I am attempting to read through keys in a dictionary and if these keys are in the values of column A then I want column B to be filled with the values of the dictionary that match that key/keys.

For example;

Dictionary:

dict = {'ks': 'Killos',
'RAS': 'Round Point System',
 'hi': 'Hello World',
 'vc': 'VVCCR'}

Resulting in the below dataframe:

I have tried to do the following:

for k,v in dict.items():
    if k.lower() in str(dataframe['Column A']).lower():
       v
       dataframe['Column B'] = [v]


    but it results in the below error: ValueError: Length of values (1) does not match length 
    of index (6483)

I wrote a code that can perform this if the input is one text but I am unable to apply it to the entire column. I have also tried to convert the dictionary to a dataframe and do it that way but still no luck.

The code I wrote for a singular text input:

text = str(input('Please insert hostname here: '))
results = list()

for k,v in codes.items():
    if k.lower() in text.lower():
       v
       results.append(v)
    else:
       continue
#print(v)
print ('The type of this device is most likely a/an: ',results)

so the question is:

Is there a way to have python read through the keys in the dictionary and return its corresponding values in Columb B of a dataframe if those keys are contained (exist in) the items in Column A?

CodePudding user response：

This is what I currently have.

dataframe = {'Column A': ['KSTRSHRASA', 'someWord-hi'], 'Column B': [np.nan, np.nan]}
df = pd.DataFrame(dataframe)

dictionary = {'ks': 'Killos',
              'RAS': 'Round Point System',
              'hi': 'Hello World',
              'vc': 'VVCCR'}

# convert dictionary key to lower case
dictionary = {k.lower(): v for k, v in dictionary.items()}

# keep a copy of original dataframe
df_original = df.copy()

# convert dataframe column A to lower case
df['Column A'] = df['Column A'].str.lower()

# if dataframe column A contains dictionary key, append it to column B as a list
df['Column B'] = df['Column A'].apply(lambda x: [dictionary[i] for i in dictionary if i in x])

# convert Column A back to original case
df['Column A'] = df_original['Column A']

OUTPUT:

      Column A                      Column B
0   KSTRSHRASA  [Killos, Round Point System]
1  someWord-hi                 [Hello World]

Hope this helps!