use*_*037 12 python dictionary pandas
我有一个csv,我正在读一个熊猫数据帧.然而,其中一列是字典形式.这是一个例子:
ColA, ColB, ColC, ColdD
20, 30, {"ab":"1", "we":"2", "as":"3"},"String"
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如何将其转换为如下所示的数据框:
ColA, ColB, AB, WE, AS, ColdD
20, 30, "1", "2", "3", "String"
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编辑 我修复了问题,它看起来像这样但是是一个需要解析的字符串,而不是dict对象.
psy*_*dia 17
根据/sf/answers/2676215601/,您可以使用.apply(pd.Series)将包含dict的列映射到新列,然后将这些新列连接回原始数据帧减去原始dict包含列:
dw=pd.DataFrame( [[20, 30, {"ab":"1", "we":"2", "as":"3"},"String"]],
columns=['ColA', 'ColB', 'ColC', 'ColdD'])
pd.concat([dw.drop(['ColC'], axis=1), dw['ColC'].apply(pd.Series)], axis=1)
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返回:
ColA ColB ColdD ab as we
20 30 String 1 3 2
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Bob*_*ner 12
所以从你的一行df开始
Col A Col B Col C Col D
0 20 30 {u'we': 2, u'ab': 1, u'as': 3} String1
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编辑:根据OP的评论,我假设我们需要先转换字符串
import ast
df["ColC"] = df["ColC"].map(lambda d : ast.literal_eval(d))
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然后我们将Col C转换为dict,转置它然后将它连接到原始df
dfNew = df.join(pd.DataFrame(df["Col C"].to_dict()).T)
dfNew
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这给了你这个
Col A Col B Col C Col D ab as we
0 20 30 {u'we': 2, u'ab': 1, u'as': 3} String1 1 3 2
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然后我们只需在dfNew中选择我们想要的列
dfNew[["Col A", "Col B", "ab", "we", "as", "Col D"]]
Col A Col B ab we as Col D
0 20 30 1 2 3 String1
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怎么样:
import pandas as pd
# Create mock dataframe
df = pd.DataFrame([
[20, 30, {'ab':1, 'we':2, 'as':3}, 'String1'],
[21, 31, {'ab':4, 'we':5, 'as':6}, 'String2'],
[22, 32, {'ab':7, 'we':8, 'as':9}, 'String2'],
], columns=['Col A', 'Col B', 'Col C', 'Col D'])
# Create dataframe where you'll store the dictionary values
ddf = pd.DataFrame(columns=['AB','WE','AS'])
# Populate ddf dataframe
for (i,r) in df.iterrows():
e = r['Col C']
ddf.loc[i] = [e['ab'], e['we'], e['as']]
# Replace df with the output of concat(df, ddf)
df = pd.concat([df, ddf], axis=1)
# New column order, also drops old Col C column
df = df[['Col A', 'Col B', 'AB', 'WE', 'AS', 'Col D']]
print(df)
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输出:
A 上校 B 上校 AB WE AS D 上校 0 20 30 1 2 3 字符串1 1 21 31 4 5 6 字符串2 2 22 32 7 8 9 字符串2
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