python基础知识之索引与切片详解

Olivia ·
更新时间:2024-09-20
· 1980 次阅读

目录

基本索引

嵌套索引

切片

numpy.array 索引 一维

numpy.array 索引 二维

pandas Series 索引

pandas DataFrame 索引

填坑

总结

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 5

numpy.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 2 pandas 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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