Python Nested List Comprehension with If Else

Question

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.

Answer 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.

Answer 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']

Answer 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]