Question

我正在使用python2.7。我有一个包含染色体位置和实验ID的文件。我目前已将这些信息存储在两个列表中：

unique_locations - containing a single value for each location
location_exp - containing lists of [location, experiment]

我没有使用字典的原因是在多个实验中找到了多个位置 - 即这是一个很多很多的关系。

我想了解每个位置的实验数量。我得到一个如下列表：

[
    [location1, [experiment1, experiment2, experiment3]], 
    [location2, [experiment2, experiment3, experiment4]]
                                                             ]

等

由于列表的长度不同，我在两个列表中使用枚举（列表）循环都失败了。我试过了：

location_experiment_sorted = []
for i, item in enumerate(unique_experiment):
    location = item[0]
    exp = item[1]
    if location not in location_experiment_sorted:
        location_experiment_sorted.append([location, exp])
    else:
        location_experiment_sorted[i].append(exp)

除此之外。我也尝试使用与多个实验列表相关的字典。有人能指出我正确的方向吗？

Answer 1

如果我理解你的话（如果位置可以用作dict键）

你可以这样做：

location_experiments={}
for location, experiment in location_exp:
    location_experiments.setdefault(location,[]).append(experiment)

Answer 2

尝试defaultdict，即：

$("#dataset\\.recordids\\.Products option")

将打印出来：

from collections import defaultdict

unique_locations = ["location1", "location2"]
location_exp = [
    ("location1", "experiment1"),
    ("location1", "experiment2"),
    ("location1", "experiment3"),
    ("location2", "experiment2"),
    ("location2", "experiment3"),
    ("location2", "experiment4")
]

location_experiment_dict = defaultdict(list)
for location, exp in location_exp:
    location_experiment_dict[location].append(exp)

print(location_experiment_dict)

Answer 3

我还没有这样做，如果失败就道歉。如果你说它是[[位置，实验]，[位置，实验]]等列表，那么：

locationList = {}
for item in unique_experiment:
    location = item[0]
    exp = item[1]
    if location not in locationList:
        locationList[location] = []
        locationList[location].append(exp)
    else:
        locationList[location].append(exp)

Answer 4

以下是另一个工作示例，使用来自dict的内置groupby和itertools：

>>> from itertools import groupby
>>> d = {}
>>> location_exp = [
    ("location1", "experiment1"),
    ("location1", "experiment2"),
    ("location1", "experiment3"),
    ("location2", "experiment2"),
    ("location2", "experiment3"),
    ("location2", "experiment4")
]
>>> for k,v in groupby(location_exp, itemgetter(0)):
        d.setdefault(k,[])
        d[k].extend([loc for _, loc in v])


[]
[]
>>> d
{'location2': ['experiment2', 'experiment3', 'experiment4'], 'location1': ['experiment1', 'experiment2', 'experiment3']}
>>> 
>>> d2 = {}
>>> location_exp2 = [
    ("location1", "experiment1"),
    ("location2", "experiment2"),
    ("location3", "experiment3"),
    ("location1", "experiment2"),
    ("location2", "experiment3"),
    ("location3", "experiment4")
]
>>> for k,v in groupby(location_exp2, itemgetter(0)):
        d2.setdefault(k,[])
        d2[k].extend([loc for _, loc in v])


[]
[]
[]
['experiment1']
['experiment2']
['experiment3']
>>> d2
{'location2': ['experiment2', 'experiment3'], 'location1': ['experiment1', 'experiment2'], 'location3': ['experiment3', 'experiment4']}

创建与列表关联的字典，并通过循环更新它

4 个答案: