我想在 pandas 中解決一個奇怪的問題。假設(shè)我有一堆對象，它們有不同的分組方式。這是我們的數(shù)據(jù)框的樣子：df=pd.DataFrame([ {'obj': 'Ball', 'group1_id': None, 'group2_id': '7' }, {'obj': 'Balloon', 'group1_id': '92', 'group2_id': '7' }, {'obj': 'Person', 'group1_id': '14', 'group2_id': '11'}, {'obj': 'Bottle', 'group1_id': '3', 'group2_id': '7' }, {'obj': 'Thought', 'group1_id': '3', 'group2_id': None},])obj group1_id group2_idBall None 7Balloon 92 7Person 14 11Bottle 3 7Thought 3 None我想根據(jù)任何組將事物分組在一起。這里注釋一下：obj group1_id group2_id # annotatedBall None 7 # group2_id = 7Balloon 92 7 # group1_id = 92 OR group2_id = 7Person 14 11 # group1_id = 14 OR group2_id = 11Bottle 3 7 # group1_id = 3 OR group2_id = 7Thought 3 None # group1_id = 3組合后，我們的輸出應(yīng)如下所示：count objs composite_id4 [Ball, Balloon, Bottle, Thought] g1=3,92|g2=71 [Person] g1=11|g2=14請注意，我們可以獲得的前三個對象group2_id=7，然后是第四個對象Thought，是因?yàn)樗梢酝ㄟ^group1_id=3為其分配group_id=7id 來與另一個項(xiàng)目匹配。注意：對于這個問題，假設(shè)一個項(xiàng)目只會屬于一個組合組（并且永遠(yuǎn)不會有可能屬于兩個組的情況）。我怎樣才能做到這一點(diǎn)pandas？
2 回答
郎朗坤
TA貢獻(xiàn)1921條經(jīng)驗(yàn) 獲得超9個贊
這一點(diǎn)也不奇怪~網(wǎng)絡(luò)問題
import networkx as nx
#we need to handle the miss value first , we fill it with same row, so that we did not calssed them into wrong group
df['key1']=df['group1_id'].fillna(df['group2_id'])
df['key2']=df['group2_id'].fillna(df['group1_id'])
# here we start to create the network
G=nx.from_pandas_edgelist(df, 'key1', 'key2')
l=list(nx.connected_components(G))
L=[dict.fromkeys(y,x) for x, y in enumerate(l)]
d={k: v for d in L for k, v in d.items()}
# we using above dict to map the same group into the same one in order to groupby them?
out=df.groupby(df.key1.map(d)).agg(objs = ('obj',list) , Count = ('obj','count'), g1= ('group1_id', lambda x : set(x[x.notnull()].tolist())), g2= ('group2_id',? lambda x : set(x[x.notnull()].tolist())))
# notice here I did not conver the composite id into string format , I keep them into different columns which more easy to understand?
Out[53]:?
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? objs? Count? ? ? ?g1? ? g2
key1? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??
0? ? ?[Ball, Balloon, Bottle, Thought]? ? ? 4? {92, 3}? ?{7}
1? ? ? ? ? ? ? ? ? ? ? ? ? ? ?[Person]? ? ? 1? ? ?{14}? {11}
紅糖糍粑
TA貢獻(xiàn)1815條經(jīng)驗(yàn) 獲得超6個贊
這里有一個更詳細(xì)的解決方案，我為分組集合構(gòu)建了“第一個鍵”的映射：
# using four id fields instead of 2
grouping_fields = ['group1_id', 'group2_id', 'group3_id', 'group4_id']
id_fields = df.loc[df[grouping_fields].notnull().any(axis=1), grouping_fields]
# build a set of all similarly-grouped items
# and use the 'first seen' as the grouping key for that
FIRST_SEEN_TO_ALL = defaultdict(set)
KEY_TO_FIRST_SEEN = {}
for row in id_fields.to_dict('records'):
? ? # why doesn't nan fall out in a boolean check?
? ? keys = [id for id in row.values() if id and (str(id) != 'nan')]
? ? row_id = keys[0]
? ? for key in keys:
? ? ? ? if (row_id != key) or (key not in KEY_TO_FIRST_SEEN):
? ? ? ? ? ? KEY_TO_FIRST_SEEN[key] = row_id
? ? ? ? ? ? first_seen_key = row_id
? ? ? ? else:
? ? ? ? ? ? first_seen_key = KEY_TO_FIRST_SEEN[key]
? ? ? ? FIRST_SEEN_TO_ALL[first_seen_key].add(key)
def fetch_group_id(row):
? ? keys = filter(None, row.to_dict().values())
? ? for key in keys:
? ? ? ? first_seen_key = KEY_TO_FIRST_SEEN.get(key)
? ? ? ? if first_seen_key:?
? ? ? ? ? ? return first_seen_key
df['group_super'] = df[grouping_fields].apply(fetch_group_id, axis=1)
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