Python Nested List Comprehension with If Else

问题

I was trying to use a list comprehension to replace multiple possible string values in a list of values.

I have a list of column names which are taken from a cursor.description;

['UNIX_Time', 'col1_MCA', 'col2_MCA', 'col3_MCA', 'col1_MCB', 'col2_MCB', 'col3_MCB']

I then have header_replace;

{'MCB': 'SourceA', 'MCA': 'SourceB'}

I would like to replace the string values for header_replace.keys() found within the column names with the values.

I have had to use the following loop;

headers = []
for header in cursor.description:
    replaced = False
    for key in header_replace.keys():
        if key in header[0]:
            headers.append(str.replace(header[0], key, header_replace[key]))
            replaced = True
            break

    if not replaced:
        headers.append(header[0])

Which gives me the correct output;

['UNIX_Time', 'col1_SourceA', 'col2_SourceA', 'col3_SourceA', 'col1_SourceB', 'col2_SourceB', 'col3_SourceB']

I tried using this list comprehension;

[str.replace(i[0],k,header_replace[k]) if k in i[0] else i[0] for k in header_replace.keys() for i in cursor.description]

But it meant that items were duplicated for the unmatched keys and I would get;

['UNIX_Time', 'col1_MCA', 'col2_MCA', 'col3_MCA', 'col1_SourceA', 'col2_SourceA', 'col3_SourceA', 
'UNIX_Time', 'col1_SourceB', 'col2_SourceB', 'col3_SourceB', 'col1_MCB', 'col2_MCB', 'col3_MCB']

But if instead I use;

[str.replace(i[0],k,header_replace[k]) for k in header_replace.keys() for i in cursor.description if k in i[0]]

@Bakuriu fixed syntax

I would get the correct replacement but then loose any items that didn't need to have an string replacement.

['col1_SourceA', 'col2_SourceA', 'col3_SourceA', 'col1_SourceB', 'col2_SourceB', 'col3_SourceB']

Is there a pythonesque way of doing this or am I over stretching list comprehensions? I certainly find them hard to read.

回答1:

[str.replace(i[0],k,header_replace[k]) if k in i[0] for k in header_replace.keys() for i in cursor.description]

this is a SyntaxError, because if expressions must contain the else part. You probably meant:

[i[0].replace(k, header_replace[k]) for k in header_replace for i in cursor.description if k in i[0]]

With the if at the end. However I must say that list-comprehension with nested loops aren't usually the way to go. I would use the expanded for loop. In fact I'd improve it removing the replaced flag:

headers = []
for header in cursor.description:
    for key, repl in header_replace.items():
        if key in header[0]:
            headers.append(header[0].replace(key, repl))
            break
    else:
        headers.append(header[0])

The else of the for loop is executed when no break is triggered during the iterations.

I don't understand why in your code you use str.replace(string, substring, replacement) instead of string.replace(substring, replacement). Strings have instance methods, so you them as such and not as if they were static methods of the class.

回答2:

If your data is exactly as you described it, you don't need nested replacements and can boil it down to this line:

l = ['UNIX_Time', 'col1_MCA', 'col2_MCA', 'col3_MCA', 'col1_MCB', 'col2_MCB', 'col3_MCB']
[i.replace('_MC', '_Source')  for i in l]

>>> ['UNIX_Time',
>>>  'col1_SourceA',
>>>  'col2_SourceA',
>>>  'col3_SourceA',
>>>  'col1_SourceB',
>>>  'col2_SourceB',
>>>  'col3_SourceB']

回答3:

I guess a function will be more readable:

def repl(key):
    for k, v in header_replace.items():
        if k in key:
            return key.replace(k, v)
    return key

print map(repl, names)

Another (less readable) option:

import re
rx = '|'.join(header_replace)
print [re.sub(rx, lambda m: header_replace[m.group(0)], name) for name in names]

来源：https://stackoverflow.com/questions/21850508/python-nested-list-comprehension-with-if-else

标签

python

list-comprehension