我有以下熊貓資料框。
d = {'id1': ['85643', '85644','85643','8564312','8564314','85645','8564316','85646','8564318','85647','85648','85649','85655','56731','34566','78931','78931'],'ID': ['G-00001', 'G-00001','G-00002','G-00002','G-00002','G-00001','vaasd','aasd','aasd','vaasd','aasd','vaasd','aasd','aasd2','aasd2','aasd2','aasd2'],'col1': [671, 2,5,3,4,5,60,0,0,6,3,2,4,32,3,1,23],'Goal': [np.nan, 56,78,np.nan,89,73,np.nan ,np.nan ,np.nan, np.nan, np.nan, 34,np.nan, 7, 84,np.nan,5 ], 'col2': [793, 4,8,32,43,55,610,0,0,16,23,72,48,3,28,5,3],'col3': [500, 22,89,33,44,55,60,1,5,6,3,2,4,13,12,14,98],'Date': ['2021-06-13', '2021-06-13','2021-06-14','2021-06-13','2021-06-14','2021-06-15','2021-06-15','2021-06-13','2021-06-16','2021-06-13','2021-06-13','2021-06-13','2021-06-16','2021-05-23','2021-05-13','2021-03-26','2021-05-13']}
dff = pd.DataFrame(data=d)
dff
id1 ID col1 Goal col2 col3 Date
0 85643 G-00001 671 NaN 793 500 2021-06-13
1 85644 G-00001 2 56.0000 4 22 2021-06-13
2 85643 G-00002 5 78.0000 8 89 2021-06-14
3 8564312 G-00002 3 NaN 32 33 2021-06-13
4 8564314 G-00002 4 89.0000 43 44 2021-06-14
5 85645 G-00001 5 73.0000 55 55 2021-06-15
6 8564316 vaasd 60 NaN 610 60 2021-06-15
7 85646 aasd 0 NaN 0 1 2021-06-13
8 8564318 aasd 0 NaN 0 5 2021-06-16
9 85647 vaasd 6 NaN 16 6 2021-06-13
10 85648 aasd 3 NaN 23 3 2021-06-13
11 85649 vaasd 2 34.0000 72 2 2021-06-13
12 85655 aasd 4 NaN 48 4 2021-06-16
13 56731 aasd2 32 7.0000 3 13 2021-05-23
14 34566 aasd2 3 84.0000 28 12 2021-05-13
15 78931 aasd2 1 NaN 5 14 2021-03-26
16 78931 aasd2 23 5.0000 3 98 2021-05-13
另外，我有下面的字典
dic = {'G-0001':{'aasd':['G-0001','85646','85648','345_2','85655','85659'],
'vaasd':['G-0001','85649','34554','85655','22183','45335','8564316']},
'G-0002':{'aasd2':['G-0002','85343','78931','45121','56731']},}
我想根據字典中的串列獲得 id1 的唯一計數。例如，如果我們考慮這個串列 'aasd2':['G-0002','85343','78931','45121','56731']。我想知道大熊貓的資料框中有多少個唯一的 id1。所以它應該用于 aasd2-2 值 ('78931','56731')。這些是 aasd2 在 pandas 中僅有的兩個值。
所以我想為串列名稱和 id 計數創建一個如下所示的表
listName count of ids
aasd 3
vaasd 2
aasd2 2
可以在python中做到這一點嗎？任何建議將不勝感激。提前致謝！！
uj5u.com熱心網友回復：
從您的 dict 創建一個資料框merge，dff然后value_counts在drop_duplicateson之后使用它('id1', 'ID')：
data = []
for d in dic.values():
for k, l in d.items():
data.extend([(v, k) for v in l])
df1 = pd.DataFrame(data, columns=['id1', 'ID'])
out = dff.merge(df1, on=['id1', 'ID']) \
.drop_duplicates(['id1', 'ID']) \
.value_counts('ID')
輸出：
>>> out
ID
aasd 3
aasd2 2
vaasd 2
dtype: int64
uj5u.com熱心網友回復：
這是另一種方式：
data = []
for d in dic.values():
for k, l in d.items():
data.extend([(v, k) for v in l])
df = df[df['id1', 'ID']].apply(tuple, axis=1).isin(data)].groupby("ID")["id1"].nunique()
uj5u.com熱心網友回復：
一種選擇是在生成最終資料幀之前在 Python 中運行整個（幾乎）程序：
from collections import ChainMap
# get the dictionary for the inner dict
content = dict(ChainMap(*dic.values()))
uniq = dff.id1.unique()
content = [(key, len(set(val).intersection(uniq)))
for key, val in content.items()]
pd.DataFrame(content, columns = ['listName', 'count of Ids'])
listName count of Ids
0 aasd2 2
1 aasd 3
2 vaasd 3
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