using .format() with a lot of replacement fields

Question

I have a long string with many replacement fields that I then format with:

firstRep  = replacementDict['firstRep']
secondRep = replacementDict['secondRep']
.
.
.
nthRep    = replacementDict['nthRep']

newString = oldString.format(firstRep = firstRep, 
                             secondRep = secondRep,...,
                             nthRep = nthRep)

Is there a way to avoid having to set every option individually and use a looped approach to this?

Thanks.

score 4 · Answer 1 · answered Mar 02 '14 at 15:54

4

You can unpack argument dicts with a ** prefix:

newString = oldString.format(**replacementDict)

answered Mar 02 '14 at 15:54

Ry-

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score 3 · Answer 2 · edited May 23 '17 at 12:28

3

Use str.format_map which is provided exactly for this usecase:

old_string.format_map(replacement_dict)

Note: format_map is provided only in python3.2+. In python2 you can use ** unpacking (see this related question), howerver this forces python to copy the dictionary and is thus a bit slower and uses more memory.

edited May 23 '17 at 12:28

Community

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1

answered Mar 02 '14 at 16:00

Bakuriu

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1

Wow, that exists? (Since Python 3, yes! Cool.) – Ry- Mar 02 '14 at 16:01

score 2 · Accepted Answer · answered Mar 02 '14 at 15:54

You can simply unpack the dictionary like this

replacementDict = {}
replacementDict["firstRep"]  = "1st, "
replacementDict["secondRep"] = "2nd, "
replacementDict["thirdRep"]  = "3rd, "

print "{firstRep}{secondRep}{thirdRep}".format(**replacementDict)
# 1st, 2nd, 3rd,

Quoting from the Format Examples,

Accessing arguments by name:

>>>
>>> 'Coordinates: {latitude}, {longitude}'.format(latitude='37.24N', longitude='-115.81W')
'Coordinates: 37.24N, -115.81W'
>>> coord = {'latitude': '37.24N', 'longitude': '-115.81W'}
>>> 'Coordinates: {latitude}, {longitude}'.format(**coord)
'Coordinates: 37.24N, -115.81W'

using .format() with a lot of replacement fields

3 Answers3