u'#']

This will make a new list that consists of every element in funds_U that is not u'#' .

Answer 2

I would do this so:

new = [item for item in funds_U if item!=u'#']

This is a list-comprehension . It goes through every item in funds_U and adds it to the new list if it's not u'#' .

Answer 3

这将修改原始对象，因此如果有其他变量指向同一对象，则其链接将保持不变。

FUNDS_U[:] = [x for x in FUNDS_U if x != "#"]