Not able to replace the string containing $ in pandas column

up vote
6
down vote

favorite

I have a dataframe

df = pd.DataFrame('a':[1,2,3], 'b':[5, '12$sell', '1$sell'])

I want to replace $sell from column b.

So I tried replace() method like below

df['b'] = df['b'].str.replace("$sell","")

but it's doesn't replace the given string and it gives me same dataframe as original.

It's working when I use it with apply

df['b'] = df['b'].apply(lambda x: str(x).replace("$sell",""))

So I want to know why it is not working in previous case?

Note: I tried replacing only $ and shockingly it works.

edited 1 hour ago

jpp

67k173984

asked 2 hours ago

Akshay Nevrekar

2,29251331

add a commentÂ |Â

up vote
6
down vote

favorite

I have a dataframe

df = pd.DataFrame('a':[1,2,3], 'b':[5, '12$sell', '1$sell'])

I want to replace $sell from column b.

So I tried replace() method like below

df['b'] = df['b'].str.replace("$sell","")

but it's doesn't replace the given string and it gives me same dataframe as original.

It's working when I use it with apply

df['b'] = df['b'].apply(lambda x: str(x).replace("$sell",""))

So I want to know why it is not working in previous case?

Note: I tried replacing only $ and shockingly it works.

edited 1 hour ago

jpp

67k173984

asked 2 hours ago

Akshay Nevrekar

2,29251331

add a commentÂ |Â

up vote
6
down vote

favorite

I have a dataframe

df = pd.DataFrame('a':[1,2,3], 'b':[5, '12$sell', '1$sell'])

I want to replace $sell from column b.

So I tried replace() method like below

df['b'] = df['b'].str.replace("$sell","")

but it's doesn't replace the given string and it gives me same dataframe as original.

It's working when I use it with apply

df['b'] = df['b'].apply(lambda x: str(x).replace("$sell",""))

So I want to know why it is not working in previous case?

Note: I tried replacing only $ and shockingly it works.

edited 1 hour ago

jpp

67k173984

asked 2 hours ago

Akshay Nevrekar

2,29251331

I have a dataframe

df = pd.DataFrame('a':[1,2,3], 'b':[5, '12$sell', '1$sell'])

I want to replace $sell from column b.

So I tried replace() method like below

df['b'] = df['b'].str.replace("$sell","")

but it's doesn't replace the given string and it gives me same dataframe as original.

It's working when I use it with apply

df['b'] = df['b'].apply(lambda x: str(x).replace("$sell",""))

So I want to know why it is not working in previous case?

Note: I tried replacing only $ and shockingly it works.

python string pandas series

edited 1 hour ago

jpp

67k173984

asked 2 hours ago

Akshay Nevrekar

2,29251331

edited 1 hour ago

jpp

67k173984

asked 2 hours ago

Akshay Nevrekar

2,29251331

edited 1 hour ago

jpp

67k173984

edited 1 hour ago

jpp

67k173984

edited 1 hour ago

jpp

67k173984

asked 2 hours ago

Akshay Nevrekar

2,29251331

asked 2 hours ago

Akshay Nevrekar

2,29251331

asked 2 hours ago

Akshay Nevrekar

2,29251331

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
5
down vote

accepted

It is regex metacharacter (end of string), escape it or add parameter regex=False:

df['b'] = df['b'].str.replace("$sell","")
print (df)
 a b
0 1 NaN
1 2 12
2 3 1

df['b'] = df['b'].str.replace("$sell","", regex=False)

If want also value 5, what is numeric, use Series.replace with regex=True for replace substrings - numeric values are not touched:

df['b'] = df['b'].replace("$sell","", regex=True)

print (df['b'].apply(type))
0 <class 'int'>
1 <class 'str'>
2 <class 'str'>
Name: b, dtype: object

Or cast to strings all data of column:

df['b'] = df['b'].astype(str).str.replace("$sell","", regex=False)

print (df['b'].apply(type))
0 <class 'str'>
1 <class 'str'>
2 <class 'str'>
Name: b, dtype: object

And for better performance if no missing values is possible use list comprehension:

df['b'] = [str(x).replace("$sell","") for x in df['b']]

print (df)
 a b
0 1 5
1 2 12
2 3 1

edited 1 hour ago

answered 1 hour ago

jezrael

290k19205283

add a commentÂ |Â

up vote
4
down vote

df['b'] = df['b'].str.replace("$sell","", regex=False)

answered 1 hour ago

Charles R

437211

add a commentÂ |Â

up vote
4
down vote

$ is a regex special character. By default, pd.Series.str.replace uses regex=True.

Instead, specify regex=False:

df['b'] = df['b'].str.replace('$sell', '', regex=False)

answered 1 hour ago

jpp

67k173984

add a commentÂ |Â

up vote
3
down vote

str.replace assumes a regex is being used. so you need to use escape i.e.

df['b'] = df['b'].str.replace("$sell","")

answered 1 hour ago

sgDysregulation

1,95611125

add a commentÂ |Â

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
5
down vote

accepted

It is regex metacharacter (end of string), escape it or add parameter regex=False:

df['b'] = df['b'].str.replace("$sell","")
print (df)
 a b
0 1 NaN
1 2 12
2 3 1

df['b'] = df['b'].str.replace("$sell","", regex=False)

If want also value 5, what is numeric, use Series.replace with regex=True for replace substrings - numeric values are not touched:

df['b'] = df['b'].replace("$sell","", regex=True)

print (df['b'].apply(type))
0 <class 'int'>
1 <class 'str'>
2 <class 'str'>
Name: b, dtype: object

Or cast to strings all data of column:

df['b'] = df['b'].astype(str).str.replace("$sell","", regex=False)

print (df['b'].apply(type))
0 <class 'str'>
1 <class 'str'>
2 <class 'str'>
Name: b, dtype: object

And for better performance if no missing values is possible use list comprehension:

df['b'] = [str(x).replace("$sell","") for x in df['b']]

print (df)
 a b
0 1 5
1 2 12
2 3 1

edited 1 hour ago

answered 1 hour ago

jezrael

290k19205283

add a commentÂ |Â

up vote
5
down vote

accepted

It is regex metacharacter (end of string), escape it or add parameter regex=False:

df['b'] = df['b'].str.replace("$sell","")
print (df)
 a b
0 1 NaN
1 2 12
2 3 1

df['b'] = df['b'].str.replace("$sell","", regex=False)

If want also value 5, what is numeric, use Series.replace with regex=True for replace substrings - numeric values are not touched:

df['b'] = df['b'].replace("$sell","", regex=True)

print (df['b'].apply(type))
0 <class 'int'>
1 <class 'str'>
2 <class 'str'>
Name: b, dtype: object

Or cast to strings all data of column:

df['b'] = df['b'].astype(str).str.replace("$sell","", regex=False)

print (df['b'].apply(type))
0 <class 'str'>
1 <class 'str'>
2 <class 'str'>
Name: b, dtype: object

And for better performance if no missing values is possible use list comprehension:

df['b'] = [str(x).replace("$sell","") for x in df['b']]

print (df)
 a b
0 1 5
1 2 12
2 3 1

edited 1 hour ago

answered 1 hour ago

jezrael

290k19205283

add a commentÂ |Â

up vote
5
down vote

accepted

It is regex metacharacter (end of string), escape it or add parameter regex=False:

df['b'] = df['b'].str.replace("$sell","")
print (df)
 a b
0 1 NaN
1 2 12
2 3 1

df['b'] = df['b'].str.replace("$sell","", regex=False)

If want also value 5, what is numeric, use Series.replace with regex=True for replace substrings - numeric values are not touched:

df['b'] = df['b'].replace("$sell","", regex=True)

print (df['b'].apply(type))
0 <class 'int'>
1 <class 'str'>
2 <class 'str'>
Name: b, dtype: object

Or cast to strings all data of column:

df['b'] = df['b'].astype(str).str.replace("$sell","", regex=False)

print (df['b'].apply(type))
0 <class 'str'>
1 <class 'str'>
2 <class 'str'>
Name: b, dtype: object

And for better performance if no missing values is possible use list comprehension:

df['b'] = [str(x).replace("$sell","") for x in df['b']]

print (df)
 a b
0 1 5
1 2 12
2 3 1

edited 1 hour ago

answered 1 hour ago

jezrael

290k19205283

It is regex metacharacter (end of string), escape it or add parameter regex=False:

df['b'] = df['b'].str.replace("$sell","")
print (df)
 a b
0 1 NaN
1 2 12
2 3 1

df['b'] = df['b'].str.replace("$sell","", regex=False)

If want also value 5, what is numeric, use Series.replace with regex=True for replace substrings - numeric values are not touched:

df['b'] = df['b'].replace("$sell","", regex=True)

print (df['b'].apply(type))
0 <class 'int'>
1 <class 'str'>
2 <class 'str'>
Name: b, dtype: object

Or cast to strings all data of column:

df['b'] = df['b'].astype(str).str.replace("$sell","", regex=False)

print (df['b'].apply(type))
0 <class 'str'>
1 <class 'str'>
2 <class 'str'>
Name: b, dtype: object

And for better performance if no missing values is possible use list comprehension:

df['b'] = [str(x).replace("$sell","") for x in df['b']]

print (df)
 a b
0 1 5
1 2 12
2 3 1

edited 1 hour ago

answered 1 hour ago

jezrael

290k19205283

edited 1 hour ago

answered 1 hour ago

jezrael

290k19205283

answered 1 hour ago

jezrael

290k19205283

answered 1 hour ago

jezrael

290k19205283

add a commentÂ |Â

up vote
4
down vote

df['b'] = df['b'].str.replace("$sell","", regex=False)

answered 1 hour ago

Charles R

437211

add a commentÂ |Â

up vote
4
down vote

df['b'] = df['b'].str.replace("$sell","", regex=False)

answered 1 hour ago

Charles R

437211

add a commentÂ |Â

up vote
4
down vote

df['b'] = df['b'].str.replace("$sell","", regex=False)

answered 1 hour ago

Charles R

437211

df['b'] = df['b'].str.replace("$sell","", regex=False)

answered 1 hour ago

Charles R

437211

answered 1 hour ago

Charles R

437211

answered 1 hour ago

Charles R

437211

answered 1 hour ago

Charles R

437211

add a commentÂ |Â

up vote
4
down vote

$ is a regex special character. By default, pd.Series.str.replace uses regex=True.

Instead, specify regex=False:

df['b'] = df['b'].str.replace('$sell', '', regex=False)

answered 1 hour ago

jpp

67k173984

add a commentÂ |Â

up vote
4
down vote

$ is a regex special character. By default, pd.Series.str.replace uses regex=True.

Instead, specify regex=False:

df['b'] = df['b'].str.replace('$sell', '', regex=False)

answered 1 hour ago

jpp

67k173984

add a commentÂ |Â

up vote
4
down vote

$ is a regex special character. By default, pd.Series.str.replace uses regex=True.

Instead, specify regex=False:

df['b'] = df['b'].str.replace('$sell', '', regex=False)

answered 1 hour ago

jpp

67k173984

$ is a regex special character. By default, pd.Series.str.replace uses regex=True.

Instead, specify regex=False:

df['b'] = df['b'].str.replace('$sell', '', regex=False)

answered 1 hour ago

jpp

67k173984

answered 1 hour ago

jpp

67k173984

answered 1 hour ago

jpp

67k173984

answered 1 hour ago

jpp

67k173984

add a commentÂ |Â

up vote
3
down vote

str.replace assumes a regex is being used. so you need to use escape i.e.

df['b'] = df['b'].str.replace("$sell","")

answered 1 hour ago

sgDysregulation

1,95611125

add a commentÂ |Â

up vote
3
down vote

str.replace assumes a regex is being used. so you need to use escape i.e.

df['b'] = df['b'].str.replace("$sell","")

answered 1 hour ago

sgDysregulation

1,95611125

add a commentÂ |Â

up vote
3
down vote

str.replace assumes a regex is being used. so you need to use escape i.e.

df['b'] = df['b'].str.replace("$sell","")

answered 1 hour ago

sgDysregulation

1,95611125

str.replace assumes a regex is being used. so you need to use escape i.e.

df['b'] = df['b'].str.replace("$sell","")

answered 1 hour ago

sgDysregulation

1,95611125

answered 1 hour ago

sgDysregulation

1,95611125

answered 1 hour ago

sgDysregulation

1,95611125

answered 1 hour ago

sgDysregulation

1,95611125

add a commentÂ |Â

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