Question

问题：

您好我是Python的新手，所以寻求一些帮助。我有多条输入线。我正在寻找一种方法来从多个列表中获取每个单词，然后将它们相加以获得多个List中该单词的每个实例的总计数。我真的很感激任何指导。

Answer 1

编辑：

"""I need a count of the word given by key(s) in multiple lists at the given index. So what I need is
>>> count(["a", "b", "c"], ["c", "b", "a"])
{"a": [1,0,1], "b": [0,2,0], "c": [1,0,1]}":
"""

我希望这是正确的：

def count (*args):
    l = len(args[0])
    end = {}
    for lst in args:
        y = 0
        for x in lst:

            end.setdefault(x, l*[0])
            end[x][y] += 1
            y += 1
    return end

如果您需要一种从RNA / DNA操纵基因组材料的方法，您可以在pypi.python.org上搜索文库。有很多好的。

Answer 2

查看dict.setdefault()方法和enumerate()函数：

def count_items(data):
    count = {}
    for datum in data:
        count[datum] = count.setdefault(datum, 0) + 1
    return count


def collate(*data):
    collated = {}
    for datum in data:
        for k, v in datum.items():
            collated[k] = datum.setdefault(k, 0) + 1
    return collated


def key_position(sequence, key):
    sequence_map = [0 for _ in sequence]
    for i, item in enumerate(sequence):
        if key == item:
            sequence_map[i] += 1
    return sequence_map


data1 = ['a', 'b', 'c', 'd', 'a']
data2 = ['a', 'b', 'c', 'd', 'a']

counted1 = count_items(data1)
counted2 = count_items(data2)

collated = collate(counted1, counted2)

a_positions = key_position(data1, 'a')

Answer 3

首先，zip您的读物：

Read_1 = ['GGGA', 'ATTA']
Read_2 = ['GATT', 'ATTA']
reads = zip(Read_1, Read_2)
# ['GGGA', 'GATT'], ['ATTA', 'ATTA']

然后，计算一下：

from collections import Counter
counters = [Counter(read) for read in reads]

然后询问给定序列的频率：

print(list(cnt['ATTA'] for cnt in counters)
# [0, 2]
print(list(cnt['GGGA'] for cnt in counters)
# [1, 0]

Python dict计数

3 个答案: