如何在python中向单个唯一键插入多个值

Question

我有df，对于第一列，我想将第二列的值捕获为list，并创建一个列表，其中key是第一列的唯一id，第二列的值列表是values。你知道吗

以下是示例：

Bra100001       Bra100001
Bra100001       Bra011864
Bra100002       Bra011842
Bra100002       Bra100002
Bra100003       Bra100003
Bra100004       Bra100004
Bra100005       Bra100005
Bra100006       Bra100006
Bra100007       Bra011656
Bra100008       Bra100007
Bra100009       Bra100008
Bra100009       Bra011638
Bra100010       Bra103178
Bra100010       Bra011635

我需要的输出是

Bra100001:(Bra100001,Bra011864),Bra100002:(Bra011842,Bra100002),Bra100003:Bra100003 and so on....

这是我的密码

with open("test_blast.txt", 'r') as fh_in:
        prev = None
        result = {}
        for line in fh_in:
            line = line.strip()
            line = line.split()
            if prev == line[0]:
                result[line[0]] = line[1]
            prev = line[0]

Answer 1

如果希望输出是元组有序字典：

import collections

result = collections.OrderedDict()

with open("test_blast.txt", 'r') as fh_in:
    for line in fh_in:
        col = line.split()
        if len(col)<2: continue
        if not col[0] in result:
            result[col[0]] = ()     
        result[col[0]] += (col[1],)

如果希望输出为示例中给出的精确格式的字符串，可以通过以下方式进一步处理结果：

out=[]
for r in result:
    s  = str(result[r]).replace(', ', ',').replace("'",'')
    if s.endswith(",)"): s = s[1:-2]
    out.append(r+':'+s)

print ",".join(out)

您可以在这里看到演示：http://repl.it/46l/3

Answer 2

有几个选项，只是简单的元组或列表，x = ('x1', 'x2')，或者将哈希与桶一起使用：

foobar = {
    'key1': [(some_vals)];
    'key2': [(some_other_vals)]
    ...
    }

那就参考foobar吧

Answer 3

with open('test_blast.txt', 'r') as f:
    lines = f.readlines()

records = [l.split() for l in lines]
records = [r for r in records if len(r) == 2]  # drop empty line at end in my tests

result = {}
for r in records:
    if not result.get(r[0]):
        result[r[0]] = []    # this is the first reference to key so initialize value
    result[r[0]].append(r[1])

# below is only needed for sorted output
keys = sorted(result.keys())
for k in keys:
    print k, ': ', result[k]