如何从多个块一直读取两个连续块的数据直到文件结束?

问题描述 投票:1回答:1

如果你能想到一个好的,请更新标题!

我有以下结构的数据:

chr    pos    A_block    A_val
  2     05       7       A,T,C
  2     11       7       T,C,G
  2     15       7       AT,C,G
  2     21       7       C,A,GT
  2     31       7       T,C,CA
  2     42       9       T,C,G
  2     55       9       C,G,GC
  2     61       9       A,GC,T
  2     05       12       AC,TG,G
  2     11       12       A,TC,TG

预期输出:为了学习,我只想重写输出文件,与输入文件相同,但使用我在下面提出的过程。

我想:step 01:一次仅读取两个连续块的值(前7和9) - > step 02:将该数据存储在字典中,其中block numbers作为主要唯一键 - > step 03:将该字典返回到预定义函数进行解析。 - >现在,读取块(9和12) - >重复相同的过程直到结束。

我想的是:

import req_packages
from collections import defaultdict

''' make a function that takes data from two blocks at a time '''
def parse_two_blocks(someData):
    for key, vals in someData:
        do ... something 
        write the obtained output
        clear memory  # to prevent memory buildup


''' Now, read the input file'''
with open('HaploBlock_toy.txt') as HaploBlocks:
    header = HaploBlocks.readline()  
    # only reads the first line as header

    ''' create a empty dict or default dict. Which ever is better?'''
    Hap_Dict = {}
    Hap_Dict = defaultdict(list)


    ''' for rest of the lines '''
    for lines in HaploBlocks:
        values = lines.strip('\n').split('\t')

        ''' append the data to the dict for unique keys on the for loop, until the number of unique keys is 2 '''
        Block = values[2]
        Hap_Dict[Block].append(values[3])

        do something to count the number of keys - how?
        if keys_count > 2:
           return parse_two_blocks(Hap_Dict)

        elif keys_count < 2 or no new keys: # This one is odd and won't work I know.
           end the program

因此,当代码被执行时,它将从块7和9读取数据,直到字典被填充并返回到预定义的函数。解析完成后,它现在可以保留前一个解析的后一个块中的数据。这样它只需要读取剩余的块。

预期输出:我现在面临的主要问题是能够一次读取两个块。我不想在`parse_two_blocks(someData)'中添加我想要解析信息的内在细节 - 这只是另一个问题。但是,让我们尝试重写与输入相同的输出。

python list dictionary for-loop defaultdict
1个回答
1
投票

将输入解析为块的动态列表(生成器)。迭代对。在评估对时,应该完成所有操作。也就是说,这些行中没有一行应该一次读取或存储整个csv文件。

#!/usr/bin/env python3

data = """chr   pos A_block A_val
2   05  7   A,T,C
2   11  7   T,C,G
2   15  7   AT,C,G
2   21  7   C,A,GT
2   31  7   T,C,CA
2   42  9   T,C,G
2   55  9   C,G,GC
2   61  9   A,GC,T
2   05  12  AC,TG,G
2   11  12  A,TC,TG"""

import csv
import io
import itertools
import collections
import operator
from pprint import pprint

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = itertools.tee(iterable)
    next(b, None)
    return zip(a, b)

def one():
    # read rows as tuples of values
    c = csv.reader(io.StringIO(data), dialect=csv.excel_tab)
    # read header row
    keys = next(c)
    block_index = keys.index('A_block')
    # group rows by block numbers
    blocks = itertools.groupby(c, key=operator.itemgetter(block_index))
    # extract just the row values for each block
    row_values = (tuple(v) for k, v in blocks)
    # rearrange the values by column
    unzipped_values = (zip(*v) for v in row_values)
    # create a dictionary for each block
    dict_blocks = (dict(zip(keys, v)) for v in unzipped_values)
    yield from pairwise(dict_blocks)


def two():
    c = csv.DictReader(io.StringIO(data), dialect=csv.excel_tab)
    blocks = itertools.groupby(c, key=lambda x: x['A_block'])
    yield from pairwise((k, list(v)) for k, v in blocks)


for a, b in one():
        pprint(a)
        pprint(b)
        print()

产量(one):

{'A_block': ('7', '7', '7', '7', '7'),
 'A_val': ('A,T,C', 'T,C,G', 'AT,C,G', 'C,A,GT', 'T,C,CA'),
 'chr': ('2', '2', '2', '2', '2'),
 'pos': ('05', '11', '15', '21', '31')}
{'A_block': ('9', '9', '9'),
 'A_val': ('T,C,G', 'C,G,GC', 'A,GC,T'),
 'chr': ('2', '2', '2'),
 'pos': ('42', '55', '61')}

{'A_block': ('9', '9', '9'),
 'A_val': ('T,C,G', 'C,G,GC', 'A,GC,T'),
 'chr': ('2', '2', '2'),
 'pos': ('42', '55', '61')}
{'A_block': ('12', '12'),
 'A_val': ('AC,TG,G', 'A,TC,TG'),
 'chr': ('2', '2'),
 'pos': ('05', '11')}

io.StringIO(string)

获取一个字符串并返回一个包含string内容的类文件对象。

来自csv.DictReader(file_object, dialect)csv module

返回每行的有序dict,其中从第一行获取的字段名称用作字段值的字典键。

groupby(iterable, key_function)

创建一个从迭代中返回连续键和组的迭代器。关键是计算每个元素的键值的函数。

lambda x: x['A_block']

一个临时函数,它接受一个名为x的输入并返回键'A_block'的值

(k, list(v)) for k, v in blocks

groupby()为值返回一个迭代器(只能使用一次)。这会将迭代器转换为列表。

pairwise(iterable) recipe

“c - >(c0,c1),(c1,c2),(c2,cz),......”

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