python基础知识之索引与切片详解
作者：编程学习网
在python的学习过程,有些同学对索引和切换会感到困惑,今天我们就来弄清楚它,下面这篇文章主要给大家介绍了关于python基础知识之索引与切片的相关资料,需要的朋友可以参考下
基本索引
In [4]: sentence = 'You are a nice girl'In [5]: L = sentence.split()In [6]: LOut[6]: ['You', 'are', 'a', 'nice', 'girl'] # 从0开始索引In [7]: L[2]Out[7]: 'a' # 负数索引，从列表右侧开始计数In [8]: L[-2]Out[8]: 'nice' # -1表示列表最后一项In [9]: L[-1]Out[9]: 'girl' # 当正整数索引超过返回时In [10]: L[100]---------------------------------------------------------------------------IndexError Traceback (most recent call last) <ipython-input-10-78da2f882365> in <module>()----> 1 L[100]IndexError: list index out of range# 当负整数索引超过返回时In [11]: L[-100]---------------------------------------------------------------------------IndexError Traceback (most recent call last) <ipython-input-11-46b47b0ecb55> in <module>()----> 1 L[-100]IndexError: list index out of range# slice 索引In [193]: sl = slice(0,-1,1)In [194]: L[sl]Out[194]: ['You', 'are', 'a', 'nice']In [199]: sl = slice(0,100)In [200]: L[sl]Out[200]: ['You', 'are', 'a', 'nice', 'girl']
嵌套索引
In [14]: L = [[1,2,3],{'I':'You are a nice girl','She':'Thank you!'},(11,22),'My name is Kyles']
In [15]: L
Out[15]:
[[1, 2, 3],
{'I': 'You are a nice girl', 'She': 'Thank you!'},
(11, 22),
'My name is Kyles']# 索引第1项，索引为0In [16]: L[0]
Out[16]: [1, 2, 3]# 索引第1项的第2子项In [17]: L[0][1]
Out[17]: 2# 索引第2项词典In [18]: L[1]
Out[18]: {'I': 'You are a nice girl', 'She': 'Thank you!'}# 索引第2项词典的 “She”In [19]: L[1]['She']
Out[19]: 'Thank you!'# 索引第3项In [20]: L[2]
Out[20]: (11, 22)# 索引第3项，第一个元组In [22]: L[2][0]
Out[22]: 11# 索引第4项In [23]: L[3]
Out[23]: 'My name is Kyles'# 索引第4项，前3个字符In [24]: L[3][:3]
Out[24]: 'My '切片
# 切片选择,从1到列表末尾In [13]: L[1:]Out[13]: ['are', 'a', 'nice', 'girl']# 负数索引，选取列表后两项In [28]: L[-2:]Out[28]: ['nice', 'girl']# 异常测试,这里没有报错！In [29]: L[-100:]Out[29]: ['You', 'are', 'a', 'nice', 'girl']# 返回空In [30]: L[-100:-200]Out[30]: []# 正向索引In [32]: L[-100:3]Out[32]: ['You', 'are', 'a']# 返回空In [33]: L[-1:3]Out[33]: []# 返回空In [41]: L[0:0]Out[41]: []
看似简单的索引，有的人不以为然，我们这里采用精准的数字索引，很容易排查错误。若索引是经过计算出的一个变量，就千万要小心了，否则失之毫厘差之千里。
numpy.array 索引 一维
In [34]: import numpy as npIn [35]: arr = np.arange(10)In [36]: arrOut[36]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])In [40]: arr.shapeOut[40]: (10,)# [0,1) In [37]: arr[0:1]Out[37]: array([0])# [0,0) In [38]: arr[0:0]Out[38]: array([], dtype=int32)# 右侧超出范围之后In [42]: arr[:1000]Out[42]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])# 左侧超出之后In [43]: arr[-100:1000]Out[43]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])# 两侧都超出In [44]: arr[100:101]Out[44]: array([], dtype=int32)# []In [45]: arr[-100:-2]Out[45]: array([0, 1, 2, 3, 4, 5, 6, 7])# []In [46]: arr[-100:-50]Out[46]: array([], dtype=int32)
numpy.array 索引 二维
In [49]: arr = np.arange(15).reshape(3,5)
In [50]: arr
Out[50]:
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]])
In [51]: arr.shape
Out[51]: (3, 5)
# axis = 0 增长的方向
In [52]: arr[0]
Out[52]: array([0, 1, 2, 3, 4])
# 选取第2行
In [53]: arr[1]
Out[53]: array([5, 6, 7, 8, 9])
# axis = 1 增长的方向，选取每一行的第1列
In [54]: arr[:,0]
Out[54]: array([ 0, 5, 10])
# axis = 1 增长的方向，选取每一行的第2列
In [55]: arr[:,1]
Out[55]: array([ 1, 6, 11])
# 选取每一行的第1,2列
In [56]: arr[:,0:2]
Out[56]:
array([[ 0, 1],
[ 5, 6],
[10, 11]])
# 右侧超出范围之后
In [57]: arr[:,0:100]
Out[57]:
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]])
# 左侧超出范围之后
In [62]: arr[:,-10:2]
Out[62]:
array([[ 0, 1],
[ 5, 6],
[10, 11]])
# []
In [58]: arr[:,0:0]
Out[58]: array([], shape=(3, 0), dtype=int32)
# []
In [59]: arr[0:0,0:1]
Out[59]: array([], shape=(0, 1), dtype=int32)
# 异常
In [63]: arr[:,-10]---------------------------------------------------------------------------IndexError Traceback (most recent call last)
<ipython-input-63-2ffa6627dc7f> in <module>()----> 1 arr[:,-10]IndexError: index -10 is out of bounds for axis 1 with size 5numpy.array 索引 三维…N维
In [67]: import numpy as np
In [68]: arr = np.arange(30).reshape(2,3,5)
In [69]: arr
Out[69]:
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]], [[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29]]])
# 根据 axis = 0 选取
In [70]: arr[0]
Out[70]:
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]])
In [71]: arr[1]
Out[71]:
array([[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29]])
# 根据 axis = 1 选取
In [72]: arr[:,0]
Out[72]:
array([[ 0, 1, 2, 3, 4],
[15, 16, 17, 18, 19]])
In [73]: arr[:,1]
Out[73]:
array([[ 5, 6, 7, 8, 9],
[20, 21, 22, 23, 24]])
# 异常指出 axis = 1 超出范围
In [74]: arr[:,4]---------------------------------------------------------------------------IndexError Traceback (most recent call last)
<ipython-input-74-9d489478e7c7> in <module>()----> 1 arr[:,4]IndexError: index 4 is out of bounds for axis 1 with size 3 # 根据 axis = 2 选取
In [75]: arr[:,:,0]
Out[75]:
array([[ 0, 5, 10],
[15, 20, 25]])
# 降维
In [76]: arr[:,:,0].shape
Out[76]: (2, 3)
In [78]: arr[:,:,0:2]
Out[78]:
array([[[ 0, 1],
[ 5, 6],
[10, 11]], [[15, 16],
[20, 21],
[25, 26]]])
In [79]: arr[:,:,0:2].shape
Out[79]: (2, 3, 2)
# 左/右侧超出范围
In [81]: arr[:,:,0:100]
Out[81]:
array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]], [[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24],
[25, 26, 27, 28, 29]]])
# 异常 axis = 0In [82]: arr[100,:,0:100]---------------------------------------------------------------------------IndexError Traceback (most recent call last)
<ipython-input-82-21efcc74439d> in <module>()----> 1 arr[100,:,0:100]IndexError: index 100 is out of bounds for axis 0 with size 2pandas Series 索引
In [84]: s = pd.Series(['You','are','a','nice','girl'])In [85]: sOut[85]:0 You1 are2 a3 nice4 girl
dtype: object# 按照索引选择In [86]: s[0]Out[86]: 'You'# []In [87]: s[0:0]Out[87]: Series([], dtype: object)In [88]: s[0:-1]Out[88]:0 You1 are2 a3 nice
dtype: object# 易错点,ix包含区间为 []In [91]: s.ix[0:0]Out[91]:0 You
dtype: objectIn [92]: s.ix[0:1]Out[92]:0 You1 are
dtype: object# ix索引不存在indexIn [95]: s.ix[400]
KeyError: 400# 按照从0开始的索引In [95]: s.iloc[0]Out[95]: 'You'In [96]: s.iloc[1]Out[96]: 'are'In [97]: s.iloc[100]
IndexError: single positional indexer is out-of-boundsIn [98]: s = pd.Series(['You','are','a','nice','girl'], index=list('abcde'))In [99]: sOut[99]:
a You
b are
c a
d nice
e girl
dtype: objectIn [100]: s.iloc[0]Out[100]: 'You'In [101]: s.iloc[1]Out[101]: 'are'# 按照 label 索引In [103]: s.loc['a']Out[103]: 'You'In [104]: s.loc['b']Out[104]: 'are'In [105]: s.loc[['b','a']]Out[105]:
b are
a You
dtype: object# loc切片索引In [106]: s.loc['a':'c']Out[106]:
a You
b are
c a
dtype: objectIn [108]: s.indexOut[108]: Index(['a', 'b', 'c', 'd', 'e'], dtype='object')pandas DataFrame 索引
In [114]: import pandas as pdIn [115]: df = pd.DataFrame({'open':[1,2,3],'high':[4,5,6],'low':[6,3,1]}, index=pd.period_range('30/12/2017',perio
...: ds=3,freq='H'))In [116]: dfOut[116]:
high low open2017-12-30 00:00 4 6 12017-12-30 01:00 5 3 22017-12-30 02:00 6 1 3# 按列索引In [117]: df['high']Out[117]:2017-12-30 00:00 42017-12-30 01:00 52017-12-30 02:00 6Freq: H, Name: high, dtype: int64In [118]: df.highOut[118]:2017-12-30 00:00 42017-12-30 01:00 52017-12-30 02:00 6Freq: H, Name: high, dtype: int64In [120]: df[['high','open']]Out[120]:
high open2017-12-30 00:00 4 12017-12-30 01:00 5 22017-12-30 02:00 6 3In [122]: df.ix[:]
D:\CodeTool\Python\Python36\Scripts\ipython:1: DeprecationWarning:
.ix is deprecated. Please use
.loc for label based indexing or.iloc for positional indexingIn [123]: df.iloc[0:0]Out[123]:Empty DataFrame
Columns: [high, low, open]Index: []In [124]: df.ix[0:0]Out[124]:Empty DataFrame
Columns: [high, low, open]Index: []
# 按照 label 索引In [127]: df.indexOut[127]: PeriodIndex(['2017-12-30 00:00', '2017-12-30 01:00', '2017-12-30 02:00'], dtype='period[H]', freq='H')In [128]: df.loc['2017-12-30 00:00']Out[128]:
high 4low 6open 1Name: 2017-12-30 00:00, dtype: int64
# 检查参数In [155]: df.loc['2017-12-30 00:00:11']Out[155]:
high 4low 6open 1Name: 2017-12-30 00:00, dtype: int64In [156]: df.loc['2017-12-30 00:00:66']
KeyError: 'the label [2017-12-30 00:00:66] is not in the [index]'填坑
In [158]: df = pd.DataFrame({'a':[1,2,3],'b':[4,5,6]}, index=[2,3,4])In [159]: dfOut[159]:
a b2 1 43 2 54 3 6# iloc 取第一行正确用法In [160]: df.iloc[0]Out[160]:
a 1b 4Name: 2, dtype: int64
# loc 正确用法In [165]: df.loc[[2,3]]Out[165]:
a b2 1 43 2 5# 注意此处 index 是什么类型In [167]: df.loc['2']
KeyError: 'the label [2] is not in the [index]'# 索引 Int64IndexOut[172]: Int64Index([2, 3, 4], dtype='int64')
# 索引为字符串In [168]: df = pd.DataFrame({'a':[1,2,3],'b':[4,5,6]}, index=list('234'))In [169]: dfOut[169]:
a b2 1 43 2 54 3 6In [170]: df.indexOut[170]: Index(['2', '3', '4'], dtype='object')
# 此处没有报错，千万注意 index 类型In [176]: df.loc['2']Out[176]:
a 1b 4Name: 2, dtype: int64
# ix 是一个功能强大的函数，但是争议却很大，往往是错误之源
# 咦，怎么输出与预想不一致！In [177]: df.ix[2]
D:\CodeTool\Python\Python36\Scripts\ipython:1: DeprecationWarning:
.ix is deprecated. Please use
.loc for label based indexing or.iloc for positional indexing
See the documentation here:
http://pandas.pydata.org/pandas-docs/stable/indexing.html#ix-indexer-is-deprecatedOut[177]:
a 3b 6Name: 4, dtype: int64
# 注意开闭区间In [180]: df.loc['2':'3']Out[180]:
a b2 1 43 2 5总结
pandas中ix是错误之源，大型项目大量使用它时，往往造成不可预料的后果。0.20.x版本也标记为抛弃该函数，二义性 和 []区间，违背 “Explicit is better than implicit.” 原则。建议使用意义明确的 iloc和loc 函数。
当使用字符串时切片时是 []区间 ，一般是 [)区间
当在numpy.ndarry、list、tuple、pandas.Series、pandas.DataFrame 混合使用时，采用变量进行索引或者切割，取值或赋值时，别太自信了，千万小心错误，需要大量的测试。
我在工程中使用matlab的矩阵和python混合使用以上对象，出现最多就是shape不对应，index，columns 错误。
最好不要混用不同数据结构，容易出错，更增加转化的性能开销
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