基本上,我的数据框如下所示:
id | refers
----------------
1 | [2,3]
2 | [1,3]
3 | []
我想添加另一列,以显示该ID被另一个ID引用了多少次。例如:
id | refers | referred_count
----------------------------------
1 | [2,3] | 1
2 | [1,3] | 1
3 | [] | 2
我当前的代码如下:
citations_dict = {}
for index, row in data_ref.iterrows():
if len(row['reference_list']) > 0:
for reference in row['reference_list']:
if reference not in citations_dict:
citations_dict[reference] = {}
d = data_ref.loc[data_ref['id'] == reference]
citations_dict[reference]['venue'] = d['venue']
citations_dict[reference]['reference'] = d['reference']
citations_dict[reference]['citation'] = 1
else:
citations_dict[reference]['citation'] += 1
问题在于,这段代码花费了很长时间。我想知道如何更快地执行此操作,也许使用熊猫?
答案 0 :(得分:1)
第1步:获取引用列中每个ID的计数并将其存储在字典中,然后将该函数应用于创建新列。
import pandas as pd
from collections import Counter
df = pd.DataFrame({'id':[1,2,3],'refers':[[2,3],[1,3],[]]})
counter = dict(Counter([item for sublist in df['refers'] for item in sublist]))
df['refer_counts'] = df['id'].apply(lambda x: counter[x])
输出
id refers refer_counts
0 1 [2, 3] 1
1 2 [1, 3] 1
2 3 [] 2
认为这正是您所需要的!
答案 1 :(得分:1)
数据
df = pd.DataFrame({'id': [1,2,3], 'refers': [[1,2,3], [1,3], []]})
id refers referred_count
0 1 [1, 2, 3] 1
1 2 [1, 3] 1
2 3 [] 2
创建引用的出现次数字典:
refer_count = df.refers.apply(pd.Series).stack()\
.reset_index(drop=True)\
.astype(int)\
.value_counts()\
.to_dict()
通过其id_count减去每个ID中的推荐人:
df['referred_count'] = df.apply(lambda x: refer_count[x['id']] - x['refers'].count(x['id']), axis = 1)
输出:
id refers referred_count
0 1 [1, 2, 3] 1
1 2 [1, 3] 1
2 3 [] 2
答案 2 :(得分:1)
首先使用numpy.hstack
和Series.value_counts
创建一个助手Series
。
这将是您的列“ referred_count”的值,其中id
为索引。
然后,您可以将df的reset_index
转换为id
,以便轻松合并本系列,最后使用reset_index
将DataFrame恢复为原始形状。
s = pd.Series(np.hstack(df['refers'])).value_counts()
df.set_index('id').assign(referred_count=s).reset_index()
[出]
id refers referred_count
0 1 [2, 3] 1
1 2 [1, 3] 1
2 3 [] 2