透视包含字符串的Pandas Dataframe - "没有要聚合的数字类型"错误

Question

关于这个错误有很多问题,但在环顾四周之后,我仍然无法找到/解决问题.我正在尝试使用字符串来转动数据框,以使某些行数据成为列,但到目前为止还没有成功.

我的df的形状

<class 'pandas.core.frame.DataFrame'>
Int64Index: 515932 entries, 0 to 515931
Data columns (total 5 columns):
id                 515932 non-null object
cc_contact_id      515932 non-null object
Network_Name       515932 non-null object
question           515932 non-null object
response_answer    515932 non-null object
dtypes: object(5)
memory usage: 23.6+ MB

样本格式

id  contact_id  question    response_answer
16  137519  2206    State   Ca
17  137520  2206    State   Ca
18  137521  2206    State   Ca
19  137522  2206    State   Ca
20  137523  2208    City    Lancaster
21  137524  2208    City    Lancaster
22  137525  2208    City    Lancaster
23  137526  2208    City    Lancaster
24  137527  2208    Trip_End Location   Home
25  137528  2208    Trip_End Location   Home
26  137529  2208    Trip_End Location   Home
27  137530  2208    Trip_End Location   Home

我想转向什么

id  contact_id      State   City       Trip_End Location
16  137519  2206    Ca      None       None None
20  137523  2208    None    Lancaster  None None
24  137527  2208    None    None       None Home
etc. etc. 

如果问题值成为列,则response_answer位于其对应的列中,并保留ID

我试过了什么

unified_df = pd.DataFrame(unified_data, columns=target_table_headers, dtype=object)

pivot_table = unified_df.pivot_table('response_answer',['id','cc_contact_id'],'question')
# OR
pivot_table = unified_df.pivot_table('response_answer','question')

DataError:无需聚合的数字类型

使用字符串值转动数据框的方法是什么？

Answer 1

默认aggfunc的pivot_table是np.sum,它不知道如何处理字符串做,你有没有指定的指数应该是什么正常.尝试类似的东西:

pivot_table = unified_df.pivot_table(index=['id', 'contact_id'],
                                     columns='question', 
                                     values='response_answer',
                                     aggfunc=lambda x: ' '.join(x))

这明确地为每id, contact_id对设置一行并在该组上创建response_answer值question.在aggfunc刚刚确保如果您有多个答案中的原始数据相同的问题,我们只是用空格串联在一起.语法pivot_table可能因您的熊猫版本而异.

这是一个简单的例子:

In [24]: import pandas as pd

In [25]: import random

In [26]: df = pd.DataFrame({'id':[100*random.randint(10, 50) for _ in range(100)], 'question': [str(random.randint(0,3)) for _ in range(100)], 'response': [str(random.randint(100,120)) for _ in range(100)]})

In [27]: df.head()
Out[27]:
     id question response
0  3100        1      116
1  4500        2      113
2  5000        1      120
3  3900        2      103
4  4300        0      117

In [28]: df.info()
<class 'pandas.core.frame.DataFrame'>
Int64Index: 100 entries, 0 to 99
Data columns (total 3 columns):
id          100 non-null int64
question    100 non-null object
response    100 non-null object
dtypes: int64(1), object(2)
memory usage: 3.1+ KB

In [29]: df.pivot_table(index='id', columns='question', values='response', aggfunc=lambda x: ' '.join(x)).head()
Out[29]:
question        0        1    2        3
id
1000      110 120      NaN  100      NaN
1100          NaN  106 108  104      NaN
1200      104 113      119  NaN      101
1300          102      NaN  116  108 120
1400          NaN      NaN  116      NaN

Answer 2

有几种方法.

1

df1 = df.groupby(["id","contact_id","Network_Name","question"])['response_answer'].aggregate(lambda x: x).unstack().reset_index()
df1.columns=df1.columns.tolist()
print (df1)

2

df1 = df.set_index(["id","contact_id","Network_Name","question"])['response_answer'].unstack().reset_index()
df1.columns=df1.columns.tolist()
print (df1)

3

df1 = df.groupby(["id","contact_id","Network_Name","question"])['response_answer'].aggregate('first').unstack().reset_index()
df1.columns=df1.columns.tolist()
print (df1)

4

df1 = df.pivot_table(index=["id","contact_id","Network_Name"], columns='question', values=['response_answer'], aggfunc='first')
df1.columns = df1.columns.droplevel()
df1 = df1.reset_index()
df1.columns=df1.columns.tolist()
print (df1)

同样的.

    id  contact_id  Network_Name       City State Trip_End_Location
0   16      137519          2206       None    Ca              None
1   17      137520          2206       None    Ca              None
2   18      137521          2206       None    Ca              None
3   19      137522          2206       None    Ca              None
4   20      137523          2208  Lancaster  None              None
5   21      137524          2208  Lancaster  None              None
6   22      137525          2208  Lancaster  None              None
7   23      137526          2208  Lancaster  None              None
8   24      137527          2208       None  None              Home
9   25      137528          2208       None  None              Home
10  26      137529          2208       None  None              Home
11  27      137530          2208       None  None              Home