(或列表清单……我刚刚编辑过)
是否存在用于转换此类结构的现有python / pandas方法
food2 = {}
food2["apple"] = ["fruit","round"]
food2["bananna"] = ["fruit","yellow","long"]
food2["carrot"] = ["veg","orange","long"]
food2["raddish"] = ["veg","red"]
进入像这样的数据透视表？
+---------+-------+-----+-------+------+--------+--------+-----+
| | fruit | veg | round | long | yellow | orange | red |
+---------+-------+-----+-------+------+--------+--------+-----+
| apple | 1 | | 1 | | | | |
+---------+-------+-----+-------+------+--------+--------+-----+
| bananna | 1 | | | 1 | 1 | | |
+---------+-------+-----+-------+------+--------+--------+-----+
| carrot | | 1 | | 1 | | 1 | |
+---------+-------+-----+-------+------+--------+--------+-----+
| raddish | | 1 | | | | | 1 |
+---------+-------+-----+-------+------+--------+--------+-----+
天真的,我可能只是循环通过字典.我看到如何在每个内部列表上使用地图,但我不知道如何在字典上加入/堆叠它们.一旦我加入了它们,我就可以使用pandas.pivot_table了
for key in food2:
attrlist = food2[key]
onefruit_pairs = map(lambda x: [key,x],attrlist)
one_fruit_frame = pd.DataFrame(onefruit_pairs,columns=['fruit','attr'])
print(one_fruit_frame)
fruit attr
0 bananna fruit
1 bananna yellow
2 bananna long
fruit attr
0 carrot veg
1 carrot orange
2 carrot long
fruit attr
0 apple fruit
1 apple round
fruit attr
0 raddish veg
1 raddish red
最佳答案
纯Python：
原文链接：https://www.f2er.com/python/438472.htmlfrom itertools import chain
def count(d):
cols = set(chain(*d.values()))
yield ['name'] + list(cols)
for row,values in d.items():
yield [row] + [(col in values) for col in cols]
测试：
>>> food2 = {
"apple": ["fruit","round"],"bananna": ["fruit","long"],"carrot": ["veg","raddish": ["veg","red"]
}
>>> list(count(food2))
[['name','long','veg','fruit','yellow','orange','round','red'],['bananna',True,False,False],['carrot',['apple',['raddish',True]]
[更新]
性能测试：
>>> from itertools import product
>>> labels = list("".join(_) for _ in product(*(["ABCDEF"] * 7)))
>>> attrs = labels[:1000]
>>> import random
>>> sample = {}
>>> for k in labels:
... sample[k] = random.sample(attrs,5)
>>> import time
>>> n = time.time(); list(count(sample)); print time.time() - n
62.0367980003
在我忙碌的机器上花了不到2分钟,因为279936行乘1000列(打开了很多镀铬标签).如果表现不可接受,请告诉我.
[更新]
从另一个答案测试性能：
>>> n = time.time(); \
... df = pd.DataFrame(dict([(k,pd.Series(v)) for k,v in sample.items()])); \
... print time.time() - n
72.0512290001
下一行(df = pd.melt(…))花了太长时间,所以我取消了测试.拿这个结果用一粒盐,因为它在繁忙的机器上运行.