dfcol1 col2['aa', 'bb', 'cc', 'dd'] [['ee', 'ff', 'gg', 'hh'], ['qq', 'ww', 'ee', 'rr']]['ss', 'dd', 'ff', 'gg'] [['mm', 'nn', 'vv', 'cc'], ['zz', 'aa', 'jj', 'kk']]['ss', 'dd'] [['mm', 'nn', 'vv', 'cc'], ['zz', 'aa', 'jj', 'kk']]我希望能够运行一个函数，将第一个列表元素连接到 中的第一个子列表元素col1（有多个子列表）col2，然后将第二个列表元素连接到 中col1的第二个子列表元素col2。结果将类似于此列：results[['aaee', 'bbff', 'ccgg', 'ddhh'],['aaqq', 'bbww', 'ccee', 'ddrr']][['ssmm', 'ddnn', 'ffvv', 'ggcc'],['sszz', 'ddaa', 'ffjj', 'ggkk']][['ssmm', 'ddnn'],['sszz', 'ddaa']]我认为这与循环遍历第一个元素有关，col1并以某种方式循环并将它们与每个子列表中的相应项目相匹配col2- 我该怎么做？转换后的代码[[[df1.agg(lambda x: get_top_matches(u,w), axis=1) for u,w in zip(x,v)]\for v in y] for x,y in zip(df1['parent_org_name_list'], df1['children_org_name_sublists'])]结果：
3 回答
慕丝7291255
TA贡献1859条经验 获得超6个赞
你可以zip在这里使用：
[[[u+w for u,w in zip(x,v)] for v in y] for x,y in zip(df['col1'], df['col2'])]
输出：
[[['aaee', 'bbff', 'ccgg', 'ddhh'], ['aaqq', 'bbww', 'ccee', 'ddrr']],
[['ssmm', 'ddnn', 'ffvv', 'ggcc'], ['sszz', 'ddaa', 'ffjj', 'ggkk']],
[['ssmm', 'ddnn'], ['sszz', 'ddaa']]]
要分配回您的数据框，您可以执行以下操作：
df['results'] = [[[u+w for u,w in zip(x,v)] for v in y]
for x,y in zip(df['col1'], df['col2'])]
holdtom
TA贡献1805条经验 获得超10个赞
Max，循环尝试这个解决方案。它允许对转换进行更精细的控制，包括处理不均匀的长度（参见len_limit示例）：
import pandas as pd
df = pd.DataFrame({'c1':[['aa', 'bb', 'cc', 'dd'],['ss', 'dd', 'ff', 'gg']],
'c2':[[['ee', 'ff', 'gg', 'hh'], ['qq', 'ww', 'ee', 'rr']],
[['mm', 'nn', 'vv', 'cc'], ['zz', 'aa', 'jj', 'kk']]],})
df ['c3'] = 'empty' # send string to 'c3' so it is object data type
print(df)
c1 c2 c3
0 [aa, bb, cc, dd] [[ee, ff, gg, hh], [qq, ww, ee, rr]] empty
1 [ss, dd, ff, gg] [[mm, nn, vv, cc], [zz, aa, jj, kk]] empty
for i, row in df.iterrows():
c3_list = []
len_limit = len (row['c1']
for c2_sublist in row['c2']:
c3_list.append([j1+j2 for j1, j2 in zip(row['c1'], c2_sublist[:len_limit])])
df.at[i, 'c3'] = c3_list
print (df['c3'])
0 [[aaee, bbff, ccgg, ddhh], [aaqq, bbww, ccee, ...
1 [[ssmm, ddnn, ffvv, ggcc], [sszz, ddaa, ffjj, ...
Name: c3, dtype: object
ITMISS
TA贡献1871条经验 获得超8个赞
尝试：
df["results"] = df[["col1", "col2"]].apply(lambda x: [list(map(''.join, zip(x["col1"], el))) for el in x["col2"]], axis=1)
输出：
>>> df["results"]
0 [[aaee, bbff, ccgg, ddhh], [aaqq, bbww, ccee, ...
1 [[ssmm, ddnn, ffvv, ggcc], [sszz, ddaa, ffjj, ...
2 [[ssmm, ddnn], [sszz, ddaa]]
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