我们说我有一个清单:
list = ['foo', 'bar', 'bak']
我想将这些字符串拆分成字符,这样我就可以为每个字符赋值(I.E. f = 5,o = 15,等等)。
我如何分解这些列表? 我正在考虑将每个元素转换为自己的列表,并将每个元素称为该列表的一个项目,但我不确定如何去做。
答案 0 :(得分:3)
字符串在Python中是可迭代的,所以你可以像这样遍历它们:
list = ['foo', 'bar', 'bak']
for item in list:
for character in item:
print(character)
答案 1 :(得分:0)
尝试为每个单词做范围功能
list = ['foo','bar','bak']
for i in range (len(list)):
for t in range(len(list[i])):
print(list[i][t])
它会打印每件物品。 如果你想创建每个字母的列表, 尝试
list = ['foo','bar','bak']
letters = []
for i in range (len(list)):
for t in range(len(list[i])):
letters.append(list[i][t])
答案 2 :(得分:0)
如果这就是你在这里寻找的东西:
package hjwt
import (
"fmt"
"time"
jwt "github.com/dgrijalva/jwt-go"
"github.com/hzwy23/hcloud/logs"
)
var (
key []byte = []byte("-jwt-hzwy23@163.com")
)
// json web token
func GenToken() string {
claims := &jwt.StandardClaims{
NotBefore: int64(time.Now().Unix()),
ExpiresAt: int64(time.Now().Unix() + 1000),
Issuer: "hzwy23",
}
token := jwt.NewWithClaims(jwt.SigningMethodHS256, claims)
ss, err := token.SignedString(key)
if err != nil {
logs.Error(err)
return ""
}
return ss
}
// Verify that token is valid
func CheckToken(token string) bool {
_, err := jwt.Parse(token, func(*jwt.Token) (interface{}, error) {
return key, nil
})
if err != nil {
fmt.Println("parase with claims failed.", err)
return false
}
return true
}
// Next, add the filter before the beego starts. The filter code is as follows:
beego.InsertFilter("/platform/*", beego.BeforeRouter, func(ctx *context.Context) {
cookie, err := ctx.Request.Cookie("Authorization")
if err != nil || !hjwt.CheckToken(cookie.Value) {
http.Redirect(ctx.ResponseWriter, ctx.Request, "/", http.StatusMovedPermanently)
}
})
// In this process, you need to set the JSON web token value to cookies, where the cookies method is set as follows in.Golang:
token := hjwt.GenToken()
cookie := http.Cookie{Name: "Authorization", Value: token, Path: "/", MaxAge: 3600}
http.SetCookie(w, &cookie)
答案 3 :(得分:0)
也许这就是你要转换成字符列表的内容:
l = ['foo', 'bar', 'bak']
list(''.join(l))
答案 4 :(得分:0)
您可以使用一行解决方案:
list1 = ['foo', 'bar', 'bak']
print([j for i in list1 for j in i])
结果:
['f', 'o', 'o', 'b', 'a', 'r', 'b', 'a', 'k']
答案 5 :(得分:-1)
将整个列表加入每个角色
my_list = [' foo',' bar',' bak']
result = list('' .join(my_list))
结果
[' f',' o',' o',' b',' a',& #39; r',' b',' a',' k']
现在您可以遍历结果以为每个字符分配值