Python dict 到 DataFrame Pandas [英] Python dict to DataFrame Pandas

问题描述

我需要帮助从这样的 dict 获取熊猫 DataFrame(2 级):

I need help for getting a pandas DataFrame from a dict like this one (2 levels):

{u'instrument': u'EUR_USD',
 u'candles': [{u'complete': True,
               u'closeMid': 1.26549,
               u'highMid': 1.27026,
               u'lowMid': 1.25006,
               u'volume': 138603,
               u'openMid': 1.26864,
               u'time': u'2014-09-29T21:00:00.000000Z'},
              ...
              {u'complete': True,
               u'closeMid': 1.244995,
               u'highMid': 1.25774,
               u'lowMid': 1.239455,
               u'volume': 167259,
               u'openMid': 1.242075,
               u'time': u'2014-11-10T22:00:00.000000Z'}
              ]
}

列标签和值应该是 instruments、Complete、CloseMid、HighMid、lowMid、音量、OpenMid、时间.

Columns labels and values should be instruments, Complete, CloseMid, HighMid, lowMid,Volume, OpenMid, time.

推荐答案

这是一个务实的解决方案.

Here is a pragmatic solution.

d = {u'instrument': u'EUR_USD', 
     u'candles': [
        {u'complete': True, u'closeMid': 1.26549, u'highMid': 1.27026, u'lowMid': 1.25006, u'volume': 138603, u'openMid': 1.26864, u'time': u'2014-09-29T21:00:00.000000Z'}, 
        {u'complete': True, u'closeMid': 1.275215, u'highMid': 1.27915, u'lowMid': 1.25838, u'volume': 164677, u'openMid': 1.265485, u'time': u'2014-10-06T21:00:00.000000Z'}, 
        {u'complete': True, u'closeMid': 1.279995, u'highMid': 1.288645, u'lowMid': 1.26249, u'volume': 207189, u'openMid': 1.27537, u'time': u'2014-10-13T21:00:00.000000Z'}, 
        {u'complete': True, u'closeMid': 1.269775, u'highMid': 1.28403, u'lowMid': 1.261385, u'volume': 125266, u'openMid': 1.280145, u'time': u'2014-10-20T21:00:00.000000Z'}, 
        {u'complete': True, u'closeMid': 1.24819, u'highMid': 1.27707, u'lowMid': 1.243775, u'volume': 210030, u'openMid': 1.270125, u'time': u'2014-10-27T21:00:00.000000Z'}, 
        {u'complete': True, u'closeMid': 1.242075, u'highMid': 1.25774, u'lowMid': 1.23582, u'volume': 246530, u'openMid': 1.24841, u'time': u'2014-11-03T22:00:00.000000Z'}, 
        {u'complete': True, u'closeMid': 1.244995, u'highMid': 1.25774, u'lowMid': 1.239455, u'volume': 167259, u'openMid': 1.242075, u'time': u'2014-11-10T22:00:00.000000Z'}
        ]}

df = pd.DataFrame.from_dict(d).join(pd.DataFrame.from_dict(d['candles'])).drop('candles', axis=1)
df

这里的问题完全不同，需要基于相同原理的新答案，但更复杂.

The problem is quite different here and requires a new answer based on the same principle, but more complex.

# Test data
d = {u'instruments': [
        {u'instrument': u'EUR_USD', 
         u'interestRate': {u'EUR': {u'ask': 0.004, u'bid': 0.1}, 
                           u'USD': {u'ask': 0.004, u'bid':0}}},
        {u'instrument': u'EUR_USD2', 
         u'interestRate': {u'EUR': {u'ask': 0.05, u'bid': 0.2}, 
                           u'USD2': {u'ask': 0.6, u'bid':0.1}}}
    ]}

# Creating an empty DataFrame
df = DataFrame()

# Iterating over the instruments list
for item in d['instruments']:
    df = pd.concat([df, pd.DataFrame.from_dict(item)
                    .join(pd.DataFrame.from_dict(item['interestRate'], orient='index'))])

# Performing some cleaning to get back a proper interestRate column   
df = df.drop('interestRate', axis=1).reset_index().rename(columns={'index':'interestRate'})

print(df)

  interestRate instrument  bid       ask
0          EUR    EUR_USD  0.1  4.00e-03
1          USD    EUR_USD  0.0  4.00e-03
2          EUR   EUR_USD2  0.2  5.00e-02
3         USD2   EUR_USD2  0.1  6.00e-01

这篇关于Python dict 到 DataFrame Pandas的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！