버미다

d={"a":8,"b":33,"c":15,"d":26,"e":12,"f":120}dict_1={}for k,v in d.items():    if v > 25:        dict_1[k]=vprint(dict_1)# 방법2ex2={k:v for k,v in d.items() if v>20}print(f' ex2 결과:{ex2}')print(dict(((k,v)for k,v in d.items() if v>=20)))#방법3ex3 = dict(filter(lambda v:v[1]>=25, d.items()))print(f'ex3 결과:{ex3}')

카테고리 없음 2024. 7. 4. 21:26

d={"a":"apple","b":"grape"}d["c"]="banana"d["d"]="kiwi"print(d)e={"a":"apple","b":"grape"}c={"c":"banana","d":"kiwi"}e.update(c)print(e)

카테고리 없음 2024. 7. 4. 20:59

d={"a":17,"b":114,"c":247,"d":362,"e":220,"f":728,"g":-283,"h":922}print(d.values())# 방법1total =0for i in d.values():    total += iprint(f'ex1 결과 {total}')#방법2print(f'ex2 결과 {sum(d.values())}')#방법3print(f'ex3 결과 {sum([d[item] for item in d])}')

카테고리 없음 2024. 7. 4. 20:40

#제거x=["a",1,"b",2,"a",3,"b",4,5,"b"]# 방법1 - 순서가 없음y=list(set(x))print(y)#순선유지from collections import OrderedDictex2 =list(OrderedDict.fromkeys(x))print(f'ex2 결과:{ex2}')# 방법3- 순서유지ex3=[]for i in x:    if i not in ex3:        ex3.append(i)print(ex3)print("-------------")

카테고리 없음 2024. 7. 4. 20:23

li=[]for i in range(1,16):    i*=10    li.append(str(i))print(li)# 빙법2print(list(map(lambda x:str(x*10),range(1,16))))# 방법3print([str(x*10) for x in range(1,16)])

카테고리 없음 2024. 7. 4. 20:06

[10, 2, 30, 4, 50, 6, 70, 8, 90, 10, 110, 12, 130, 14, 150, 16, 170, 18, 190, 20] [10, 2, 30, 4, 50, 6, 70, 8, 90, 10, 110, 12, 130, 14, 150, 16, 170, 18, 190, 20] li=[]for i in range(1,21):    if i%2==0:        li.append(i)    else:        i=i*10        li.append(i)print(li)x=[x *10 if x %2 !=0 else x for x in range(1,21)]print(x)      print([[j for j in range(5)]for i in range(5)])  [[0, 1, 2,..

카테고리 없음 2024. 7. 4. 19:48

#방법1ex1=[]for i in range(30,-12,-2):    ex1.append(i)print(ex1)#방법2ex2=list(range(30,-12,-2))print(ex2)

카테고리 없음 2024. 7. 4. 19:39

x=["grapes","mango","orange","peach","kiwi","apple","lime","cherry","tomato","blueberry","watermelon"]a=[]for i in x:    if i == "apple" or  i=="kiwi":        a.append(i.upper())print(a)# 추가 방법# 방법1ex1=[]for i in range(len(x)):    if x[i] == "apple" or x[i] =="kiwi":        ex1.append(x[i].upper())print(f'ex1 결과:{ex1}')# 방법2ex2=map(lambda b:b.upper(),(filter(lambda a:a=="apple" or a=="kiwi",x)))..

카테고리 없음 2024. 7. 4. 19:23