實現詞云的繪制
步驟：
1.繪制詞云的形狀
from wordcloud import WordCloud
import jieba
import imageio
mask = imageio.imread('./china.jpg') #要繪制詞云的形狀
2.讀取小說內容
with open('./novel/threekingdom.txt', 'r', encoding='utf-8') as f:
words = f.read()
counts = {} # {‘曹操’:234，‘回寨’:56}
excludes = {"將軍", "卻說", "丞相", "二人", "不可", "荊州", "不能", "如此", "商議",
"如何", "主公", "軍士", "軍馬", "左右", "次日", "引兵", "大喜", "天下",
"東吳", "于是", "今日", "不敢", "魏兵", "陛下", "都督", "人馬", "不知",
"孔明曰","玄德曰","劉備","云長"}
3.分詞
words_list = jieba.lcut(words)
# print(words_list)
for word in words_list:
if len(word) <= 1:
continue
else:
# 更新字典中的值
# counts[word] = 取出字典中原來鍵對應的值 + 1
# counts[word] = counts[word] + 1 # counts[word]如果沒有就要報錯
# 字典。get（k） 如果字典中沒有這個鍵 返回 NONE
counts[word] = counts.get(word, 0) + 1
4.詞語過濾,刪除無關詞，重復詞
counts['孔明'] = counts['孔明'] + counts['孔明曰']
counts['玄德'] = counts['玄德'] + counts['玄德曰'] +counts['劉備']
counts['關公'] = counts['關公'] +counts['云長']
for word in excludes:
del counts[word]
5.排序
items = list(counts.items())
print(items)
def sort_by_count(x):
return x[1]
items.sort(key=sort_by_count, reverse=True)
6.序列解包
li=[]
for i in range(10):
# 序列解包
role, count = items[i]
print(role, count)
for _ in range(count): #_是告訴看代碼的人循環里不需要使用臨時變量
li.append(role)
7.結論
text=' '.join(li)
WordCloud(
font_path='msyh.ttc',
background_color='white',
width=800,
height=600,
mask=mask,
#相鄰兩個值的重復
collocations=False
).generate(text).to_file('Top10.png')
匿名函數
匿名函數：lambda函數是一種快速定義單行的最小函數，可以用在任何需要用到函數的地方。
匿名函數返回值是一個函數對象，可以使用變量去接收這個對象。例如 x = lambda x,y:x*y ，代表x是一個計算兩數想成的函數，調用時可以寫成x(3,5)
匿名函數的優點：
-使代碼精簡
-有些只使用一次的函數，不用為其命名
-讓代碼更容易理解
格式：
lambda 參數:返回值
例子
sum=lambda x1,x2:x1+x2
print(sum (2,3))
參數可以有無數個，但是表達式只能有一個
下例函數改為為匿名函數
name_info_list = [
('張三',4500),
('李四',9900),
('王五',2000),
('趙六',5500),
]
def sort_by_gz(x):
return x[1]
name_info_list.sort(key = sort_by_gz)
print('排序后', name_info_list)
修改之后
name_info_list = [
('張三',4500),
('李四',9900),
('王五',2000),
('趙六',5500),
]
name_info_list.sort(key=lambda x:x[1],reverse=True)
print(name_info_list)
lambda匿名函數是python語言的一種特色，當我們不需要定義一個函數的時候，可以使用匿名函數來做。
匿名函數的限制：
就是只能有一個表達式，不用寫return，返回值就是該表達式的結果。
匿名函數的好處：
即函數沒有名字，不用擔心函數名沖突，此外，匿名函數也是一個函數對象，也可以把匿名函數賦值給一個變量，再利用變量來調用該函數。
列表推導式，列表解析和字典解析
推導式comprehensions（又稱解析式），是Python的一種獨有特性。推導式是可以從一個數據序列構建另一個新的數據序列的結構體。
1.列表推導式
之前我們創建列表是利用for循環
li=[]
for i in range(10):
li.append(i)
print(li)
列表推導式，也叫列表解析式，英文名稱為list comprehension，可以使用非常簡潔的方式來快速生成滿足特定需求的列表，代碼具有非常強的可讀性。另外，Python的內部實現對列表推導式做了大量優化，可以保證很快的運行速度。
格式：[表達式 for 臨時變量 in 可迭代對象 可以追加條件]
使用列表推導式
print([i for i in range(10)])
利用列表推導式只用一條語句就可以創建列表了。
列表解析
篩選出列表中所以的偶數
li=[]
for i in range(10):
if i % 2 ==0:
li.append(i)
print(li)
使用列表解析
print([i for i in range(10) if i%2==0])
篩選出列表中大于0的數(隨機產生10個數)
from random import randint
num_list=[randint(-10,10) for _ in range(10)]
print(num_list)
print([i for i in num_list if i>0])
字典解析
生成100給學生的成績
stu_grades={'student{}'.format(i):randint(50,100) for i in range (1,101)}
print(stu_grades)
篩選大于60分的所以學生
print({k:v for k,v in stu_grades.items() if v>60})
matplotlib庫
1.曲線圖
from matplotlib import pyplot as plt
用100個點繪制正弦曲線圖[0,2pi]
import numpy as np
plt.rcParams["font.sans-serif"] = ['SimHei']
plt.rcParams['axes.unicode_minus'] = False
x=np.linspace(0,2*np.pi,num=100)
print(x)
y=np.sin(x)
plt.plot(x,y,color='g',linestyle='--',label='sin(x)')
cosy=np.cos(x)
plt.plot(x,cosy, color='r',label='cos(x)')
plt.xlabel('時間(s)')
plt.ylabel('電壓(v)')
plt.title('歡迎來到python世界')
plt.legend()
plt.show()
2.柱狀圖
import string
from random import randint
# print(string.ascii_uppercase[0:6])
# ['A', 'B', 'C'...]
x = ['口紅{}'.format(x) for x in string.ascii_uppercase[0:5]]
y = [randint(200, 500) for _ in range(5)]
print(x)
print(y)
plt.xlabel('口紅品牌')
plt.ylabel('價格（元）')
plt.bar(x, y)
plt.show()
3.餅圖
隨機產生6個員工的工資范圍在(3500, 9000),并畫出餅圖
from random import randint
import string
counts = [randint(3500, 9000) for _ in range(6)]
labels = ['員工{}'.format(x) for x in string.ascii_lowercase[:6] ]
# 距離圓心點距離
explode = [0.1,0,0, 0, 0,0]
colors = ['red', 'purple','blue', 'yellow','gray','green']
plt.pie(counts,explode = explode,shadow=True, labels=labels, autopct = '%1.1f%%',colors=colors)
plt.legend(loc=2)
plt.axis('equal')
plt.show()
4.散點圖
均值為0，標準差為1的正太分布數據
x=np.random.normal(0,1,100)
y=np.random.normal(0,1,100)
plt.scatter(x,y,alpha=0.5)
plt.show()
x=np.random.normal(0,1,100000)
y=np.random.normal(0,1,100000)
plt.scatter(x,y,alpha=0.1)
plt.show()
將之前做好的三國top10人物以餅圖的方式展示
li=[]
peo_li=[]
for i in range(10):
# 序列解包
role, count = items[i]
a={'name':'','count':0}
a['name']=role
a['count']=count
peo_li.append(a)
print(role, count)
for _ in range(count): #_是告訴看代碼的人循環里不需要使用臨時變量
li.append(role)
#在解包的同時將前十的人物即出現次數存放當peo_list列表中
counts = []
labels = []
for i in range(len(peo_li)):
counts.append(peo_li[i]['count'])
labels.append(peo_li[i]['name'])
# 距離圓心點距離
explode = [0.1, 0, 0, 0, 0, 0,0,0,0,0]
#colors = ['red', 'purple', 'blue', 'yellow', 'gray', 'green']
plt.pie(counts, explode=explode, shadow=True, labels=labels, autopct = '%1.1f%%')
plt.legend(loc=2)
plt.axis('equal')
plt.show()
完整代碼：
from wordcloud import WordCloud
import jieba
import imageio
mask = imageio.imread('./china.jpg')
1.讀取小說內容
with open('./novel/threekingdom.txt', 'r', encoding='utf-8') as f:
words = f.read()
counts = {} # {‘曹操’:234，‘回寨’:56}
excludes = {"將軍", "卻說", "丞相", "二人", "不可", "荊州", "不能", "如此", "商議",
"如何", "主公", "軍士", "軍馬", "左右", "次日", "引兵", "大喜", "天下",
"東吳", "于是", "今日", "不敢", "魏兵", "陛下", "都督", "人馬", "不知",
"孔明曰","玄德曰","劉備","云長"}
2. 分詞
words_list = jieba.lcut(words)
# print(words_list)
for word in words_list:
if len(word) <= 1:
continue
else:
# 更新字典中的值
# counts[word] = 取出字典中原來鍵對應的值 + 1
# counts[word] = counts[word] + 1 # counts[word]如果沒有就要報錯
# 字典。get（k） 如果字典中沒有這個鍵 返回 NONE
counts[word] = counts.get(word, 0) + 1
print(len(counts))
3. 詞語過濾,刪除無關詞，重復詞
counts['孔明'] = counts['孔明'] + counts['孔明曰']
counts['玄德'] = counts['玄德'] + counts['玄德曰'] +counts['劉備']
counts['關公'] = counts['關公'] +counts['云長']
for word in excludes:
del counts[word]
4.排序 [(), ()]
items = list(counts.items())
print(items)
# def sort_by_count(x):
# return x[1]
# items.sort(key=sort_by_count, reverse=True)
items.sort(key=lambda i: i[1], reverse=True)
li=[]
peo_li=[]
for i in range(10):
# 序列解包
role, count = items[i]
a={'name':'','count':0}
a['name']=role
a['count']=count
peo_li.append(a)
print(role, count)
for _ in range(count): #_是告訴看代碼的人循環里不需要使用臨時變量
li.append(role)
5.得出結論
text=' '.join(li)
WordCloud(
font_path='msyh.ttc',
background_color='white',
width=800,
height=600,
mask=mask,
#相鄰兩個值的重復
collocations=False
).generate(text).to_file('Top10.png')
#用餅圖顯示人物
from random import randint
import string
from matplotlib import pyplot as plt
plt.rcParams["font.sans-serif"] = ['SimHei']
plt.rcParams['axes.unicode_minus'] = False
counts = []
labels = []
for i in range(len(peo_li)):
counts.append(peo_li[i]['count'])
labels.append(peo_li[i]['name'])
# 距離圓心點距離
explode = [0.1, 0, 0, 0, 0, 0,0,0,0,0]
#colors = ['red', 'purple', 'blue', 'yellow', 'gray', 'green']
plt.pie(counts, explode=explode, shadow=True, labels=labels, autopct = '%1.1f%%')
plt.legend(loc=2)
plt.axis('equal')
plt.show()
練習：將紅樓夢的top10人物繪制餅圖
完整代碼
from wordcloud import WordCloud
import jieba
import imageio
mask = imageio.imread('./china.jpg')
1.讀取小說內容
with open('./novel/all.txt', 'r', encoding='utf-8') as f:
words = f.read()
#print(words)
counts = {}
excludes = {"什么", "一個", "我們", "你們", "如今", "說道", "知道", "起來", "這里",
"出來", "眾人", "那里", "自己", "一面", "只見", "太太", "兩個", "沒有",
"怎么", "不是", "不知", "這個", "聽見", "這樣", "進來", "咱們", "就是",
"老太太", "東西", "告訴", "回來", "只是", "大家", "姑娘", "奶奶", "鳳姐兒","分節"}
2. 分詞
words_list = jieba.lcut(words)
for word in words_list:
if len(word) <= 1:
continue
else:
# 更新字典中的值
# counts[word] = 取出字典中原來鍵對應的值 + 1
# counts[word] = counts[word] + 1 # counts[word]如果沒有就要報錯
# 字典。get（k） 如果字典中沒有這個鍵 返回 NONE
counts[word] = counts.get(word, 0) + 1
print(len(counts))
3. 詞語過濾,刪除無關詞，重復詞
counts['賈母'] = counts['賈母'] + counts['老太太']
counts['寶釵'] = counts['寶釵'] + counts['薛寶釵']
counts['鳳姐'] = counts['鳳姐兒'] + counts['王熙鳳'] +counts['鳳姐']
counts['寶玉'] = counts['賈寶玉'] +counts['寶玉']
counts['王夫人'] = counts['王夫人'] + counts['太太']
counts['黛玉'] = counts['黛玉'] + counts['林黛玉']
counts['賈政']=counts['賈政']+counts['老爺']
for word in excludes:
del counts[word]
4.排序 [(), ()]
items = list(counts.items())
#print(items)
items.sort(key=lambda i: i[1], reverse=True)
li=[]
peo_li=[]
for i in range(10):
# 序列解包
role, count = items[i]
a={'name':'','count':0}
a['name']=role
a['count']=count
peo_li.append(a)
print(role, count)
for _ in range(count): #_是告訴看代碼的人循環里不需要使用臨時變量
li.append(role)
5.得出結論
text=' '.join(li)
WordCloud(
font_path='msyh.ttc',
background_color='white',
width=800,
height=600,
mask=mask,
#相鄰兩個值的重復
collocations=False
).generate(text).to_file('紅樓Top10.png')
#用餅圖顯示人物
from random import randint
import string
from matplotlib import pyplot as plt
plt.rcParams["font.sans-serif"] = ['SimHei']
plt.rcParams['axes.unicode_minus'] = False
counts = []
labels = []
for i in range(len(peo_li)):
counts.append(peo_li[i]['count'])
labels.append(peo_li[i]['name'])
# 距離圓心點距離
explode = [0.1, 0, 0, 0, 0, 0,0,0,0,0]
#colors = ['red', 'purple', 'blue', 'yellow', 'gray', 'green']
plt.pie(counts, explode=explode, shadow=True, labels=labels, autopct = '%1.1f%%')
plt.legend(loc=2)
plt.axis('equal')
plt.show()
