Remap Values in Pandas Column With a Dict, Preserve Nans

Remap values in pandas column with a dict, preserve NaNs

You can use .replace. For example:

>>> df = pd.DataFrame({'col2': {0: 'a', 1: 2, 2: np.nan}, 'col1': {0: 'w', 1: 1, 2: 2}})
>>> di = {1: "A", 2: "B"}
>>> df
  col1 col2
0    w    a
1    1    2
2    2  NaN
>>> df.replace({"col1": di})
  col1 col2
0    w    a
1    A    2
2    B  NaN

or directly on the Series, i.e. df["col1"].replace(di, inplace=True).

Remap values in a Pandas column based on dictionary key/value pairs using RegEx in replace() function

One approach would be using columns attribute:

regex_patterns = {
    'last_name' : '[^A-Za-z \/\-\.\']',
    'first_name' : '[^A-Za-z \/\-\.\']',
    'salary' : '[^0-9 ]'
}
for column in df.columns:
  df[column] = df[[column]].replace(regex_pattern[column], np.NaN, regex=True)

Replace Pandas DataFrame column values based on containing dictionary keys

You could use DataFrame.replace with regex parameter set to True and pass the mapping dictionary.

df.replace(dictionary, regex=True)

#   col2
# 0    5
# 1  abc
# 2    8

This usage of df.replace is less known. You can read more about it here.

Replace values in columns of dataframe based on dictionary not working

Try this method based on your example.

df1['A'] = df1['A'].map(newVals)

Remap values in pandas using a dict produces one column series instead of the full DataFrame

I think what you are trying to do is this:

data['last'] = data['last'].map(my_dict)

Updating based on comment with relation to the link:

In [1]: di = {1: "A", 2: "B"}

In [5]: from numpy import NaN

In [6]: df = DataFrame({'col1':['w', 1, 2], 'col2': ['a', 2, NaN]})

In [7]: df
Out[7]:
  col1 col2
0    w    a
1    1    2
2    2  NaN

In [8]: df['col1'].map(di)
Out[8]:
0    NaN
1      A
2      B
Name: col1, dtype: object

In [9]: df
Out[9]:
  col1 col2
0    w    a
1    1    2
2    2  NaN

In [10]: df['col1'] = df['col1'].map(di)

In [11]: df
Out[11]:
  col1 col2
0  NaN    a
1    A    2
2    B  NaN

If you want this to happen in data3 instead of data then you could assign the Series result of the map to a column in data3.

pandas: replace column value with keys and values in a dictionary of list values

The best is to change the logic and try to minimize the pandas steps.

You can craft a dictionary that will directly contain your ideal output:

dic2 = {v:k for k,l in dic.items() for v in l}
# {'can': 'Should', 'could': 'Should', 'shall': 'Could', 'will': 'Would'}

# or if not yet formatted:
# dic2 = {v.lower():k.capitalize() for k,l in dic.items() for v in l}

import re
regex = '|'.join(map(re.escape, dic2))

df['text'] = df['text'].str.replace(f'\b({regex})\b',
                                    lambda m: dic2.get(m.group()),
                                    case=False, # only if case doesn't matter
                                    regex=True)

output (as text2 column for clarity):

                           text                         text2
0        can you open the door?     Should you open the door?
1  shall you write the address?  Could you write the address?

Remapping values in a Pandas column when the column is a list or set

If want replace if match values in lists use get in list comprehension with filter list by isinstance:

f = lambda x: [di.get(y,y) for y in x] if isinstance(x, list) else x
df['col1'] = df['col1'].apply(f)
print (df)
     col1 col2
0       w    a
1  [A, B]    2
2  [B, B]  NaN

There are some another solutions whats happens if no match, added 3 to list:

print (df)
        col1 col2
0          w    a
1     [1, 2]    2
2  [2, 2, 3]  NaN

#if no match return original, here 3
f1 = lambda x: [di.get(y,y) for y in x] if isinstance(x, list) else x
df['col11'] = df['col1'].apply(f1)

#if no match return None
f2 = lambda x: [di.get(y,None) for y in x] if isinstance(x, list) else x
df['col12'] = df['col1'].apply(f2)

#if no match remove not match value
f3 = lambda x: [di[y] for y in x if y in di] if isinstance(x, list) else x
df['col13'] = df['col1'].apply(f3)
print (df)
        col1 col2      col11         col12   col13
0          w    a          w             w       w
1     [1, 2]    2     [A, B]        [A, B]  [A, B]
2  [2, 2, 3]  NaN  [B, B, 3]  [B, B, None]  [B, B]

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