Question

我有以下代码从pdb文件打印预定义序列。现在我想为每个chain_id保存单独的输出文件。

如何为每个chain_id保存单独的输出？

预期产出：

我想为每个chain_id保存输出文件。

如果输入文件名是1AHI.PDB，则在此文件中如果我们有四个链ID A，B，C，D，那么我想要输出文件：1AHIA.txt，1AHIB.txt，{ {1}}，1AHIC.txt。这将适用于每个输入文件。我的输入目录中有2000多个输入文件。

代码：

1AHID.txt

*在Ans *之后编辑

错误：

from glob import glob

in_loc = r'C:/Users/Documents/NAD/NAD/result/test_result_file/'
out_loc = r'C:/Users/Documents/NAD/NAD/result/test_result_file/Final_result/'


def test():
    fnames = glob(in_loc+'*.pdb')

    for each in fnames:
    # This is the new generated file out of input file (.txt).
        formatted_file = each.replace('pdb', 'txt')

        formatted_file = formatted_file.replace(in_loc, out_loc)

    # This is the input file
        in_f = open(each, 'r')

    # A new file to be opened.
        out_f = open(formatted_file, "w")

    # Filtering results from input file
        try:
            out_chain_list = filter_file(in_f)
            for each_line in out_chain_list:
                out_f.write(each_line)

        # Closing all the opened files.
            out_f.close()
            in_f.close()

        except Exception as e:
            print('Exception for file: ', each, '\n', e)
            out_f.close()
            in_f.close()


def filter_file(in_f):
    atom_ids = ['C4B', 'O4B', 'C1B', 'C2B', 'C3B']
    chain_ids = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
    order = [0, 1, 4, 3, 2]

    previous_chain_id = None
    chain_list = []
    out_chain_list = []

    for line in in_f:

        if line.startswith('HETATM '):
            line = line.replace('HETATM ', 'HETATM')

        if line.startswith('HETATM'):
            line_list = line.split()
            chain_id = line_list[3][0]
            atom_id = line_list[1]

            if atom_id in atom_ids and chain_id in chain_ids:

                if chain_id != previous_chain_id:
                    c_ls = []
                    if chain_list:
                        c_l = chain_list[-5:]
                        c_l = [c_l[i] for i in order]

                        for i in range(5):
                            c_ls += c_l[:4]
                            c_ls.append('\n')
                            c_l = c_l[-4:] + c_l[:1]

                    out_chain_list += c_ls
                    chain_list.append('\n')

                chain_list.append(line)
                previous_chain_id = chain_id

    c_ls = []
    if chain_list:
        c_l = chain_list[-5:]
        c_l = [c_l[i] for i in order]

        for i in range(5):
            c_ls += c_l[:4]
            c_ls.append('\n')
            c_l = c_l[-4:] + c_l[:1]
    out_chain_list += c_ls

    return out_chain_list

test()

Answer 1

您当前的代码会为每个输入文件打开一个输出。但是你想要每个out_chain项的输出文件，每个输入文件中可以有多个out_chain项。因此，您需要在处理out_chain项目的内部循环中打开和关闭输出文件。这是你可以做到的一种方式：

for(int i = 0; i < v.size; i++){
    float[] vertices = new float[v.size()]
    vertices[i - 1] = v.get(i - 1);
    vertices[i] = v.get(i);
}
sr.polygon(v);

Answer 2

您可以修改filter_file您可以收到以chain_id为键的字典。如果您的out_chain_dict格式为{'chain_id':out_chain_list}，则可以轻松为每个chain_id创建一个不同的文件：

def test():
    fnames = glob(in_loc+'*.pdb')

    for each in fnames:
    # This is the path for new generated file.
        path_file = each.replace(in_loc, out_loc)

    # This is the input file and iltering results from input file
        with open(each, 'r') as in_f:
            try:
                out_chain_dict = filter_file(in_f)

            except Exception as e:
                print('Exception for file: ', each, '\n', e)
                continue

            for (chain_id, out_chain_list) in out_chain_dict.items():
                # This is the new generated file out of input file (.txt).
                formatted_file = path_file.replace('.pdb', chain_id + '.txt')

                # A new file to be opened.
                with open(formatted_file, "w") as out_f:
                    for each_line in out_chain_list:
                        out_f.write(each_line)

filter_file的编辑：

def filter_file(in_f):
    atom_ids = ['C4B', 'O4B', 'C1B', 'C2B', 'C3B']
    chain_ids = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z']
    order = [0, 1, 4, 3, 2]

    previous_chain_id = None
    chain_list = []
    out_chain_dict = {}     # Change out_chain_list to dict

    for line in in_f:

        if line.startswith('HETATM '):
            line = line.replace('HETATM ', 'HETATM')

        if line.startswith('HETATM'):
            line_list = line.split()
            chain_id = line_list[3][0]
            atom_id = line_list[1]

            if atom_id in atom_ids and chain_id in chain_ids:

                if chain_id != previous_chain_id:
                    c_ls = []
                    if chain_list:
                        c_l = chain_list[-5:]
                        c_l = [c_l[i] for i in order]

                        for i in range(5):
                            c_ls += c_l[:4]
                            c_ls.append('\n')
                            c_l = c_l[-4:] + c_l[:1]

                    try:                                            # Here add c_ls to an existing key chain_id
                        out_chain_dict[chain_id] += c_ls   #
                    except KeyError:                                # or create new chain_id key
                        out_chain_dict[chain_id] = c_ls    # if it appears at the first time
                    chain_list.append('\n')

                chain_list.append(line)
                previous_chain_id = chain_id

    c_ls = []
    if chain_list:
        c_l = chain_list[-5:]
        c_l = [c_l[i] for i in order]

        for i in range(5):
            c_ls += c_l[:4]
            c_ls.append('\n')
            c_l = c_l[-4:] + c_l[:1]
    # I guess here we add the last chain_id which corresponds to `chain_id` key
    try:
        out_chain_dict[chain_id] += c_ls
    except KeyError:
        out_chain_dict[chain_id] = c_ls

    return out_chain_dict

如何为每个chain_id保存单独的输出文件？

2 个答案: