创建 python function 以迭代 pandas Z6A8064B5DF479455500553C47C50 列并将其字符串值ZC550

Question

I have the following dataframe:

symbol  Open  close
SPY  34,2  33,2
AMZN 30.2  10,2
.................

and I want to create a function that will convert each character in a string into ASCII. This function will than be applied on symbol field and its value will be placed in a new field (called 'id'). The final dataframe will look:

symbol  Open  close id
SPY  34,2  33,2  838089
AMZN 30.2  10,2  65779078

this is what I have done

def symbolid(x):
    strAscii = ''
    for i in range (len(x)):
        strAscii =  strAscii + str(ord(x[i]))
        print(x)
    return strAscii

df['id'] = df.apply(lambda x: symbolid(df['symbol']), axis=1)

I get TypeError: ord() expected a character, but string of length 3 found

any help would be appreciate/ Using python 3.6

Answer 1

You were not that far. But in your lambda function you still refere df while you should use x :

df['id'] = df.apply(lambda x: symbolid(x['symbol']), axis=1)

shows

SPY
SPY
SPY
AMZN
AMZN
AMZN
AMZN

and gives as expected

df
  symbol  Open close        id
0    SPY  34,2  33,2    838089
1   AMZN  30.2  10,2  65779078

But your code is not very efficient because:

1. in your function you add to a string when you could join an iterator
2. you convert each row into a Series when you could simply transform one column

In the end,

df['symbol'].transform(lambda x: ''.join(str(ord(i)) for i in x))

Also gives the expect dataframe

Answer 2

In your lambda function you use whole column as argument, that's source of error, you can only use symbol column.

    import pandas as pd

data = {
    "symbol": ["SPY","AMZN"],
     "Open": [34.2 , 33.2],
     "close":[30.2  ,10.2]
}
df = pd.DataFrame(data)
def symbolid(x):
    strAscii = ''
    for i in range (len(x)):
        strAscii =  strAscii + str(ord(x[i]))
        print(x)
    return strAscii
df['id'] = df['symbol'].apply(lambda x: symbolid(x))
print(df)