10分鐘入門已經(jīng)寫完了孔飒，那么基本的東西大概都了解憔儿。
入門以后我們的目標(biāo)變成了要玩的6,666666666666666666666666
所以
cookbook (short and sweet example)
官網(wǎng)上的描述是short and sweet example罪治，是用python3.4的，其他版本的python可能會需要一點(diǎn)小修改。
idioms(怎么翻譯)
if-then/if-then-else on one column, and assignment to another one or more columns:
In [1]: df = pd.DataFrame(
...: {'AAA' : [4,5,6,7], 'BBB' : [10,20,30,40],'CCC' : [100,50,-30,-50]}); df
...:
Out[1]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
if-then
In [2]: df.ix[df.AAA >= 5,'BBB'] = -1; df
Out[2]:
AAA BBB CCC
0 4 10 100
1 5 -1 50
2 6 -1 -30
3 7 -1 -50
In [3]: df.ix[df.AAA >= 5,['BBB','CCC']] = 555; df
Out[3]:
AAA BBB CCC
0 4 10 100
1 5 555 555
2 6 555 555
3 7 555 555
In [4]: df.ix[df.AAA < 5,['BBB','CCC']] = 2000; df
Out[4]:
AAA BBB CCC
0 4 2000 2000
1 5 555 555
2 6 555 555
3 7 555 555
或者可以使用 where
In [5]: df_mask = pd.DataFrame({'AAA' : [True] * 4, 'BBB' : [False] * 4,'CCC' : [True,False] * 2})
In [6]: df.where(df_mask,-1000)
Out[6]:
AAA BBB CCC
0 4 -1000 2000
1 5 -1000 -1000
2 6 -1000 555
3 7 -1000 -1000
或者使用np的where可以if-then-else
In [7]: df = pd.DataFrame(
...: {'AAA' : [4,5,6,7], 'BBB' : [10,20,30,40],'CCC' : [100,50,-30,-50]}); df
...:
Out[7]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [8]: df['logic'] = np.where(df['AAA'] > 5,'high','low'); df
Out[8]:
AAA BBB CCC logic
0 4 10 100 low
1 5 20 50 low
2 6 30 -30 high
3 7 40 -50 high
split
In [9]: df = pd.DataFrame(
...: {'AAA' : [4,5,6,7], 'BBB' : [10,20,30,40],'CCC' : [100,50,-30,-50]}); df
...:
Out[9]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [10]: dflow = df[df.AAA <= 5]
In [11]: dfhigh = df[df.AAA > 5]
In [12]: dflow; dfhigh
Out[12]:
AAA BBB CCC
2 6 30 -30
3 7 40 -50
Building Criteria （構(gòu)造規(guī)范/這翻譯水平真是夠了）
Select with multi-column criteria
In [13]: df = pd.DataFrame(
....: {'AAA' : [4,5,6,7], 'BBB' : [10,20,30,40],'CCC' : [100,50,-30,-50]}); df
....:
Out[13]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
...and (without assignment returns a Series) 不賦值返回的就是一個Series
In [14]: newseries = df.loc[(df['BBB'] < 25) & (df['CCC'] >= -40), 'AAA']; newseries
Out[14]:
0 4
1 5
Name: AAA, dtype: int64
...or (without assignment returns a Series)
In [15]: newseries = df.loc[(df['BBB'] > 25) | (df['CCC'] >= -40), 'AAA']; newseries;
...or (with assignment modifies the DataFrame.)
In [16]: df.loc[(df['BBB'] > 25) | (df['CCC'] >= 75), 'AAA'] = 0.1; df
Out[16]:
AAA BBB CCC
0 0.1 10 100
1 5.0 20 50
2 0.1 30 -30
3 0.1 40 -50
Select rows with data closest to certain value using argsort
In [17]: df = pd.DataFrame(
....: {'AAA' : [4,5,6,7], 'BBB' : [10,20,30,40],'CCC' : [100,50,-30,-50]}); df
....:
Out[17]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [18]: aValue = 43.0
In [19]: df.ix[(df.CCC-aValue).abs().argsort()]
Out[19]:
AAA BBB CCC
1 5 20 50
0 4 10 100
2 6 30 -30
3 7 40 -50
Dynamically reduce a list of criteria using a binary operators
In [20]: df = pd.DataFrame(
....: {'AAA' : [4,5,6,7], 'BBB' : [10,20,30,40],'CCC' : [100,50,-30,-50]}); df
....:
Out[20]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [21]: Crit1 = df.AAA <= 5.5
In [22]: Crit2 = df.BBB == 10.0
In [23]: Crit3 = df.CCC > -40.0
# One could hard code:
In [24]: AllCrit = Crit1 & Crit2 & Crit3
# ...Or it can be done with a list of dynamically built criteria
In [25]: CritList = [Crit1,Crit2,Crit3]
In [26]: AllCrit = functools.reduce(lambda x,y: x & y, CritList)
In [27]: df[AllCrit]
Out[27]:
AAA BBB CCC
0 4 10 100
selection
The indexing docs.
Using both row labels and value conditionals
In [28]: df = pd.DataFrame(
....: {'AAA' : [4,5,6,7], 'BBB' : [10,20,30,40],'CCC' : [100,50,-30,-50]}); df
....:
Out[28]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [29]: df[(df.AAA <= 6) & (df.index.isin([0,2,4]))]
Out[29]:
AAA BBB CCC
0 4 10 100
2 6 30 -30
Use loc for label-oriented slicing and iloc positional slicing
In [30]: data = {'AAA' : [4,5,6,7], 'BBB' : [10,20,30,40],'CCC' : [100,50,-30,-50]}
In [31]: df = pd.DataFrame(data=data,index=['foo','bar','boo','kar']); df
Out[31]:
AAA BBB CCC
foo 4 10 100
bar 5 20 50
boo 6 30 -30
kar 7 40 -50
There are 2 explicit slicing methods, with a third general case
- Positional-oriented (Python slicing style : exclusive of end)
- Label-oriented (Non-Python slicing style : inclusive of end)
- General (Either slicing style : depends on if the slice contains labels or positions)
In [32]: df.loc['bar':'kar'] #Label
Out[32]:
AAA BBB CCC
bar 5 20 50
boo 6 30 -30
kar 7 40 -50
# Generic
In [33]: df.ix[0:3] #Same as .iloc[0:3]
Out[33]:
AAA BBB CCC
foo 4 10 100
bar 5 20 50
boo 6 30 -30
In [34]: df.ix['bar':'kar'] #Same as .loc['bar':'kar']
Out[34]:
AAA BBB CCC
bar 5 20 50
boo 6 30 -30
kar 7 40 -50
Ambiguity arises when an index consists of integers with a non-zero start or non-unit increment.(所以還是盡量不要這么用叁温，徒增麻煩)
In [35]: df2 = pd.DataFrame(data=data,index=[1,2,3,4]); #Note index starts at 1.
In [36]: df2.iloc[1:3] #Position-oriented
Out[36]:
AAA BBB CCC
2 5 20 50
3 6 30 -30
In [37]: df2.loc[1:3] #Label-oriented
Out[37]:
AAA BBB CCC
1 4 10 100
2 5 20 50
3 6 30 -30
In [38]: df2.ix[1:3] #General, will mimic loc (label-oriented)
Out[38]:
AAA BBB CCC
1 4 10 100
2 5 20 50
3 6 30 -30
In [39]: df2.ix[0:3] #General, will mimic iloc (position-oriented), as loc[0:3] would raise a KeyError
Out[39]:
AAA BBB CCC
1 4 10 100
2 5 20 50
3 6 30 -30
Using inverse operator (~) to take the complement of a mask
In [40]: df = pd.DataFrame(
....: {'AAA' : [4,5,6,7], 'BBB' : [10,20,30,40], 'CCC' : [100,50,-30,-50]}); df
....:
Out[40]:
AAA BBB CCC
0 4 10 100
1 5 20 50
2 6 30 -30
3 7 40 -50
In [41]: df[~((df.AAA <= 6) & (df.index.isin([0,2,4])))]
Out[41]:
AAA BBB CCC
1 5 20 50
3 7 40 -50
下面是panel江掩，就不學(xué)了学辱，跳過去，下一篇學(xué)習(xí)下面的東西环形。
