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How to convert value's in DF by loop?
How to merge two dictionaries in a single expression?How do I check if a list is empty?How do I check whether a file exists without exceptions?How can I safely create a nested directory?Converting string into datetimeAccessing the index in 'for' loops?Convert bytes to a string?How do I sort a dictionary by value?How do I list all files of a directory?Iterating over dictionaries using 'for' loops
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I have data set with string column and I had to convert them into int value the problem is some columns have 90+ different value and to convert them
manually and take time, there is a function to solve it automatically?
I did it, but manually:
price_dict = 91:0, 65:1, 0:1, 20:1, 35:1, 32:1, 41:1, 36:1, 15:1, 90:1, 6:1, 67:1, 2:1, 57:1, 39:1, 1:1, 79:1, 34:1, 85:1 # not all
app[`Price`] = app[`Price`].apply(lambda a: price_dict[a]) # 0 = free, 1 = not free
app[`Price`].value_counts() # To check.
I have tried :
for x in app[`Price`]:
if x == 1:
price_dict = 91:0
else:
price_dict = x:1
app[`Price`] = app[`Price`].apply(lambda a: price_dict[a])# 0 = free, 1 = not free
app[`Price`].value_counts()
python
edited Mar 24 at 5:11
Miroslav Glamuzina
2,85721223
asked Mar 24 at 2:31
mim alamlkimim alamlki
31
add a comment |
I have data set with string column and I had to convert them into int value the problem is some columns have 90+ different value and to convert them
manually and take time, there is a function to solve it automatically?
I did it, but manually:
price_dict = 91:0, 65:1, 0:1, 20:1, 35:1, 32:1, 41:1, 36:1, 15:1, 90:1, 6:1, 67:1, 2:1, 57:1, 39:1, 1:1, 79:1, 34:1, 85:1 # not all
app[`Price`] = app[`Price`].apply(lambda a: price_dict[a]) # 0 = free, 1 = not free
app[`Price`].value_counts() # To check.
I have tried :
for x in app[`Price`]:
if x == 1:
price_dict = 91:0
else:
price_dict = x:1
app[`Price`] = app[`Price`].apply(lambda a: price_dict[a])# 0 = free, 1 = not free
app[`Price`].value_counts()
python
edited Mar 24 at 5:11
Miroslav Glamuzina
2,85721223
asked Mar 24 at 2:31
mim alamlkimim alamlki
31
add a comment |
I have data set with string column and I had to convert them into int value the problem is some columns have 90+ different value and to convert them
manually and take time, there is a function to solve it automatically?
I did it, but manually:
price_dict = 91:0, 65:1, 0:1, 20:1, 35:1, 32:1, 41:1, 36:1, 15:1, 90:1, 6:1, 67:1, 2:1, 57:1, 39:1, 1:1, 79:1, 34:1, 85:1 # not all
app[`Price`] = app[`Price`].apply(lambda a: price_dict[a]) # 0 = free, 1 = not free
app[`Price`].value_counts() # To check.
I have tried :
for x in app[`Price`]:
if x == 1:
price_dict = 91:0
else:
price_dict = x:1
app[`Price`] = app[`Price`].apply(lambda a: price_dict[a])# 0 = free, 1 = not free
app[`Price`].value_counts()
python
edited Mar 24 at 5:11
Miroslav Glamuzina
2,85721223
asked Mar 24 at 2:31
mim alamlkimim alamlki
31
I have data set with string column and I had to convert them into int value the problem is some columns have 90+ different value and to convert them
manually and take time, there is a function to solve it automatically?
I did it, but manually:
price_dict = 91:0, 65:1, 0:1, 20:1, 35:1, 32:1, 41:1, 36:1, 15:1, 90:1, 6:1, 67:1, 2:1, 57:1, 39:1, 1:1, 79:1, 34:1, 85:1 # not all
app[`Price`] = app[`Price`].apply(lambda a: price_dict[a]) # 0 = free, 1 = not free
app[`Price`].value_counts() # To check.
I have tried :
for x in app[`Price`]:
if x == 1:
price_dict = 91:0
else:
price_dict = x:1
app[`Price`] = app[`Price`].apply(lambda a: price_dict[a])# 0 = free, 1 = not free
app[`Price`].value_counts()
python
python
edited Mar 24 at 5:11
Miroslav Glamuzina
2,85721223
asked Mar 24 at 2:31
mim alamlkimim alamlki
31
edited Mar 24 at 5:11
Miroslav Glamuzina
2,85721223
asked Mar 24 at 2:31
mim alamlkimim alamlki
31
edited Mar 24 at 5:11
Miroslav Glamuzina
2,85721223
edited Mar 24 at 5:11
Miroslav Glamuzina
2,85721223
edited Mar 24 at 5:11
Miroslav Glamuzina
2,85721223
2,85721223
asked Mar 24 at 2:31
mim alamlkimim alamlki
31
asked Mar 24 at 2:31
mim alamlkimim alamlki
31
asked Mar 24 at 2:31
mim alamlkimim alamlki
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You can use map to transform a column using a dictionary to define the transformation:
price_dict = 91:0, 65:1, 0:1, 20:1, 35:1, 32:1, 41:1, 36:1, 15:1, 90:1, 6:1, 67:1, 2:1, 57:1, 39:1, 1:1, 79:1, 34:1, 85:1 # not all
app['Price'] = app['Price'].map(price_dict)
app['Price'].value_counts()
You can also replace values in columns like this:
app['Price_new'] = 1
app.loc[app.Price == 91, 'Price_new'] = 0
answered Mar 24 at 2:35
NathanielNathaniel
2,225314
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use map to transform a column using a dictionary to define the transformation:
price_dict = 91:0, 65:1, 0:1, 20:1, 35:1, 32:1, 41:1, 36:1, 15:1, 90:1, 6:1, 67:1, 2:1, 57:1, 39:1, 1:1, 79:1, 34:1, 85:1 # not all
app['Price'] = app['Price'].map(price_dict)
app['Price'].value_counts()
You can also replace values in columns like this:
app['Price_new'] = 1
app.loc[app.Price == 91, 'Price_new'] = 0
answered Mar 24 at 2:35
NathanielNathaniel
2,225314
add a comment |
You can use map to transform a column using a dictionary to define the transformation:
price_dict = 91:0, 65:1, 0:1, 20:1, 35:1, 32:1, 41:1, 36:1, 15:1, 90:1, 6:1, 67:1, 2:1, 57:1, 39:1, 1:1, 79:1, 34:1, 85:1 # not all
app['Price'] = app['Price'].map(price_dict)
app['Price'].value_counts()
You can also replace values in columns like this:
app['Price_new'] = 1
app.loc[app.Price == 91, 'Price_new'] = 0
answered Mar 24 at 2:35
NathanielNathaniel
2,225314
add a comment |
You can use map to transform a column using a dictionary to define the transformation:
price_dict = 91:0, 65:1, 0:1, 20:1, 35:1, 32:1, 41:1, 36:1, 15:1, 90:1, 6:1, 67:1, 2:1, 57:1, 39:1, 1:1, 79:1, 34:1, 85:1 # not all
app['Price'] = app['Price'].map(price_dict)
app['Price'].value_counts()
You can also replace values in columns like this:
app['Price_new'] = 1
app.loc[app.Price == 91, 'Price_new'] = 0
answered Mar 24 at 2:35
NathanielNathaniel
2,225314
You can use map to transform a column using a dictionary to define the transformation:
price_dict = 91:0, 65:1, 0:1, 20:1, 35:1, 32:1, 41:1, 36:1, 15:1, 90:1, 6:1, 67:1, 2:1, 57:1, 39:1, 1:1, 79:1, 34:1, 85:1 # not all
app['Price'] = app['Price'].map(price_dict)
app['Price'].value_counts()
You can also replace values in columns like this:
app['Price_new'] = 1
app.loc[app.Price == 91, 'Price_new'] = 0
answered Mar 24 at 2:35
NathanielNathaniel
2,225314
edited Mar 24 at 2:41
answered Mar 24 at 2:35
NathanielNathaniel
2,225314
answered Mar 24 at 2:35
NathanielNathaniel
2,225314
answered Mar 24 at 2:35
NathanielNathaniel
2,225314
2,225314
add a comment |
add a comment |
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