如何根据列值删除行,其中某行的列值是另一行的子集？

概述假设我有一个数据帧df如下： - index company url address 0 A . www.abc.contact.com 16D Bayberry Rd, New Bedford, MA, 02740, USA 1 A . www.abc.c

假设我有一个数据帧df如下： –

index company  url                          address  0     A .    www.abc.contact.com         16D BayBerry Rd,New bedford,MA,02740,USA 1     A .    www.abc.contact.com .       MA,USA 2     A .    www.abc.about.com .         USA 3     B .    www.pqr.com .               New bedford,USA 4     B.     www.pqr.com/about .         MA,USA

我想从数据框中删除所有行,其中地址是另一个地址的子集,公司是相同的.例如,我希望这5行中的这两行.

index  company  url                          address  0     A .    www.abc.contact.com         16D BayBerry Rd,USA 3     B .    www.pqr.com .               New bedford,USA

最佳答案也许它不是最佳解决方案,但它可以在这个小型数据框架上工作：

EDIT添加了对公司名称的检查,假设我们删除了标点符号

df = pd.DataFrame({"company": ['A','A','B','B'],"address": ['16D BayBerry Rd,USA','MA,'USA','New bedford,USA']})# Splitting addresses by column and making sets from every address to use "issubset" lateraddresses = List(df['address'].apply(lambda x: set(x.split(','))).values)companIEs = List(df['company'].values)rows_to_drop = []  # Storing row indexes to drop here# Iterating by every addressfor i,(address,company) in enumerate(zip(addresses,companIEs)):    # Iteraing by the remaining addresses    rem_addr = addresses[:i] + addresses[(i + 1):]    rem_comp = companIEs[:i] + companIEs[(i + 1):]    for other_addr,other_comp in zip(rem_addr,rem_comp):        # If address is a subset of another address,add it to drop        if address.issubset(other_addr) and company == other_comp:            rows_to_drop.append(i)            breakdf = df.drop(rows_to_drop)print(df)company address0   A   16D BayBerry Rd,USA3   B   New bedford,USA

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