‘‘‘ 【课程2.14】 数值计算和统计基础 常用数学、统计方法 ‘‘‘
# 基本参数：axis、skipna import numpy as np import pandas as pd df = pd.DataFrame({‘key1‘:[4,5,3,np.nan,2], ‘key2‘:[1,2,np.nan,4,5], ‘key3‘:[1,2,3,‘j‘,‘k‘]}, index = [‘a‘,‘b‘,‘c‘,‘d‘,‘e‘]) print(df) print(df[‘key1‘].dtype,df[‘key2‘].dtype,df[‘key3‘].dtype) print(‘-----‘) m1 = df.mean() print(m1,type(m1)) print(‘单独统计一列:‘,df[‘key2‘].mean()) print(‘-----‘) # np.nan ：空值 # .mean()计算均值 # 只统计数字列 # 可以通过索引单独统计一列 m2 = df.mean(axis=1) print(m2) print(‘-----‘) # axis参数：默认为0，以列来计算，axis=1，以行来计算，这里就按照行来汇总了 m3 = df.mean(skipna=False) print(m3) print(‘-----‘) # skipna参数：是否忽略NaN，默认True，如False，有NaN的列统计结果仍未NaN
输出：
key1 key2 key3 a 4.0 1.0 1 b 5.0 2.0 2 c 3.0 NaN 3 d NaN 4.0 j e 2.0 5.0 k float64 float64 object ----- key1 3.5 key2 3.0 dtype: float64 <class ‘pandas.core.series.Series‘> 单独统计一列: 3.0 ----- a 2.5 b 3.5 c 3.0 d 4.0 e 3.5 dtype: float64 ----- key1 NaN key2 NaN dtype: float64 -----
# 主要数学计算方法，可用于Series和DataFrame（1） df = pd.DataFrame({‘key1‘:np.arange(10), ‘key2‘:np.random.rand(10)*10}) print(df) print(‘-----‘) print(df.count(),‘→ count统计非Na值的数量\n‘) print(df.min(),‘→ min统计最小值\n‘,df[‘key2‘].max(),‘→ max统计最大值\n‘) print(df.quantile(q=0.75),‘→ quantile统计分位数，参数q确定位置\n‘) print(df.sum(),‘→ sum求和\n‘) print(df.mean(),‘→ mean求平均值\n‘) print(df.median(),‘→ median求算数中位数，50%分位数\n‘) print(df.std(),‘\n‘,df.var(),‘→ std,var分别求标准差，方差\n‘) print(df.skew(),‘→ skew样本的偏度\n‘) print(df.kurt(),‘→ kurt样本的峰度\n‘)
输出：
key1 key2 0 0 4.667989 1 1 4.336625 2 2 0.746852 3 3 9.670919 4 4 8.732045 5 5 0.013751 6 6 8.963752 7 7 0.279303 8 8 8.586821 9 9 8.899657 ----- key1 10 key2 10 dtype: int64 → count统计非Na值的数量 key1 0.000000 key2 0.013751 dtype: float64 → min统计最小值 9.67091932107 → max统计最大值 key1 6.750000 key2 8.857754 dtype: float64 → quantile统计分位数，参数q确定位置 key1 45.000000 key2 54.897714 dtype: float64 → sum求和 key1 4.500000 key2 5.489771 dtype: float64 → mean求平均值 key1 4.500000 key2 6.627405 dtype: float64 → median求算数中位数，50%分位数 key1 3.027650 key2 3.984945 dtype: float64 key1 9.166667 key2 15.879783 dtype: float64 → std,var分别求标准差，方差 key1 0.000000 key2 -0.430166 dtype: float64 → skew样本的偏度 key1 -1.200000 key2 -1.800296 dtype: float64 → kurt样本的峰度
# 主要数学计算方法，可用于Series和DataFrame（2） df[‘key1_s‘] = df[‘key1‘].cumsum() df[‘key2_s‘] = df[‘key2‘].cumsum() print(df,‘→ cumsum样本的累计和\n‘) df[‘key1_p‘] = df[‘key1‘].cumprod() df[‘key2_p‘] = df[‘key2‘].cumprod() print(df,‘→ cumprod样本的累计积\n‘) print(df.cummax(),‘\n‘,df.cummin(),‘→ cummax,cummin分别求累计最大值，累计最小值\n‘) # 会填充key1，和key2的值
输出：
key1 key2 key1_s key2_s 0 0 4.667989 0 4.667989 1 1 4.336625 1 9.004614 2 2 0.746852 3 9.751466 3 3 9.670919 6 19.422386 4 4 8.732045 10 28.154431 5 5 0.013751 15 28.168182 6 6 8.963752 21 37.131934 7 7 0.279303 28 37.411236 8 8 8.586821 36 45.998057 9 9 8.899657 45 54.897714 → cumsum样本的累计和 key1 key2 key1_s key2_s key1_p key2_p 0 0 4.667989 0 4.667989 0 4.667989 1 1 4.336625 1 9.004614 0 20.243318 2 2 0.746852 3 9.751466 0 15.118767 3 3 9.670919 6 19.422386 0 146.212377 4 4 8.732045 10 28.154431 0 1276.733069 5 5 0.013751 15 28.168182 0 17.556729 6 6 8.963752 21 37.131934 0 157.374157 7 7 0.279303 28 37.411236 0 43.955024 8 8 8.586821 36 45.998057 0 377.433921 9 9 8.899657 45 54.897714 0 3359.032396 → cumprod样本的累计积 key1 key2 key1_s key2_s key1_p key2_p 0 0.0 4.667989 0.0 4.667989 0.0 4.667989 1 1.0 4.667989 1.0 9.004614 0.0 20.243318 2 2.0 4.667989 3.0 9.751466 0.0 20.243318 3 3.0 9.670919 6.0 19.422386 0.0 146.212377 4 4.0 9.670919 10.0 28.154431 0.0 1276.733069 5 5.0 9.670919 15.0 28.168182 0.0 1276.733069 6 6.0 9.670919 21.0 37.131934 0.0 1276.733069 7 7.0 9.670919 28.0 37.411236 0.0 1276.733069 8 8.0 9.670919 36.0 45.998057 0.0 1276.733069 9 9.0 9.670919 45.0 54.897714 0.0 3359.032396 key1 key2 key1_s key2_s key1_p key2_p 0 0.0 4.667989 0.0 4.667989 0.0 4.667989 1 0.0 4.336625 0.0 4.667989 0.0 4.667989 2 0.0 0.746852 0.0 4.667989 0.0 4.667989 3 0.0 0.746852 0.0 4.667989 0.0 4.667989 4 0.0 0.746852 0.0 4.667989 0.0 4.667989 5 0.0 0.013751 0.0 4.667989 0.0 4.667989 6 0.0 0.013751 0.0 4.667989 0.0 4.667989 7 0.0 0.013751 0.0 4.667989 0.0 4.667989 8 0.0 0.013751 0.0 4.667989 0.0 4.667989 9 0.0 0.013751 0.0 4.667989 0.0 4.667989 → cummax,cummin分别求累计最大值，累计最小值
# 唯一值：.unique() s = pd.Series(list(‘asdvasdcfgg‘)) sq = s.unique() print(s) print(sq,type(sq)) print(pd.Series(sq)) # 得到一个唯一值数组 # 通过pd.Series重新变成新的Series sq.sort() print(sq) # 重新排序
输出：
0 a 1 s 2 d 3 v 4 a 5 s 6 d 7 c 8 f 9 g 10 g dtype: object [‘a‘ ‘s‘ ‘d‘ ‘v‘ ‘c‘ ‘f‘ ‘g‘] <class ‘numpy.ndarray‘> 0 a 1 s 2 d 3 v 4 c 5 f 6 g dtype: object [‘a‘ ‘c‘ ‘d‘ ‘f‘ ‘g‘ ‘s‘ ‘v‘]
# 值计数：.value_counts() sc = s.value_counts(sort = False) # 也可以这样写：pd.value_counts(sc, sort = False) print(sc) # 得到一个新的Series，计算出不同值出现的频率 # sort参数：排序，默认为True
输出：
s 2 d 2 v 1 c 1 a 2 g 2 f 1 dtype: int64
# 成员资格：.isin() s = pd.Series(np.arange(10,15)) df = pd.DataFrame({‘key1‘:list(‘asdcbvasd‘), ‘key2‘:np.arange(4,13)}) print(s) print(df) print(‘-----‘) print(s.isin([5,14])) print(df.isin([‘a‘,‘bc‘,‘10‘,8])) # 用[]表示 # 得到一个布尔值的Series或者Dataframe
输出：
0 10 1 11 2 12 3 13 4 14 dtype: int32 key1 key2 0 a 4 1 s 5 2 d 6 3 c 7 4 b 8 5 v 9 6 a 10 7 s 11 8 d 12 ----- 0 False 1 False 2 False 3 False 4 True dtype: bool key1 key2 0 True False 1 False False 2 False False 3 False False 4 False True 5 False False 6 True False 7 False False 8 False False
原文：https://www.cnblogs.com/654321cc/p/9351806.html