如何._replace() python3 namedtuple object 的多个元素的值？

Question

Below is how I create and use a namedtuple object:

>>> from collections import namedtuple
>>> labels = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']
>>> data = namedtuple('data', labels)
>>> my_data1 = data(a=0, b=1, c=2, d=3, e=4, f=5, g=6, h=7, i=8, j=9)
>>> my_data1
data(a=0, b=1, c=2, d=3, e=4, f=5, g=6, h=7, i=8, j=9)

Instead of having to manually type this multiple times:

my_data1 = data(a=0, b=1, c=2, d=3, e=4, f=5, g=6, h=7, i=8, j=9)

How do I write a for-loop or use a comprehension approach to write this? Eg if values is a list object, eg list(range(10)) or a list of any values, how do I create a new instance of data with a for-loop and different list of values?

Answer 1

You can zip() the labels and values into a tuple list, then dict() it and pass to your data() constructor as keyword arguments:

labels = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']
values = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
kwargs = dict(zip(labels, values)))
my_data1 = data(**kwargs))

Answer 2

If you want a one-liner you can do something like this:

from inspect import signature
my_data1 = data(**{i: j for i, j in zip(signature(data).parameters, range(10))})

This basically gets the parameters of the namedtuple and gets the numbers 0-9 and zips them together. With some dictionary comprehension magic I created a dictionary with the keys being the parameters of the named tuple and the values being the actual values (the numbers). At the end I unpacked the dictionary and put everything inside of the data namedtuple.