From left to right, applying list operations in python3

Question

Is there any possible way to achieve nonsense treatment from left to right for operations in list in python?

eg. scala

 val a = ((1 to 50)
  .map(_ * 4)
  .filter( _ <= 170)
  .filter(_.toString.length == 2)
  .filter (_ % 20 == 0)
  .zipWithIndex
  .map{ case(x,n) => s"Result[$n]=$x"}
  .mkString("  .. "))

  a: String = Result[0]=20  .. Result[1]=40  .. Result[2]=60  .. Result[3]=80

As long as I understand that many people do not prefer the above syntax, I like the ability to move from left to right and add arbitrary operations along the way.

Understanding python is fornot so easy to read when there are three or more operations. It seems the result is that we should break everything into pieces.

[f(a) for a in g(b) for b in h(c) for ..]

Is there any chance for the mentioned approach?

Note. I tried several libraries, including toolz.functoolz. This is a difficult assessment python3lazy: each level returns an object map. In addition, it is not obvious that it can work at the entrance list.

Answer 1

@JohanL , , Python.

, , Pipe https://github.com/JulienPalard/Pipe .

: . . , . - , . @Pipe , @Pipe , .

, :

rawLabels="""Country: Name of country
Agr: Percentage employed in agriculture
Min: Percentage employed in mining
Man: Percentage employed in manufacturing
PS: Percentage employed in power supply industries
Con: Percentage employed in construction
SI: Percentage employed in service industries
Fin: Percentage employed in finance
SPS: Percentage employed in social and personal services
TC: Percentage employed in transport and communications"""

:

mylabs = "Country Agriculture Mining Manufacturing Power Construction Service Finance Social Transport"

( ):

@Pipe
def split(iterable, delim= ' '):
    for s in iterable: yield s.split(delim)

@Pipe
def trim(iterable):
    for s in iterable: yield s.strip()

@Pipe
def pzip(iterable,coll):
    for s in zip(list(iterable),coll): yield s

@Pipe
def slice(iterable, dim):
  if len(dim)==1:
    for x in iterable:
      yield x[dim[0]]
  elif len(dim)==2:
    for x in iterable:
      for y in x[dim[0]]:
        yield y[dim[1]]

@Pipe
def toMap(iterable):
  return dict(list(iterable))

:

labels = (rawLabels.split('\n') 
     | trim 
     | split(':')
     | slice([0])
     | pzip(mylabs.split(' '))
     | toMap )

:

print('labels=%s' % repr(labels))

labels={'PS': 'Power', 'Min': 'Mining', 'Country': 'Country', 'SPS': 'Social', 'TC': 'Transport', 'SI': 'Service', 'Con': 'Construction', 'Fin': 'Finance', 'Agr': 'Agriculture', 'Man': 'Manufacturing'}

2 2018

https://github.com/dwt/fluent

3 ( 2019 .) : matthagy/scalaps https://github.com/matthagy/scalaps ( , ):

from scalaps import *
a = (Seq(range(1,51))
     .map(lambda x: x * 4)
     .filter(lambda x: x <= 170)
     .filter(lambda x: len(str(x)) == 2)
     .filter( lambda x: x % 20 ==0)
     .enumerate()
     .map(lambda x: 'Result[%d]=%s' %(x[0],x[1]))
     .mkstring(' .. '))
print(a)

Result[0]=20 .. Result[1]=40 .. Result[2]=60 .. Result[3]=80

Answer 2

Pythonic, Python - map filter reduce functools. , , scala, ( , ):

from functools import reduce
a = reduce(lambda f,s: f'{f} .. {s}',
    map(lambda nx: f'Result[{nx[0]}]: {nx[1]}',
    enumerate(
    filter(lambda n: n%20 == 0,
    filter(lambda n: len(str(n)) == 2,
    filter(lambda n: n <= 170,
    map(lambda n: n*4,
    range(1,51))))))))

, , , . , reduce, .

map filter:

a = reduce(lambda f,s: f'{f} .. {s}',
    list(map(lambda nx: f'Result[{nx[0]}]: {nx[1]}',
    list(enumerate(
    list(filter(lambda n: n%20 == 0,
    list(filter(lambda n: len(str(n)) == 2,
    list(filter(lambda n: n <= 170,
    list(map(lambda n: n*4,
    list(range(1,51)))))))))))))))

, , , , . / :

n4 = [n*4 for n in range(1,51)]
fn4 = [n for n in n4 if n <= 170 if len(str(n))==2 if n%20 == 0]
rfn4 = [f'Result[{n}]: {x}' for n, x in enumerate(fn4)]
a = ' .. '.join(rfn4)

( , ) , , scala. , ( ), . , :

n4 = (n*4 for n in range(1,51))
fn4 = (n for n in n4 if n <= 170 if len(str(n))==2 if n%20 == 0)
rfn4 = (f'Result[{n}]: {x}' for n, x in enumerate(fn4))
a = ' .. '.join(rfn4)

, , . , ; , .

Answer 3

SSPipe.

, , , map, filter, str, len, enumerate, str.format, str.join p px , API. , , p px:

from sspipe import p, px
a = (
    range(1, 50+1)
    | p(map, px * 4)
    | p(filter, px <= 170)
    | p(filter, p(str) | p(len) | (px == 2))
    | p(filter, px % 20 == 0)
    | p(enumerate)
    | p(map, p('Result[{0[0]}]={0[1]}'.format)) 
    | p('  ..  '.join)
)
print(a)