Pandas进阶修炼120题系列一共涵盖了数据处理、计算、可视化等常用操作，希望通过120道精心挑选的习题吃透pandas。并且针对部分习题给出了多种解法与注解，动手敲一遍代码一定会让你有所收获！
data = {'grammer':['Python','C','Java','GO',np.nan,'SQL','PHP','Python'],
'score':[1,2,np.nan,4,5,6,7,10]}import numpy as np
import pandas as pd
df = pd.DataFrame(data)
# 假如是直接创建
df = pd.DataFrame({
'grammer': ['Python','C','Java','GO',np.nan,'SQL','PHP','Python'],
'score': [1,2,np.nan,4,5,6,7,10]})grammer score
0 Python 1.0
7 Python 10.0#> 1
df[df['grammer'] == 'Python']
#> 2
results = df['grammer'].str.contains('Python')
results.fillna(value=False,inplace = True)
df[results]Index(['grammer', 'score'], dtype='object')df.columnsdf.rename(columns={'score':'popularity'}, inplace = True)df['grammer'].value_counts()# pandas里有一个插值方法，就是计算缺失值上下两数的均值
df['popularity'] = df['popularity'].fillna(df['popularity'].interpolate())df[df['popularity'] > 3]df.drop_duplicates(['grammer'])df['popularity'].mean()
# 4.75df['grammer'].to_list()
# ['Python', 'C', 'Java', 'GO', nan, 'SQL', 'PHP', 'Python']df.to_excel('filename.xlsx')df.shape
# (8, 2)df[(df['popularity'] > 3) & (df['popularity'] < 7)]temp = df['popularity']
df.drop(labels=['popularity'], axis=1,inplace = True)
df.insert(0, 'popularity', temp)df[df['popularity'] == df['popularity'].max()]df.tail()df = df.drop(labels=df.shape[0]-1)row = {'grammer':'Perl','popularity':6.6}
df = df.append(row,ignore_index=True)df.sort_values('popularity',inplace=True)df['grammer'] = df['grammer'].fillna('R')
df['len_str'] = df['grammer'].map(lambda x: len(x))第二期：数据处理基础
import pandas as pd
import numpy as np
df = pd.read_excel(r'C:\Users\chenx\Documents\Data Analysis\pandas120.xlsx')df.head()# 方法一：apply + 自定义函数
def func(df):
lst = df['salary'].split('-')
smin = int(lst[0].strip('k'))
smax = int(lst[1].strip('k'))
df['salary'] = int((smin + smax) / 2 * 1000)
return df
df = df.apply(func,axis=1)
# 方法二：iterrows + 正则
import re
for index,row in df.iterrows():
nums = re.findall('\d+',row[2])
df.iloc[index,2] = int(eval(f'({nums[0]} + {nums[1]}) / 2 * 1000'))education salary
不限 19600.000000
大专 10000.000000
本科 19361.344538
硕士 20642.857143df.groupby('education').mean()for index,row in df.iterrows():
df.iloc[index,0] = df.iloc[index,0].to_pydatetime().strftime('%m-%d')<class 'pandas.core.frame.DataFrame'>
RangeIndex: 135 entries, 0 to 134
Data columns (total 4 columns):
createTime 135 non-null object
education 135 non-null object
salary 135 non-null int64
categories 135 non-null category
dtypes: category(1), int64(1), object(2)
memory usage: 3.5+ KBdf.info()df.describe()summary(df)
bins = [0,5000, 20000, 50000]
group_names = ['低', '中', '高']
df['categories'] = pd.cut(df['salary'], bins, labels=group_names)df.sort_values('salary', ascending=False)df.iloc[32]np.median(df['salary'])
# 17500.0# Jupyter运行matplotlib成像需要运行魔术命令
%matplotlib inline
plt.rcParams['font.sans-serif'] = ['SimHei'] # 解决中文乱码
plt.rcParams['axes.unicode_minus'] = False # 解决符号问题
import matplotlib.pyplot as plt
plt.hist(df.salary)
# 也可以用原生pandas方法绘图
df.salary.plot(kind='hist')df.salary.plot(kind='kde',xlim = (0,70000))del df['categories']
# 等价于
df.drop(columns=['categories'], inplace=True)df['test'] = df['education'] + df['createTime']df['test1'] = df['salary'].map(str) + df['education']df[['salary']].apply(lambda x: x.max() - x.min())
# salary 41500
# dtype: int64pd.concat([df[1:2], df[-1:]])df.append(df.iloc[7])createTime object
education object
salary int64
test object
test1 object
dtype: objectdf.dtypes
# createTime object
# education object
# salary int64
# test object
# test1 object
# dtype: objectdf.set_index('createTime')df1 = pd.DataFrame(pd.Series(np.random.randint(1, 10, 135)))df= pd.concat([df,df1],axis=1)df['new'] = df['salary'] - df[0]df.isnull().values.any()
# Falsedf['salary'].astype(np.float64)len(df[df['salary'] > 10000])
# 119本科 119
硕士 7
不限 5
大专 4
Name: education, dtype: int64df.education.value_counts()df['education'].nunique()
# 4rowsums = df[['salary','new']].apply(np.sum, axis=1)
res = df.iloc[np.where(rowsums > 60000)[0][-3:], :]import pandas as pd
import numpy as np
df = pd.read_excel(r'C:\Users\chenx\Documents\Data Analysis\Pandas51-80.xls')备注
请将答案中路径替换为自己机器存储数据的绝对路径，51—80相关习题与该数据有关
df.head(3)代码 1
简称 2
日期 2
前收盘价(元) 2
开盘价(元) 2
最高价(元) 2
最低价(元) 2
收盘价(元) 2
成交量(股) 2
成交金额(元) 2
.................df.isnull().sum()df[df['日期'].isnull()]列名：'代码', 第[327]行位置有缺失值
列名：'简称', 第[327, 328]行位置有缺失值
列名：'日期', 第[327, 328]行位置有缺失值
列名：'前收盘价(元)', 第[327, 328]行位置有缺失值
列名：'开盘价(元)', 第[327, 328]行位置有缺失值
列名：'最高价(元)', 第[327, 328]行位置有缺失值
列名：'最低价(元)', 第[327, 328]行位置有缺失值
列名：'收盘价(元)', 第[327, 328]行位置有缺失值
................for i in df.columns:
if df[i].count() != len(df):
row = df[i][df[i].isnull().values].index.tolist()
print('列名：'{}', 第{}行位置有缺失值'.format(i,row))df.dropna(axis=0, how='any', inplace=True)备注
axis：0-行操作（默认），1-列操作
how：any-只要有空值就删除（默认），all-全部为空值才删除
inplace：False-返回新的数据集（默认），True-在原数据集上操作# Jupyter运行matplotlib
%matplotlib inline
df['收盘价(元)'].plot()
# 等价于
import matplotlib.pyplot as plt
plt.plot(df['收盘价(元)'])plt.rcParams['font.sans-serif'] = ['SimHei'] # 解决中文乱码
plt.rcParams['axes.unicode_minus'] = False # 解决符号问题
df[['收盘价(元)','开盘价(元)']].plot()plt.hist(df['涨跌幅(%)'])
# 等价于
df['涨跌幅(%)'].hist()df['涨跌幅(%)'].hist(bins = 30)temp = pd.DataFrame(columns = df.columns.to_list())for index,row in df.iterrows():
if type(row[13]) != float:
temp = temp.append(df.loc[index])df[df['换手率(%)'] == '--']备注
通过上一题我们发现换手率的异常值只有--
df = df.reset_index(drop=True)备注
有时我们修改数据会导致索引混乱
lst = []
for index,row in df.iterrows():
if type(row[13]) != float:
lst.append(index)
df.drop(labels=lst,inplace=True)df['换手率(%)'].plot(kind='kde',xlim=(0,0.6))df['收盘价(元)'].diff()data['收盘价(元)'].pct_change()df.set_index('日期')题目：以5个数据作为一个数据滑动窗口，在这个5个数据上取均值(收盘价)
df['收盘价(元)'].rolling(5).mean()题目：以5个数据作为一个数据滑动窗口，计算这五个数据总和(收盘价)
df['收盘价(元)'].rolling(5).sum()题目：将收盘价5日均线、20日均线与原始数据绘制在同一个图上
df['收盘价(元)'].plot()
df['收盘价(元)'].rolling(5).mean().plot()
df['收盘价(元)'].rolling(20).mean().plot()题目：按周为采样规则，取一周收盘价最大值
df = df.set_index('日期')
df['收盘价(元)'].resample('W').max()题目：绘制重采样数据与原始数据
df['收盘价(元)'].plot()
df['收盘价(元)'].resample('7D').max().plot()df.shift(5)df.shift(-5)df['开盘价(元)'].expanding(min_periods=1).mean()df['expanding Open mean']=df['开盘价(元)'].expanding(min_periods=1).mean()
df[['开盘价(元)', 'expanding Open mean']].plot(figsize=(16, 6))df['former 30 days rolling Close mean']=df['收盘价(元)'].rolling(20).mean()
df['upper bound']=df['former 30 days rolling Close mean']+2*df['收盘价(元)'].rolling(20).std()
df['lower bound']=df['former 30 days rolling Close mean']-2*df['收盘价(元)'].rolling(20).std()df[['收盘价(元)', 'former 30 days rolling Close mean','upper bound','lower bound' ]].plot(figsize=(16, 6))import pandas as pd
import numpy as np
print(np.__version__)
# 1.16.5
print(pd.__version__)
# 0.25.1tem = np.random.randint(1,100,20)
df1 = pd.DataFrame(tem)tem = np.arange(0,100,5)
df2 = pd.DataFrame(tem)tem = np.random.normal(0, 1, 20)
df3 = pd.DataFrame(tem)df = pd.concat([df1,df2,df3],axis=0,ignore_index=True)0 1 2
0 95 0 0.022492
1 22 5 -1.209494
2 3 10 0.876127
3 21 15 -0.162149
4 51 20 -0.815424
5 30 25 -0.303792
...............df = pd.concat([df1,df2,df3],axis=1,ignore_index=True)np.percentile(df, q=[0, 25, 50, 75, 100])df.columns = ['col1','col2','col3']df['col1'][~df['col1'].isin(df['col2'])]temp = df['col1'].append(df['col2'])
temp.value_counts()[:3]np.argwhere(df['col1'] % 5==0)df['col1'].diff().tolist()df.iloc[:, ::-1]df['col1'].take([1,10,15])
# 等价于
df.iloc[[1,10,15],0]res = np.diff(np.sign(np.diff(df['col1'])))
np.where(res== -2)[0] + 1
# array([ 2, 4, 7, 9, 12, 15], dtype=int64)df[['col1','col2','col3']].mean(axis=1)np.convolve(df['col2'], np.ones(3)/3, mode='valid')df.sort_values('col3',inplace=True)df.col1[df['col1'] > 50] = '高'np.linalg.norm(df['col1']-df['col2'])
# 194.29873905921264df1 = pd.read_csv(r'C:\Users\chenx\Documents\Data Analysis\数据1.csv',encoding='gbk', usecols=['positionName', 'salary'],nrows = 10)df2 = pd.read_csv(r'C:\Users\chenx\Documents\Data Analysis\数据2.csv',
converters={'薪资水平': lambda x: '高' if float(x) > 10000 else '低'} )期望结果
df2.iloc[::20, :][['薪资水平']]df = pd.DataFrame(np.random.random(10)**10, columns=['data'])期望结果
Python解法
df = pd.DataFrame(np.random.random(10)**10, columns=['data'])
df.round(3)df.style.format({'data': '{0:.2%}'.format})df['data'].argsort()[len(df)-3]df.iloc[::-1, :]df1= pd.DataFrame({'key1': ['K0', 'K0', 'K1', 'K2'],
'key2': ['K0', 'K1', 'K0', 'K1'],
'A': ['A0', 'A1', 'A2', 'A3'],
'B': ['B0', 'B1', 'B2', 'B3']})
df2= pd.DataFrame({'key1': ['K0', 'K1', 'K1', 'K2'],
'key2': ['K0', 'K0', 'K0', 'K0'],
'C': ['C0', 'C1', 'C2', 'C3'],
'D': ['D0', 'D1', 'D2', 'D3']})pd.merge(df1, df2, on=['key1', 'key2'])备注
只保存df1的数据
pd.merge(df1, df2, how='left', on=['key1', 'key2'])left_join(df1,df2,by = c('key1','key2'))df = pd.read_csv(r'C:\Users\chenx\Documents\Data Analysis\数据1.csv',encoding='gbk')
pd.set_option('display.max.columns', None)np.where(df.secondType == df.thirdType)np.argwhere(df['salary'] > df['salary'].mean())[2]
# array([5], dtype=int64)df[['salary']].apply(np.sqrt)df['split'] = df['linestaion'].str.split('_')df.shape[1]
# 54df[df['industryField'].str.startswith('数据')]pd.pivot_table(df,values=['salary','score'],index='positionId')df[['salary','score']].agg([np.sum,np.mean,np.min])df.agg({'salary':np.sum,'score':np.mean})df[['district','salary']].groupby(by='district').mean().sort_values(
'salary',ascending=False).head(1)联系客服
