Python 将共享同一密钥的行合并为一行
我有一个数据框,希望创建另一个列,将名称以Python 将共享同一密钥的行合并为一行,python,python-3.x,pandas,dataframe,Python,Python 3.x,Pandas,Dataframe,我有一个数据框,希望创建另一个列,将名称以Answer和QID中相同的值开头的列组合在一起 也就是说,具有以下数据帧 QID Category Text QType Question: Answer0 Answer1 Country 0 16 Automotive Access to car Single Do you have access to a car? I own a car/cars I own a car
Answer
和QID
中相同的值开头的列组合在一起
也就是说,具有以下数据帧
QID Category Text QType Question: Answer0 Answer1 Country
0 16 Automotive Access to car Single Do you have access to a car? I own a car/cars I own a car/cars UK
1 16 Automotive Access to car Single Do you have access to a car? I lease/ have a company car I lease/have a company car UK
2 16 Automotive Access to car Single Do you have access to a car? I have access to a car/cars I have access to a car/cars UK
3 16 Automotive Access to car Single Do you have access to a car? No, I don’t have access to a car/cars No, I don't have access to a car UK
4 16 Automotive Access to car Single Do you have access to a car? Prefer not to say Prefer not to say UK
因此,我希望得到以下结果:
QID Category Text QType Question: Answer0 Answer1 Answer2 Answer3 Country Answers
0 16 Automotive Access to car Single Do you have access to a car? I own a car/cars I lease/ have a company car I have access to a car/cars No, I don’t have access to a car/cars UK ['I own a car/cars', 'I lease/ have a company car' ,'I have access to a car/cars', 'No, I don’t have access to a car/cars', 'Prefer not to say Prefer not to say']
到目前为止,我尝试了以下方法:
previous_qid = None
i = 0
j = 0
answers = []
new_row = {}
new_df = pd.DataFrame(columns=df.columns)
for _, row in df.iterrows():
# get QID
qid = row['QID']
if qid == previous_qid:
i+=1
new_row['Answer'+str(i)]=row['Answer0']
answers.append(row['Answer0'])
elif new_row != {}:
# we moved to a new row
new_row['QID'] = qid
new_row['Question'] = row['Question']
new_row['Answers'] = answers
# we create a new row in the new_dataframe
new_df.append(new_row, ignore_index=True)
# we clean up everything to receive the next row
answers = []
i=0
j+=1
new_row = {}
# we add the information of the current row
new_row['Answer'+str(i)]=row['Answer0']
answers.append(row['Answer0'])
previous_qid = qid
但是new\u df
结果为空。这是通过QID逻辑分组得到答案列表,然后将列表拆分回列
重新导入
data=“”QID类别文本QType问题:回答0回答1国家/地区
你有车吗?我有车我有车英国
1 16汽车使用汽车单人您有汽车使用权吗?我租赁/有公司汽车我租赁/有公司汽车英国
2 16汽车进入汽车单人你有进入汽车的权利吗?我有进入汽车的权利我有进入汽车的权利
你有车吗?没有,我没有车/没有,我没有车
4 16汽车进入汽车单人你有进入汽车的途径吗?宁愿不说宁愿不说英国
a=[[t.strip()表示重新拆分中的t(“,l)如果t!=”“]表示重新拆分中的l(([0-9]?[])*(.*),r“\2”,l)表示数据拆分中的l(“\n”)]]
df=pd.DataFrame(data=a[1:],columns=a[0])
#lazy-除了QID和Answer列之外,首先需要所有属性
agg={col:“first”表示列表(df.columns)中的col,如果col!=“QID”和“Answer”不在col}
#获取QID答案0中所有答案的列表
agg={**agg,***{“Answer0”:lambda s:list}
#行调用的助手函数。不需要,但更具可读性
def ans(r,i):
返回“”如果i>=len(r[“AnswerT”])否则r[“AnswerT”][i]
#使用assign将列表从聚合中拆分回列
#将Answer0从聚合重命名为AnserT,以便可以引用它。
#当你不再需要它的时候,不要放弃它
dfgrouped=df.groupby(“QID”).agg(agg).reset_index().rename(columns={“Answer0”:“AnswerT”}.assign(
Answer0=λdfa:dfa.apply(λr:ans(r,0),轴=1),
答案1=λdfa:dfa.apply(λr:ans(r,1),轴=1),
回答2=λdfa:dfa.apply(λr:ans(r,2),轴=1),
答案3=λdfa:dfa.apply(λr:ans(r,3),轴=1),
回答4=λdfa:dfa.apply(λr:ans(r,4),轴=1),
回答5=λdfa:dfa.apply(λr:ans(r,5),轴=1),
回答6=λdfa:dfa.apply(λr:ans(r,6),轴=1),
).下降(“应答”,轴=1)
打印(dfgrouped.to_字符串(index=False))
输出
QID Category Text QType Question: Country Answer0 Answer1 Answer2 Answer3 Answer4 Answer5 Answer6
16 Automotive Access to car Single Do you have access to a car? UK I own a car/cars I lease/ have a company car I have access to a car/cars No, I don’t have access to a car/cars Prefer not to say
更有活力
这对高级python
有了更深入的了解。使用**kwargs
和functools.partial
。实际上它仍然是静态的,列被定义为常量MAXANS
导入工具
MAXANS=8
def ansassign(dfa,行=0):
返回dfa.apply(lambda r:“如果行>=len(r[“AnswerT”]),否则r[“AnswerT”][row],axis=1)
dfgrouped=df.groupby(“QID”).agg(agg).reset_index().rename(columns={“Answer0”:“AnswerT”}.assign(
**{f“Answer{i}”:functools.partial(ansassign,row=i)表示范围内的i(MAXANS)}
).下降(“应答”,轴=1)
发布更多基本示例和预期结果。以上预期结果对我来说毫无意义。非常感谢!事实上,我的答案可能不止7个,我怎样才能使它变得动态,以获得与具有相同(QID
,问题:
)的行一样多的答案?@revolutionormonica我想不出一种真正动态的方法,因为无法找到列表中逐行的项目数。已经更新了,但是要注意很少有人是这类编码的专家。你的更新真的很酷!太先进了!我在考虑和大家分享这个故事。也许这能帮你弄到号码dynamically@RevolucionforMonica在许多方面,我更喜欢第一种方法——它更透明。80%的时间用于维护代码。。。使用很多高级概念的代码维护起来非常昂贵。是的,实际上我认为你是对的。我正在使用我共享的数据帧使您的代码动态化。我还不知道,但我相信我能做点什么。我发布了,而不是在这里用这个动态问题困扰你,如果你想要更多的点^^