pandas基礎(chǔ)

1、請(qǐng)利用多種方式將截圖內(nèi)容轉(zhuǎn)成DataFrame。

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導(dǎo)入包：

import pandas as pd
import numpy as np

function1:

pd.DataFrame(data=[['Pandas',1.0],
                   ['numpy',2.0],
                   ['flask',np.NaN],
                   ['django',4.0],
                   [np.NaN,5.0],
                   ['EasticSearch',6.0],
                   ['Pandas',7.0],
                   ['pymysql',10.0]],
             columns = ['python','score'],
             )

function2:

python_dict = { 'python':['Pandas','numpy','flask','django','NaN','EasticSearch','Pandas','pymysql'],
                'score': [1.0, 2.0, np.NaN, 4.0, 5.0,6.0,7.0,10.0]  
               }
df = pd.DataFrame(python_dict)
df

2、提取含有字符串“Pandas”的行

df[df['python'] == 'Pandas']

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3、輸出df的所有列名

print(df.columns)

Index(['python', 'score'], dtype='object')

4、修改第二列列名為‘popularity’

df.rename(columns={'score':'popularity'},inplace=True)
df

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5、統(tǒng)計(jì)python列中每種模塊包出現(xiàn)的次數(shù)

df['python'].value_counts()

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6、將空值用上下值的平均值填充

df['popularity'] = df['popularity'].fillna(df['popularity'].interpolate())
df

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python算法

1、求三位數(shù)組合

lst = [3, 6, 2, 7]

這四個(gè)數(shù)字能組成多少個(gè)互不相同且無重復(fù)數(shù)字的三位數(shù)？比如362算一個(gè)，326算一個(gè)，請(qǐng)逐個(gè)輸出他們

lst = [3, 6, 2, 7]
lst2 = []
for i in lst:
   for j in lst:
       for k in lst:
           if i != j and i != k and j != k:
               result = i*100+j*10+k
               lst2.append(result)
print(lst2)     

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2、生成矩陣

已知兩個(gè)列表

lst_1 = [1, 2, 3, 4]
lst_2 = ['a', 'b', 'c', 'd']

請(qǐng)寫算法，將兩個(gè)列表交叉相乘，生成如下的矩陣

[['1a', '2a', '3a', '4a'],
 ['1b', '2b', '3b', '4b'],
 ['1c', '2c', '3c', '4c'],
 ['1d', '2d', '3d', '4d']]

lst_1 = [1, 2, 3, 4]
lst_2 = ['a', 'b', 'c', 'd']
lst_4 = []
for i in lst_2:
    lst_3 = []
    for j in lst_1:
        lst_3.append(str(j) + i)
    lst_4.append(lst_3)
print(lst_4)

3、列表偏移

已知列表

lst = [1,2,3,4,5]

要求列表向右偏移兩位后，變成

lst = [4,5,1,2,3]

lst = [1,2,3,4,5]
result = lst[len(lst)-2:]+lst[:len(lst)-2]
print(result)