诚然,我有一个关系表(思考人有一个物品),我成功地把这个表装成一个ool子。 我想走下一个步骤,并为任何特定用户之间的距离找到一个矩阵。
我知道,我是否想在距离外走一条路,我基本上可以做以下工作:
df = ...
df["val"] = 1
pivot = df.pivot(index="person", columns="hasa", values="val").fillna(0)
# compute one difference
(pivot["Alice"] - pivot["Carol"]).abs().sum()
我不知道如何从这里到完整的数据框架。
| person | hasa |
|---|---|
| Alice | Apple |
| Bob | Banana |
| Carol | Carrot |
| Bob | Apple |
| Apple | Banana | Carrot | |
|---|---|---|---|
| Alice | 1 | 0 | 0 |
| Bob | 1 | 1 | 0 |
| Carol | 0 | 0 | 1 |
| Alice | Bob | Carol | |
|---|---|---|---|
| Alice | 0 | 1 | 2 |
| Bob | 1 | 0 | 3 |
| Carol | 2 | 3 | 0 |
让我们开始简单。 可以通过创建另一个数据框架,储存这些距离,来计算贵纸桌上各行之间的距离。 这里的主要逻辑是,通过纸张表的每行翻开,计算与其他各行的距离。
# Create initial DataFrame
df = pd.DataFrame({
person : [ Alice , Bob , Carol , Bob ],
hasa : [ Apple , Banana , Carrot , Apple ]
})
# Pivot DataFrame
pivot_df = pd.pivot_table(df, index= person , columns= hasa , aggfunc=len, fill_value=0)
# Create empty DataFrame for distances
distance_df = pd.DataFrame(index=pivot_df.index, columns=pivot_df.index)
# Fill DataFrame with distances
for person1 in pivot_df.index:
for person2 in pivot_df.index:
distance_df.loc[person1, person2] = (pivot_df.loc[person1] - pivot_df.loc[person2]).abs().sum()
现在,distance_df 应当有正确的价值观。 请注意,由此产生的数据基是主分校一带的元数据(因为从Alice到Bob的距离与Bobo到Alice的距离相同)。 主要分校为零(因为一个人与自己之间的距离总是零)。
您可直接使用< 或者,你也可以使用 或人工使用
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Crosstab/code>, 则与
from scipy.spatial.distance import cdist
pivot = pd.crosstab(df[ person ], df[ hasa ])
out = pd.DataFrame(cdist(pivot, pivot)**2,
index=pivot.index,
columns=pivot.index,
)
corr,其习俗功能是第二步:out = pivot.T.corr(lambda a,b: sum(a!=b))
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