CODEWITHSUNDEEP
pythongenerator
ssy
ssy

Reputation: 41

how to return multiple lines with python each time next() is called?

hello i am a beginner of python, there is a simple question. i'm asked to write a generator to traverse a txt file, each row in the file is 3 coordinates of a point (x,y,z) how to return 5 points(5 lines) each time next() is called?

here is my code, i can only generate one line each time thx a lot!

import itertools

def gen_points(name):
    f=open(name)
    for l in f:
            clean_string=l.rstrip()
            x,y,z=clean_string.split()
            x= float(x)
            y= float(y)
            z= float(z)
            yield x,y,z
    f.close()

file_name=r"D:\master ppt\Spatial Analysis\data\22.txt"
a=gen_points(file_name)
g=itertools.cycle(a)

print(next(g))

Upvotes: 4

Views: 148

Answers (2)

Reid Ballard
Reid Ballard

Reputation: 1525

You don't have to yield immediately, so hold onto the output and then yield it later:

## Adding batchsize so you can change it on the fly if you need to
def gen_points(name, batchsize = 5):
    ## The "with" statement is better practice
    with open(name,'r') as f:
        ## A container to hold the output
        output = list()
        for l in f:
            clean_string=l.rstrip()
            x,y,z=clean_string.split()
            x= float(x)
            y= float(y)
            z= float(z)
            ## Store instead of yielding
            output.append([x,y,z])
            ## Check if we have enough lines
            if len(output) >= batchsize:
                yield output
                ## Clear list for next batch
                output = list()

        ## In case you have an "odd" number of points (i.e.- 23 points in this situation)
        if output: yield output

Upvotes: 1

Jean-François Fabre
Jean-François Fabre

Reputation: 140276

just wait until you have five items in your list of triplets, and yield that instead:

def gen_points(name):
    with open(name) as f:
        five_items = []
        for l in f:
            five_items.append(tuple(map(float,l.split())))
            if len(five_items) == 5:
                yield five_items
                # create another empty object
                five_items = []
        if five_items:
            yield five_items

also yield at the end of the loop if not empty to avoid losing the last elements if the number of elements isn't dividable by 5.

Aside: clean_string=l.rstrip() was useless since split already takes care of linefeeds and whatnot.

Upvotes: 5

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