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decode pandas data frame with sklearn
How to join (merge) data frames (inner, outer, left, right)?Drop data frame columns by nameValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()Renaming columns in pandasDelete column from pandas DataFrame by column nameHow to iterate over rows in a DataFrame in Pandas?Select rows from a DataFrame based on values in a column in pandasPython Pandas add column for row-wise max value of selected columnsNumPy creation by fromfunction errorCheck if string is in a pandas dataframe
I have a data frame with many columns. some of them are string and some other are integer.
I used this code to encode my data frame:
le = LabelEncoder()
for col in df.columns:
df_encoded[col] = df.apply(le.fit_transform)
it worked! but when I want to decode it with this code:
for col in df.columns:
df_decoded[col] = df_encoded.apply(le.inverse_transform)
I receive this error:
ValueError: ('The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()', 'occurred at index MYCOLUMNNAME')
python dataframe sklearn-pandas
asked Mar 8 at 4:34
CFDCFD
966
add a comment |
I have a data frame with many columns. some of them are string and some other are integer.
I used this code to encode my data frame:
le = LabelEncoder()
for col in df.columns:
df_encoded[col] = df.apply(le.fit_transform)
it worked! but when I want to decode it with this code:
for col in df.columns:
df_decoded[col] = df_encoded.apply(le.inverse_transform)
I receive this error:
ValueError: ('The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()', 'occurred at index MYCOLUMNNAME')
python dataframe sklearn-pandas
asked Mar 8 at 4:34
CFDCFD
966
provide complete error message and sample df too
– AkshayNevrekar
Mar 8 at 4:37
This is the complete error. I just removed the name of column for not making confusion
– CFD
Mar 8 at 6:02
add a comment |
I have a data frame with many columns. some of them are string and some other are integer.
I used this code to encode my data frame:
le = LabelEncoder()
for col in df.columns:
df_encoded[col] = df.apply(le.fit_transform)
it worked! but when I want to decode it with this code:
for col in df.columns:
df_decoded[col] = df_encoded.apply(le.inverse_transform)
I receive this error:
ValueError: ('The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()', 'occurred at index MYCOLUMNNAME')
python dataframe sklearn-pandas
asked Mar 8 at 4:34
CFDCFD
966
I have a data frame with many columns. some of them are string and some other are integer.
I used this code to encode my data frame:
le = LabelEncoder()
for col in df.columns:
df_encoded[col] = df.apply(le.fit_transform)
it worked! but when I want to decode it with this code:
for col in df.columns:
df_decoded[col] = df_encoded.apply(le.inverse_transform)
I receive this error:
ValueError: ('The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()', 'occurred at index MYCOLUMNNAME')
python dataframe sklearn-pandas
python dataframe sklearn-pandas
asked Mar 8 at 4:34
CFDCFD
966
asked Mar 8 at 4:34
CFDCFD
966
asked Mar 8 at 4:34
CFDCFD
966
asked Mar 8 at 4:34
CFDCFD
966
asked Mar 8 at 4:34
CFDCFD
966
966
provide complete error message and sample df too
– AkshayNevrekar
Mar 8 at 4:37
This is the complete error. I just removed the name of column for not making confusion
– CFD
Mar 8 at 6:02
add a comment |
provide complete error message and sample df too
– AkshayNevrekar
Mar 8 at 4:37
This is the complete error. I just removed the name of column for not making confusion
– CFD
Mar 8 at 6:02
provide complete error message and sample df too
– AkshayNevrekar
Mar 8 at 4:37
provide complete error message and sample df too
– AkshayNevrekar
Mar 8 at 4:37
This is the complete error. I just removed the name of column for not making confusion
– CFD
Mar 8 at 6:02
This is the complete error. I just removed the name of column for not making confusion
– CFD
Mar 8 at 6:02
add a comment |
1 Answer
1
active
oldest
votes
The type of data differs from column to column, so using apply with fit_transform won't work here. It will seem to work properly but the LabelEncoder will be fitted to the rightmost column at the end of the operation, so when you'll try to apply the inverse_transform, the LabelEncoder will replace all the elements in the other columns with the ones it saw in the rightmost column. E.g.:
df = pd.DataFrame(['A': 1, 'B': 'p', 'A': 1, 'B': 'q', 'A': 2, 'B': 'o', 'A': 3, 'B': 'p'])
df
A B
0 1 p
1 1 q
2 2 o
3 3 p
df = df.apply(le.fit_transform)
df
A B
0 0 1
1 0 2
2 1 0
3 2 1 # Looks fine
df.apply(le.inverse_transform)
A B
0 o p
1 o q
2 p o
3 q p # Whoops
You will see the same result even if you iterate over the columns one by one and perform the fit_transform and inverse_transform.
You need to fit the encoder to the correct column before inversing:
le = LabelEncoder()
df_encoded = pd.DataFrame(columns=df.columns)
df_decoded = pd.DataFrame(columns=df.columns)
for col in df.columns:
df_encoded[col] = le.fit_transform(df[col])
df_encoded
A B
0 0 1
1 0 2
2 1 0
3 2 1
for col in df.columns:
le = le.fit(df[col])
df_decoded[col] = le.inverse_transform(df_encoded[col])
df_decoded
A B
0 1 p
1 1 q
2 2 o
3 3 p # Yeay
answered Mar 8 at 7:46
suicidalteddysuicidalteddy
610114
Thanks for your answer. It worked! ...another question...What if we have something like this "P,Q'' in one of our cells in column B? I want to encode P and Q separately...I think it is a list inside the cell
– CFD
Mar 8 at 16:05
add a comment |
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The type of data differs from column to column, so using apply with fit_transform won't work here. It will seem to work properly but the LabelEncoder will be fitted to the rightmost column at the end of the operation, so when you'll try to apply the inverse_transform, the LabelEncoder will replace all the elements in the other columns with the ones it saw in the rightmost column. E.g.:
df = pd.DataFrame(['A': 1, 'B': 'p', 'A': 1, 'B': 'q', 'A': 2, 'B': 'o', 'A': 3, 'B': 'p'])
df
A B
0 1 p
1 1 q
2 2 o
3 3 p
df = df.apply(le.fit_transform)
df
A B
0 0 1
1 0 2
2 1 0
3 2 1 # Looks fine
df.apply(le.inverse_transform)
A B
0 o p
1 o q
2 p o
3 q p # Whoops
You will see the same result even if you iterate over the columns one by one and perform the fit_transform and inverse_transform.
You need to fit the encoder to the correct column before inversing:
le = LabelEncoder()
df_encoded = pd.DataFrame(columns=df.columns)
df_decoded = pd.DataFrame(columns=df.columns)
for col in df.columns:
df_encoded[col] = le.fit_transform(df[col])
df_encoded
A B
0 0 1
1 0 2
2 1 0
3 2 1
for col in df.columns:
le = le.fit(df[col])
df_decoded[col] = le.inverse_transform(df_encoded[col])
df_decoded
A B
0 1 p
1 1 q
2 2 o
3 3 p # Yeay
answered Mar 8 at 7:46
suicidalteddysuicidalteddy
610114
Thanks for your answer. It worked! ...another question...What if we have something like this "P,Q'' in one of our cells in column B? I want to encode P and Q separately...I think it is a list inside the cell
– CFD
Mar 8 at 16:05
add a comment |
The type of data differs from column to column, so using apply with fit_transform won't work here. It will seem to work properly but the LabelEncoder will be fitted to the rightmost column at the end of the operation, so when you'll try to apply the inverse_transform, the LabelEncoder will replace all the elements in the other columns with the ones it saw in the rightmost column. E.g.:
df = pd.DataFrame(['A': 1, 'B': 'p', 'A': 1, 'B': 'q', 'A': 2, 'B': 'o', 'A': 3, 'B': 'p'])
df
A B
0 1 p
1 1 q
2 2 o
3 3 p
df = df.apply(le.fit_transform)
df
A B
0 0 1
1 0 2
2 1 0
3 2 1 # Looks fine
df.apply(le.inverse_transform)
A B
0 o p
1 o q
2 p o
3 q p # Whoops
You will see the same result even if you iterate over the columns one by one and perform the fit_transform and inverse_transform.
You need to fit the encoder to the correct column before inversing:
le = LabelEncoder()
df_encoded = pd.DataFrame(columns=df.columns)
df_decoded = pd.DataFrame(columns=df.columns)
for col in df.columns:
df_encoded[col] = le.fit_transform(df[col])
df_encoded
A B
0 0 1
1 0 2
2 1 0
3 2 1
for col in df.columns:
le = le.fit(df[col])
df_decoded[col] = le.inverse_transform(df_encoded[col])
df_decoded
A B
0 1 p
1 1 q
2 2 o
3 3 p # Yeay
answered Mar 8 at 7:46
suicidalteddysuicidalteddy
610114
Thanks for your answer. It worked! ...another question...What if we have something like this "P,Q'' in one of our cells in column B? I want to encode P and Q separately...I think it is a list inside the cell
– CFD
Mar 8 at 16:05
add a comment |
The type of data differs from column to column, so using apply with fit_transform won't work here. It will seem to work properly but the LabelEncoder will be fitted to the rightmost column at the end of the operation, so when you'll try to apply the inverse_transform, the LabelEncoder will replace all the elements in the other columns with the ones it saw in the rightmost column. E.g.:
df = pd.DataFrame(['A': 1, 'B': 'p', 'A': 1, 'B': 'q', 'A': 2, 'B': 'o', 'A': 3, 'B': 'p'])
df
A B
0 1 p
1 1 q
2 2 o
3 3 p
df = df.apply(le.fit_transform)
df
A B
0 0 1
1 0 2
2 1 0
3 2 1 # Looks fine
df.apply(le.inverse_transform)
A B
0 o p
1 o q
2 p o
3 q p # Whoops
You will see the same result even if you iterate over the columns one by one and perform the fit_transform and inverse_transform.
You need to fit the encoder to the correct column before inversing:
le = LabelEncoder()
df_encoded = pd.DataFrame(columns=df.columns)
df_decoded = pd.DataFrame(columns=df.columns)
for col in df.columns:
df_encoded[col] = le.fit_transform(df[col])
df_encoded
A B
0 0 1
1 0 2
2 1 0
3 2 1
for col in df.columns:
le = le.fit(df[col])
df_decoded[col] = le.inverse_transform(df_encoded[col])
df_decoded
A B
0 1 p
1 1 q
2 2 o
3 3 p # Yeay
answered Mar 8 at 7:46
suicidalteddysuicidalteddy
610114
The type of data differs from column to column, so using apply with fit_transform won't work here. It will seem to work properly but the LabelEncoder will be fitted to the rightmost column at the end of the operation, so when you'll try to apply the inverse_transform, the LabelEncoder will replace all the elements in the other columns with the ones it saw in the rightmost column. E.g.:
df = pd.DataFrame(['A': 1, 'B': 'p', 'A': 1, 'B': 'q', 'A': 2, 'B': 'o', 'A': 3, 'B': 'p'])
df
A B
0 1 p
1 1 q
2 2 o
3 3 p
df = df.apply(le.fit_transform)
df
A B
0 0 1
1 0 2
2 1 0
3 2 1 # Looks fine
df.apply(le.inverse_transform)
A B
0 o p
1 o q
2 p o
3 q p # Whoops
You will see the same result even if you iterate over the columns one by one and perform the fit_transform and inverse_transform.
You need to fit the encoder to the correct column before inversing:
le = LabelEncoder()
df_encoded = pd.DataFrame(columns=df.columns)
df_decoded = pd.DataFrame(columns=df.columns)
for col in df.columns:
df_encoded[col] = le.fit_transform(df[col])
df_encoded
A B
0 0 1
1 0 2
2 1 0
3 2 1
for col in df.columns:
le = le.fit(df[col])
df_decoded[col] = le.inverse_transform(df_encoded[col])
df_decoded
A B
0 1 p
1 1 q
2 2 o
3 3 p # Yeay
answered Mar 8 at 7:46
suicidalteddysuicidalteddy
610114
edited Mar 8 at 10:38
answered Mar 8 at 7:46
suicidalteddysuicidalteddy
610114
answered Mar 8 at 7:46
suicidalteddysuicidalteddy
610114
answered Mar 8 at 7:46
suicidalteddysuicidalteddy
610114
610114
Thanks for your answer. It worked! ...another question...What if we have something like this "P,Q'' in one of our cells in column B? I want to encode P and Q separately...I think it is a list inside the cell
– CFD
Mar 8 at 16:05
add a comment |
Thanks for your answer. It worked! ...another question...What if we have something like this "P,Q'' in one of our cells in column B? I want to encode P and Q separately...I think it is a list inside the cell
– CFD
Mar 8 at 16:05
Thanks for your answer. It worked! ...another question...What if we have something like this "P,Q'' in one of our cells in column B? I want to encode P and Q separately...I think it is a list inside the cell
– CFD
Mar 8 at 16:05
Thanks for your answer. It worked! ...another question...What if we have something like this "P,Q'' in one of our cells in column B? I want to encode P and Q separately...I think it is a list inside the cell
– CFD
Mar 8 at 16:05
add a comment |
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provide complete error message and sample df too
– AkshayNevrekar
Mar 8 at 4:37
This is the complete error. I just removed the name of column for not making confusion
– CFD
Mar 8 at 6:02