使用json数据集中的行值,在特定条件下更改行值

您可以尝试:

mask = df['sentiment'].notnull()
df.loc[mask, 'sentiment'] = df.loc[mask, 'sentiment'].apply(lambda x: x['basic'])

您可以这样做:

df = pd.read_json(path)  # creates the dataframe with dict objects in sentiment column 
pd.concat([df.drop(['sentiment'], axis=1), df['sentiment'].apply(pd.Series)], axis=1)  # create new columns for each sentiment type

[{
    "date": "2018-01-01", 
    "body": "some txt", 
    "id": 111, 
    "sentiment": null
}, 
{
    "date": "2018-01-02", 
    "body": "some txt", 
    "id": 112, 
    "sentiment": {
        "basic": "Bearish"
    }
},
{
    "date": "2018-01-03", 
    "body": "some other txt", 
    "id": 113, 
    "sentiment": {
        "basic" : "Bullish",
        "non_basic" : "Bearish"
    }
}]

1号线后的df:

             body       date   id                                     sentiment
0        some txt 2018-01-01  111                                          None
1        some txt 2018-01-02  112                          {'basic': 'Bearish'}
2  some other txt 2018-01-03  113  {'basic': 'Bullish', 'non_basic': 'Bearish'}

2号线后的df:

             body       date   id    basic non_basic
0        some txt 2018-01-01  111      NaN       NaN
1        some txt 2018-01-02  112  Bearish       NaN
2  some other txt 2018-01-03  113  Bullish   Bearish

嗯。

fillna pop + join

这里有一个可扩展的解决方案,它可以避免按行执行 apply 并将任意数量的键转换为序列:

df = pd.DataFrame({'body': [0, 1],
                   'sentiment': [None, {u'basic': u'Bullish'}]})

df['sentiment'] = df['sentiment'].fillna(pd.Series([{}]*len(df.index), index=df.index))

df = df.join(pd.DataFrame(df.pop('sentiment').values.tolist()))

print(df)

   body    basic
0     0      NaN
1     1  Bullish