Clever Geek Handbook

Finding the number of equal elements at the beginning of a list

Given a list in python, I would like to know how many equal elements are at the beginning of the list.

Input example:

x1 = ['a','a','b','c','a','a','a','c']
x2 = [1, 1, 1, 3, 1, 1, 1, 8]
x3 = ['foo','bar','foobar']

Some magic function (or one liner) will appear:

f(x1) = 2 # There are 2 'a' values in the beginning.
f(x2) = 3 # There are 3 1-values in the beginning.
f(x3) = 1 # Only 1 'foo' in beginning.

If I do this:

sum([1 if x=='a' else 0 for x in x1])

I just get the number of occurrences of "a" in x1, not the number of leading values ​​per string. It would be nice to have one liner that doesn't have to know the first meaning.

+3


source to share


3 answers


itertools.groupby

may I help...

from itertools import groupby

def f(lst):
    if_empty = ('ignored_key', ())
    k, v = next(groupby(lst), if_empty)
    return sum(1 for _ in v)

And of course we can turn that into a 1-liner (no import):

sum(1 for _ in next(groupby(lst), ('ignored', ()))[1])


But I would not recommend it.

demo:

>>> from itertools import groupby
>>> 
>>> def f(lst):
...     if_empty = ('ignored_key', ())
...     k, v = next(groupby(lst), if_empty)
...     return sum(1 for _ in v)
... 
>>> f(x1)
2
>>> f(x2)
3
>>> f(x3)
1
>>> f([])
0
+5


source


You can use takewhile

.

import itertools

xs = [1, 1, 1, 3, 1, 1, 1, 8]

sum(1 for _ in itertools.takewhile(lambda x: x == xs[0], xs))


In function:

def count_first(iterable):
    i = iter(iterable)
    first = next(i)
    return 1 + sum(1 for _ in itertools.takewhile(lambda x: x == first, i))
+3


source


It might be better to check for the first occurrence of something that is not equal to the first value:

x1 = ['a','a','b','c','a','a','a','c']
x2 = [1, 1, 1, 3, 1, 1, 1, 8]
x3 = ['foo','bar','foobar']
x4 = []
x5 = [1,1,1,1,1,1]

def f(x):
    pos = -1
    for pos,a in enumerate(x):
        if a!=x[0]:
            return pos
    return pos+1


print(f(x1))
print(f(x2))
print(f(x3))
print(f(x4))
print(f(x5))

2
3
1
0
6
+1


source






More articles:


All Articles