How to select lines and print in a list python

I have a list that looks like this:

1 f
2 f
3 f
4 g
5 g
6 g


and I want to print out the first two lines of f and g, as:

f

g

1 f
2 f
4 g
5 g


any idea how to approach this?

2 Answers
2

You could to consider grouping your list. An example:

l = [['1', 'f'], ['2', 'f'], ['3', 'f'], ['4', 'g'], ['5', 'g'], ['6', 'g']]


To group the list, use a key. In this case, the key would be the second value of each list item, so x[1]:

x[1]

from itertools import groupby
for _, g in groupby(l, lambda x: x[1]):
print(list(g)[:2])


Slice each group, to only print the first two items, list(g)[:2].

list(g)[:2]

Output:

[['1', 'f'], ['2', 'f']]
[['4', 'g'], ['5', 'g']]


Note: this is similar to uniq in the shell. And it depends or the key values being sorted. You might have to sort the data before using this approach.

uniq

Have added a list of dict, with elem to print along with a count key
Simply increment the value of count to make sure no more than max_limit elements are print.

myList = ['1 f', '2 f', '3 f', '4 g', '5 g', '6 g']
elems_to_print = ['elem': 'f', 'count': 0, 'elem': 'g', 'count': 0]
max_limit = 2

for myL in myList:
for elem in elems_to_print:
if elem['elem'] in myL and elem['count'] < max_limit:
#Elem found
elem['count'] += 1
print("%s"%(myL))


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