0

我需要打印一些p 使用print这个循环的值。

for prediction in pred:
            j=0
            if(prediction == 1):
                print ("Titolo: " + target_playlist["song_title"][i] + ",  Artista:  "+ target_playlist["artist"][i] + ",   Percentuali:  "+ p[i])
                likedSongs= likedSongs + 1
            i = i +1

如果我尝试p在循环 fow 示例之外打印

for i in range(len(p)):
            print("Predicted=%s" % (p[i]))

它可以工作，但是当我p[i]在 for 循环中放入 print 时，我遇到了类型问题。我该如何解决？

 if(algoritmo_scelto==1):
            pred = c.predict(target_playlist[features])
            p = c.predict_proba(target_playlist[features])
        if(algoritmo_scelto==2):
            pred = knn.predict(target_playlist[features])
            p = knn.predict_proba(target_playlist[features])
        if(algoritmo_scelto==3):
            pred = forest.predict(target_playlist[features])
            p = forest.predict_proba(target_playlist[features])
        if(algoritmo_scelto==4):
            pred = k_means.predict(target_playlist[features])
            p = k_means.predict_proba(target_playlist[features])

        likedSongs = 0
        i = 0

        for prediction in pred:
            j=0
            if(prediction == 1):
                print ("Titolo: " + target_playlist["song_title"][i] + ",  Artista:  "+ target_playlist["artist"][i] + ",   Percentuali:  "+ p[i])
                likedSongs= likedSongs + 1
            i = i +1   





    Exception in Tkinter callback
        Traceback (most recent call last):
          File "C:\Users\david\AppData\Local\Programs\Python\Python36-32\lib\tkinter\__init__.py", line 1702, in __call__
            return self.func(*args)
          File "C:\Users\david\eclipse-workspace1\project\src\changp.py", line 179, in <lambda>
            command=lambda: par())
          File "C:\Users\david\eclipse-workspace1\project\src\changp.py", line 173, in par
            estraiPreferite(self.entry1.get(), var.get())
          File "C:\Users\david\eclipse-workspace1\project\src\changp.py", line 550, in estraiPreferite
            print ("Titolo: " + target_playlist["song_title"][i] + ",  Artista:  "+ target_playlist["artist"][i] + ",   Percentuali:  "+ p[j])
        TypeError: ufunc 'add' did not contain a loop with signature matching types dtype('<U72') dtype('<U72') dtype('<U72')

通过为某些对象属性分配值来过滤唯一值要快得多 - 不会有重复项。对于初始数组的每个 +1 成员，这种方法会变得越来越好，因为循环会导致快速算法复杂化

let arr = [
  {value: 'L7-LO', name: 'L7-LO'},
  {value: '%L7-LO', name: '%L7-LO'},
  {value: 'L7-LO', name: 'L7-LO'},
  {value: '%L7-LO', name: '%L7-LO'},
  {value: 'L7-L3', name: 'L7-L3'},
  {value: '%L7-L3', name: '%L7-L3'},
  {value: 'LO-L3', name: 'LO-L3'},
  {value: '%LO-L3', name: '%LO-L3'}
];
let obj = {};

const unique = () => {
  let result = [];
  
  arr.forEach((item, i) => {
    obj[item['value']] = i;
  });
  
  for (let key in obj) {
    let index = obj[key];
    result.push(arr[index])
  }
  
  return result;
}

arr = unique(); // for example; 

console.log(arr);

4

1 回答 1

0

创建一个字符串数组：

In [148]: arr = np.array(['a','bcd','ef'])
In [149]: arr
Out[149]: array(['a', 'bcd', 'ef'], dtype='<U3')

我可以索引元素并将它们与 Python 字符串 `+' 连接起来：

In [150]: arr[0]+arr[1]
Out[150]: 'abcd'

但是我不能以这种方式将数组与字符串 dtype 连接起来：

In [151]: arr[[0]]+arr[[1]]
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-151-3186795057b5> in <module>()
----> 1 arr[[0]]+arr[[1]]

TypeError: ufunc 'add' did not contain a loop with signature matching types dtype('<U3') dtype('<U3') dtype('<U3')

In [156]: arr[[0]],arr[[1]]
Out[156]: (array(['a'], dtype='<U3'), array(['bcd'], dtype='<U3'))
In [157]: 'foo %s %s'%(arr[[0]], arr[[1]])
Out[157]: "foo ['a'] ['bcd']"
于 2019-01-08T19:51:56.870 回答