迭代行时，如何使用掩码更新DataFrame中的值

Question

With the below code I'm trying to update the column df_test['placed'] to = 1 when the if statement is triggered and a prediction is placed. I haven't been able to get this to update correctly though, the code compiles but doesn't update to = 1 for the respective predictions placed.

df_test['placed'] = np.zeros(len(df_test))
for i in set(df_test['id']) :
    mask = df_test['id']==i
    predictions = lm.predict(X_test[mask])
    j = np.argmax(predictions)
    if predictions[j] > 0 :
        df_test['placed'][mask][j] = 1
        print(df_test['placed'][mask][j])

Answer 1

Answering your question

Edit: changed suggestion based on comments

The assignment part of your code, df_test['placed'][mask][j] = 1 , uses what is called chained indexing . In short, your assignment only changes a temporary copy of the DataFrame that gets immediately thrown away, and never changes the original DataFrame.

To avoid this, the rule of thumb when doing assignment is: use only one set of square braces on a single DataFrame. For your problem, that should look like:

df_test.loc[mask.nonzero()[0][j], 'placed'] = 1

(I know the mask.nonzero() uses two sets of square brackets; actually nonzero() returns a tuple, and the first element of that tuple is an ndarray. But the dataframe only uses one set, and that's the important part.)

Some other notes

There are a couple notes I have on using pandas (& numpy ).

• Pandas & NumPy both have a feature called broadcasting . Basically, if you're assigning a single value to an entire array, you don't need to make an array of the same size first; you can just assign the single value, and pandas/NumPy automagically figures out for you how to apply it. So the first line of your code can be replaced with df_test['placed'] = 0 , and it accomplishes the same thing.

• Generally speaking when working with pandas & numpy objects, loops are bad ; usually you can find a way to use some combination of broadcasting , element-wise operations and boolean indexing to do what a loop would do. And because of the way those features are designed, it'll run a lot faster too. Unfortunately I'm not familiar enough with the lm.predict method to say, but you might be able to avoid the whole for -loop entirely for this code.