import?pandas?as?pd

df?=?pd.DataFrame({"name":?["Satori",?"Koishi",?"Marisa"],
???????????????????"score":?[99,?98,?100],
???????????????????"rank":?[2,?3,?1]})

print(df)
"""
?????name??score??rank
0??Satori?????99?????2
1??Koishi?????98?????3
2??Marisa????100?????1
"""

print(df.to_records())
"""
[(0,?'Satori',??99,?2)?(1,?'Koishi',??98,?3)?(2,?'Marisa',?100,?1)]
"""
#?返回的時(shí)候?qū)⑺饕矌狭?，我們可以去?
print(df.to_records(index=False))
"""
[('Satori',??99,?2)?('Koishi',??98,?3)?('Marisa',?100,?1)]
"""
#?df.to_records?返回的是?numpy?的數(shù)組，可以再轉(zhuǎn)成列表
print(df.to_records(index=False).tolist())
"""
[('Satori',?99,?2),?('Koishi',?98,?3),?('Marisa',?100,?1)]
"""

#?返回?Python?的字典，key?是字段名，value?是對(duì)應(yīng)的每一列
print(df.to_dict())
"""
{'name':?{0:?'Satori',?1:?'Koishi',?2:?'Marisa'},
?'rank':?{0:?2,?1:?3,?2:?1},
?'score':?{0:?99,?1:?98,?2:?100}}
"""

#?但這里的?value?有一些問(wèn)題，就是它把索引也包含在里面了
#?我們可以去掉它
print(
????{k:?tuple(v.values())?for?k,?v?in?df.to_dict().items()}
)
"""
{'name':?('Satori',?'Koishi',?'Marisa'),
?'rank':?(2,?3,?1),
?'score':?(99,?98,?100)}
"""

#?當(dāng)然啦，to_dict()?還可以手動(dòng)實(shí)現(xiàn)
print(
????{col:?tuple(df[col])?for?col?in?df.columns}
)
"""
{'name':?('Satori',?'Koishi',?'Marisa'),
?'rank':?(2,?3,?1),
?'score':?(99,?98,?100)}
"""

print(df.to_dict(orient="records"))
"""
[{'name':?'Satori',?'rank':?2,?'score':?99},
?{'name':?'Koishi',?'rank':?3,?'score':?98},
?{'name':?'Marisa',?'rank':?1,?'score':?100}]
"""

import?pandas?as?pd

data?=?[{'name':?'Satori',?'rank':?2,?'score':?99},
????????{'name':?'Koishi',?'rank':?3,?'score':?98},
????????{'name':?'Marisa',?'rank':?1,?'score':?100}]

#?對(duì)于這種數(shù)據(jù)，可以通過(guò)?DataFrame?的?from_records?方法
#?列表里的字典代表了?DataFrame?的每一行，每個(gè)字典都具有相同的?key
#?而這些?key?則表示?DataFrame?的列
print(pd.DataFrame.from_records(data))
"""
?????name??rank??score
0??Satori?????2?????99
1??Koishi?????3?????98
2??Marisa?????1????100
"""
#?或者更簡(jiǎn)單的，直接調(diào)用?pd.DataFrame?即可
print(pd.DataFrame(data))
"""
?????name??rank??score
0??Satori?????2?????99
1??Koishi?????3?????98
2??Marisa?????1????100
"""
#?如果列表里面的字典，不具備相同的 key，會(huì)怎么樣呢？
data[2]["length"]?=?155
print(pd.DataFrame(data))
"""
?????name??rank??score??length
0??Satori?????2?????99?????NaN
1??Koishi?????3?????98?????NaN
2??Marisa?????1????100???155.0
"""
#?很簡(jiǎn)單，會(huì)將所有的?key?都考慮在內(nèi)
#?如果某一行沒(méi)有指定的?key，那么對(duì)應(yīng)的值就是空

import?pandas?as?pd

data?=?{'2020-01-01':?{'name':?'Satori',?'rank':?2,?'score':?99},
????????'2020-01-02':?{'name':?'Koishi',?'rank':?3,?'score':?98},
????????'2020-01-03':?{'name':?'Marisa',?'rank':?1,?'score':?100}}

print(pd.DataFrame.from_dict(data,?orient="index"))
"""
??????????????name??rank??score
2020-01-01??Satori?????2?????99
2020-01-02??Koishi?????3?????98
2020-01-03??Marisa?????1????100
"""

import?pandas?as?pd

data?=?{'name':?['Satori',?'Koishi',?'Marisa'],
????????'rank':?[2,?3,?1],
????????'score':?[99,?98,?100]}
#?直接調(diào)用?DataFrame?即可
print(pd.DataFrame(data))
"""
?????name??rank??score
0??Satori?????2?????99
1??Koishi?????3?????98
2??Marisa?????1????100
"""