Python - Swap values in multiple dataframes

up vote
6
down vote

favorite

I have a DataFrame like this

id val1 val2
0 A B
1 B B
2 A A
3 A A

And I would like swap values such as:

id val1 val2
0 B A
1 A A
2 B B
3 B B

I need to consider that the df could have other columns that I would like to keep unchanged.

asked 1 hour ago

EGM8686

1294

add a commentÂ |Â

up vote
6
down vote

favorite

I have a DataFrame like this

id val1 val2
0 A B
1 B B
2 A A
3 A A

And I would like swap values such as:

id val1 val2
0 B A
1 A A
2 B B
3 B B

I need to consider that the df could have other columns that I would like to keep unchanged.

asked 1 hour ago

EGM8686

1294

add a commentÂ |Â

up vote
6
down vote

favorite

I have a DataFrame like this

id val1 val2
0 A B
1 B B
2 A A
3 A A

And I would like swap values such as:

id val1 val2
0 B A
1 A A
2 B B
3 B B

I need to consider that the df could have other columns that I would like to keep unchanged.

asked 1 hour ago

EGM8686

1294

I have a DataFrame like this

id val1 val2
0 A B
1 B B
2 A A
3 A A

And I would like swap values such as:

id val1 val2
0 B A
1 A A
2 B B
3 B B

I need to consider that the df could have other columns that I would like to keep unchanged.

python pandas dataframe

asked 1 hour ago

EGM8686

1294

asked 1 hour ago

EGM8686

1294

asked 1 hour ago

EGM8686

1294

asked 1 hour ago

EGM8686

1294

asked 1 hour ago

EGM8686

1294

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
4
down vote

You can use pd.DataFrame.applymap with a dictionary:

d = 'B': 'A', 'A': 'B'

df = df.applymap(d.get).fillna(df)

print(df)

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

For performance, in particular memory usage, you may wish to use categorical data:

for col in df.columns[1:]:
 df[col] = df[col].astype('category')
 df[col] = df[col].cat.rename_categories(d)

answered 1 hour ago

jpp

74.7k184390

add a commentÂ |Â

up vote
4
down vote

Try stacking, mapping, and then unstacking:

df[['val1', 'val2']] = (
 df[['val1', 'val2']].stack().map('B': 'A', 'A': 'B').unstack())

df
 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

For a (much) faster solution, use a nested list comprehension.

mapping = 'B': 'A', 'A': 'B'
df[['val1', 'val2']] = [
 [mapping.get(x, x) for x in row] for row in df[['val1', 'val2']].values]

df
 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

edited 1 hour ago

answered 1 hour ago

coldspeed

108k1796166

add a commentÂ |Â

up vote
4
down vote

You can swap two values efficiently using numpy.where. However, if there are more than two values, this method stops working.

a = df[['val1', 'val2']].values
df[['val1', 'val2']] = np.where(a=='A', 'B', 'A')

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

To adapt this keep other values the same, you can use np.select:

c1 = a=='A'
c2 = a=='B'
np.select([c1, c2], ['B', 'A'], a)

edited 1 hour ago

answered 1 hour ago

user3483203

27.2k72250

add a commentÂ |Â

up vote
4
down vote

Use factorize and roll the corresponding values

def swaparoo(col):
 i, r = col.factorize()
 return pd.Series(r[(i + 1) % len(r)], col.index)

df[['id']].join(df[['val1', 'val2']].apply(swaparoo))

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

Alternative gymnastics using the same function. This incorporates the whole dataframe into the factorization.

df.set_index('id').stack().pipe(swaparoo).unstack().reset_index()

Examples

df = pd.DataFrame(dict(id=range(4), val1=[*'ABAA'], val2=[*'BBAA']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2
0 0 A B
1 1 B B
2 2 A A
3 3 A A

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

df = pd.DataFrame(dict(id=range(4), val1=[*'AAAA'], val2=[*'BBBB']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2
0 0 A B
1 1 A B
2 2 A B
3 3 A B

 id val1 val2
0 0 B A
1 1 B A
2 2 B A
3 3 B A

df = pd.DataFrame(dict(id=range(4), val1=[*'AAAA'], val2=[*'BBBB'], val3=[*'CCCC']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2 val3
0 0 A B C
1 1 A B C
2 2 A B C
3 3 A B C

 id val1 val2 val3
0 0 B C A
1 1 B C A
2 2 B C A
3 3 B C A

df = pd.DataFrame(dict(id=range(4), val1=[*'ABCD'], val2=[*'BCDA'], val3=[*'CDAB']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2 val3
0 0 A B C
1 1 B C D
2 2 C D A
3 3 D A B

 id val1 val2 val3
0 0 B C D
1 1 C D A
2 2 D A B
3 3 A B C

edited 1 hour ago

answered 1 hour ago

piRSquared

145k19127259

1

this remind me this question ;-) stackoverflow.com/questions/52506862/â€¦
â€“Â Wen
1 hour ago

I thought of the same thing.
â€“Â piRSquared
1 hour ago

add a commentÂ |Â

up vote
4
down vote

Using replace : why we need a C here , check this

df[['val1','val2']].replace('A':'C','B':'A','C':'B')
Out[263]: 
 val1 val2
0 B A
1 A A
2 B B
3 B B

edited 1 hour ago

answered 1 hour ago

Wen

87.5k72653

2

This is a funny way to go about it :). I once looked into that awful replace function... so needlessly complex the mind boggles. The docs have this "gem": You are encouraged to experiment and play with this method to gain intuition about how it works.
â€“Â jpp
1 hour ago

@jpp yep ,I am still waiting for them to fix the bug /;-(
â€“Â Wen
1 hour ago

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
4
down vote

You can use pd.DataFrame.applymap with a dictionary:

d = 'B': 'A', 'A': 'B'

df = df.applymap(d.get).fillna(df)

print(df)

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

For performance, in particular memory usage, you may wish to use categorical data:

for col in df.columns[1:]:
 df[col] = df[col].astype('category')
 df[col] = df[col].cat.rename_categories(d)

answered 1 hour ago

jpp

74.7k184390

add a commentÂ |Â

up vote
4
down vote

You can use pd.DataFrame.applymap with a dictionary:

d = 'B': 'A', 'A': 'B'

df = df.applymap(d.get).fillna(df)

print(df)

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

For performance, in particular memory usage, you may wish to use categorical data:

for col in df.columns[1:]:
 df[col] = df[col].astype('category')
 df[col] = df[col].cat.rename_categories(d)

answered 1 hour ago

jpp

74.7k184390

add a commentÂ |Â

up vote
4
down vote

You can use pd.DataFrame.applymap with a dictionary:

d = 'B': 'A', 'A': 'B'

df = df.applymap(d.get).fillna(df)

print(df)

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

For performance, in particular memory usage, you may wish to use categorical data:

for col in df.columns[1:]:
 df[col] = df[col].astype('category')
 df[col] = df[col].cat.rename_categories(d)

answered 1 hour ago

jpp

74.7k184390

You can use pd.DataFrame.applymap with a dictionary:

d = 'B': 'A', 'A': 'B'

df = df.applymap(d.get).fillna(df)

print(df)

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

For performance, in particular memory usage, you may wish to use categorical data:

for col in df.columns[1:]:
 df[col] = df[col].astype('category')
 df[col] = df[col].cat.rename_categories(d)

answered 1 hour ago

jpp

74.7k184390

answered 1 hour ago

jpp

74.7k184390

answered 1 hour ago

jpp

74.7k184390

answered 1 hour ago

jpp

74.7k184390

add a commentÂ |Â

up vote
4
down vote

Try stacking, mapping, and then unstacking:

df[['val1', 'val2']] = (
 df[['val1', 'val2']].stack().map('B': 'A', 'A': 'B').unstack())

df
 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

For a (much) faster solution, use a nested list comprehension.

mapping = 'B': 'A', 'A': 'B'
df[['val1', 'val2']] = [
 [mapping.get(x, x) for x in row] for row in df[['val1', 'val2']].values]

df
 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

edited 1 hour ago

answered 1 hour ago

coldspeed

108k1796166

add a commentÂ |Â

up vote
4
down vote

Try stacking, mapping, and then unstacking:

df[['val1', 'val2']] = (
 df[['val1', 'val2']].stack().map('B': 'A', 'A': 'B').unstack())

df
 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

For a (much) faster solution, use a nested list comprehension.

mapping = 'B': 'A', 'A': 'B'
df[['val1', 'val2']] = [
 [mapping.get(x, x) for x in row] for row in df[['val1', 'val2']].values]

df
 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

edited 1 hour ago

answered 1 hour ago

coldspeed

108k1796166

add a commentÂ |Â

up vote
4
down vote

Try stacking, mapping, and then unstacking:

df[['val1', 'val2']] = (
 df[['val1', 'val2']].stack().map('B': 'A', 'A': 'B').unstack())

df
 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

For a (much) faster solution, use a nested list comprehension.

mapping = 'B': 'A', 'A': 'B'
df[['val1', 'val2']] = [
 [mapping.get(x, x) for x in row] for row in df[['val1', 'val2']].values]

df
 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

edited 1 hour ago

answered 1 hour ago

coldspeed

108k1796166

Try stacking, mapping, and then unstacking:

df[['val1', 'val2']] = (
 df[['val1', 'val2']].stack().map('B': 'A', 'A': 'B').unstack())

df
 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

For a (much) faster solution, use a nested list comprehension.

mapping = 'B': 'A', 'A': 'B'
df[['val1', 'val2']] = [
 [mapping.get(x, x) for x in row] for row in df[['val1', 'val2']].values]

df
 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

edited 1 hour ago

answered 1 hour ago

coldspeed

108k1796166

edited 1 hour ago

answered 1 hour ago

coldspeed

108k1796166

answered 1 hour ago

coldspeed

108k1796166

answered 1 hour ago

coldspeed

108k1796166

add a commentÂ |Â

up vote
4
down vote

You can swap two values efficiently using numpy.where. However, if there are more than two values, this method stops working.

a = df[['val1', 'val2']].values
df[['val1', 'val2']] = np.where(a=='A', 'B', 'A')

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

To adapt this keep other values the same, you can use np.select:

c1 = a=='A'
c2 = a=='B'
np.select([c1, c2], ['B', 'A'], a)

edited 1 hour ago

answered 1 hour ago

user3483203

27.2k72250

add a commentÂ |Â

up vote
4
down vote

You can swap two values efficiently using numpy.where. However, if there are more than two values, this method stops working.

a = df[['val1', 'val2']].values
df[['val1', 'val2']] = np.where(a=='A', 'B', 'A')

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

To adapt this keep other values the same, you can use np.select:

c1 = a=='A'
c2 = a=='B'
np.select([c1, c2], ['B', 'A'], a)

edited 1 hour ago

answered 1 hour ago

user3483203

27.2k72250

add a commentÂ |Â

up vote
4
down vote

You can swap two values efficiently using numpy.where. However, if there are more than two values, this method stops working.

a = df[['val1', 'val2']].values
df[['val1', 'val2']] = np.where(a=='A', 'B', 'A')

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

To adapt this keep other values the same, you can use np.select:

c1 = a=='A'
c2 = a=='B'
np.select([c1, c2], ['B', 'A'], a)

edited 1 hour ago

answered 1 hour ago

user3483203

27.2k72250

You can swap two values efficiently using numpy.where. However, if there are more than two values, this method stops working.

a = df[['val1', 'val2']].values
df[['val1', 'val2']] = np.where(a=='A', 'B', 'A')

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

To adapt this keep other values the same, you can use np.select:

c1 = a=='A'
c2 = a=='B'
np.select([c1, c2], ['B', 'A'], a)

edited 1 hour ago

answered 1 hour ago

user3483203

27.2k72250

edited 1 hour ago

answered 1 hour ago

user3483203

27.2k72250

answered 1 hour ago

user3483203

27.2k72250

answered 1 hour ago

user3483203

27.2k72250

add a commentÂ |Â

up vote
4
down vote

Use factorize and roll the corresponding values

def swaparoo(col):
 i, r = col.factorize()
 return pd.Series(r[(i + 1) % len(r)], col.index)

df[['id']].join(df[['val1', 'val2']].apply(swaparoo))

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

Alternative gymnastics using the same function. This incorporates the whole dataframe into the factorization.

df.set_index('id').stack().pipe(swaparoo).unstack().reset_index()

Examples

df = pd.DataFrame(dict(id=range(4), val1=[*'ABAA'], val2=[*'BBAA']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2
0 0 A B
1 1 B B
2 2 A A
3 3 A A

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

df = pd.DataFrame(dict(id=range(4), val1=[*'AAAA'], val2=[*'BBBB']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2
0 0 A B
1 1 A B
2 2 A B
3 3 A B

 id val1 val2
0 0 B A
1 1 B A
2 2 B A
3 3 B A

df = pd.DataFrame(dict(id=range(4), val1=[*'AAAA'], val2=[*'BBBB'], val3=[*'CCCC']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2 val3
0 0 A B C
1 1 A B C
2 2 A B C
3 3 A B C

 id val1 val2 val3
0 0 B C A
1 1 B C A
2 2 B C A
3 3 B C A

df = pd.DataFrame(dict(id=range(4), val1=[*'ABCD'], val2=[*'BCDA'], val3=[*'CDAB']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2 val3
0 0 A B C
1 1 B C D
2 2 C D A
3 3 D A B

 id val1 val2 val3
0 0 B C D
1 1 C D A
2 2 D A B
3 3 A B C

edited 1 hour ago

answered 1 hour ago

piRSquared

145k19127259

1

this remind me this question ;-) stackoverflow.com/questions/52506862/â€¦
â€“Â Wen
1 hour ago

I thought of the same thing.
â€“Â piRSquared
1 hour ago

add a commentÂ |Â

up vote
4
down vote

Use factorize and roll the corresponding values

def swaparoo(col):
 i, r = col.factorize()
 return pd.Series(r[(i + 1) % len(r)], col.index)

df[['id']].join(df[['val1', 'val2']].apply(swaparoo))

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

Alternative gymnastics using the same function. This incorporates the whole dataframe into the factorization.

df.set_index('id').stack().pipe(swaparoo).unstack().reset_index()

Examples

df = pd.DataFrame(dict(id=range(4), val1=[*'ABAA'], val2=[*'BBAA']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2
0 0 A B
1 1 B B
2 2 A A
3 3 A A

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

df = pd.DataFrame(dict(id=range(4), val1=[*'AAAA'], val2=[*'BBBB']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2
0 0 A B
1 1 A B
2 2 A B
3 3 A B

 id val1 val2
0 0 B A
1 1 B A
2 2 B A
3 3 B A

df = pd.DataFrame(dict(id=range(4), val1=[*'AAAA'], val2=[*'BBBB'], val3=[*'CCCC']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2 val3
0 0 A B C
1 1 A B C
2 2 A B C
3 3 A B C

 id val1 val2 val3
0 0 B C A
1 1 B C A
2 2 B C A
3 3 B C A

df = pd.DataFrame(dict(id=range(4), val1=[*'ABCD'], val2=[*'BCDA'], val3=[*'CDAB']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2 val3
0 0 A B C
1 1 B C D
2 2 C D A
3 3 D A B

 id val1 val2 val3
0 0 B C D
1 1 C D A
2 2 D A B
3 3 A B C

edited 1 hour ago

answered 1 hour ago

piRSquared

145k19127259

1

this remind me this question ;-) stackoverflow.com/questions/52506862/â€¦
â€“Â Wen
1 hour ago

I thought of the same thing.
â€“Â piRSquared
1 hour ago

add a commentÂ |Â

up vote
4
down vote

Use factorize and roll the corresponding values

def swaparoo(col):
 i, r = col.factorize()
 return pd.Series(r[(i + 1) % len(r)], col.index)

df[['id']].join(df[['val1', 'val2']].apply(swaparoo))

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

Alternative gymnastics using the same function. This incorporates the whole dataframe into the factorization.

df.set_index('id').stack().pipe(swaparoo).unstack().reset_index()

Examples

df = pd.DataFrame(dict(id=range(4), val1=[*'ABAA'], val2=[*'BBAA']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2
0 0 A B
1 1 B B
2 2 A A
3 3 A A

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

df = pd.DataFrame(dict(id=range(4), val1=[*'AAAA'], val2=[*'BBBB']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2
0 0 A B
1 1 A B
2 2 A B
3 3 A B

 id val1 val2
0 0 B A
1 1 B A
2 2 B A
3 3 B A

df = pd.DataFrame(dict(id=range(4), val1=[*'AAAA'], val2=[*'BBBB'], val3=[*'CCCC']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2 val3
0 0 A B C
1 1 A B C
2 2 A B C
3 3 A B C

 id val1 val2 val3
0 0 B C A
1 1 B C A
2 2 B C A
3 3 B C A

df = pd.DataFrame(dict(id=range(4), val1=[*'ABCD'], val2=[*'BCDA'], val3=[*'CDAB']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2 val3
0 0 A B C
1 1 B C D
2 2 C D A
3 3 D A B

 id val1 val2 val3
0 0 B C D
1 1 C D A
2 2 D A B
3 3 A B C

edited 1 hour ago

answered 1 hour ago

piRSquared

145k19127259

Use factorize and roll the corresponding values

def swaparoo(col):
 i, r = col.factorize()
 return pd.Series(r[(i + 1) % len(r)], col.index)

df[['id']].join(df[['val1', 'val2']].apply(swaparoo))

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

Alternative gymnastics using the same function. This incorporates the whole dataframe into the factorization.

df.set_index('id').stack().pipe(swaparoo).unstack().reset_index()

Examples

df = pd.DataFrame(dict(id=range(4), val1=[*'ABAA'], val2=[*'BBAA']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2
0 0 A B
1 1 B B
2 2 A A
3 3 A A

 id val1 val2
0 0 B A
1 1 A A
2 2 B B
3 3 B B

df = pd.DataFrame(dict(id=range(4), val1=[*'AAAA'], val2=[*'BBBB']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2
0 0 A B
1 1 A B
2 2 A B
3 3 A B

 id val1 val2
0 0 B A
1 1 B A
2 2 B A
3 3 B A

df = pd.DataFrame(dict(id=range(4), val1=[*'AAAA'], val2=[*'BBBB'], val3=[*'CCCC']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2 val3
0 0 A B C
1 1 A B C
2 2 A B C
3 3 A B C

 id val1 val2 val3
0 0 B C A
1 1 B C A
2 2 B C A
3 3 B C A

df = pd.DataFrame(dict(id=range(4), val1=[*'ABCD'], val2=[*'BCDA'], val3=[*'CDAB']))

print(
 df,
 df.set_index('id').stack().pipe(swaparoo).unstack().reset_index(),
 sep='nn'
)

 id val1 val2 val3
0 0 A B C
1 1 B C D
2 2 C D A
3 3 D A B

 id val1 val2 val3
0 0 B C D
1 1 C D A
2 2 D A B
3 3 A B C

edited 1 hour ago

answered 1 hour ago

piRSquared

145k19127259

edited 1 hour ago

answered 1 hour ago

piRSquared

145k19127259

answered 1 hour ago

piRSquared

145k19127259

answered 1 hour ago

piRSquared

145k19127259

1

this remind me this question ;-) stackoverflow.com/questions/52506862/â€¦
â€“Â Wen
1 hour ago

I thought of the same thing.
â€“Â piRSquared
1 hour ago

add a commentÂ |Â

1

this remind me this question ;-) stackoverflow.com/questions/52506862/â€¦
â€“Â Wen
1 hour ago

I thought of the same thing.
â€“Â piRSquared
1 hour ago

this remind me this question ;-) stackoverflow.com/questions/52506862/â€¦
â€“Â Wen
1 hour ago

I thought of the same thing.
â€“Â piRSquared
1 hour ago

add a commentÂ |Â

up vote
4
down vote

Using replace : why we need a C here , check this

df[['val1','val2']].replace('A':'C','B':'A','C':'B')
Out[263]: 
 val1 val2
0 B A
1 A A
2 B B
3 B B

edited 1 hour ago

answered 1 hour ago

Wen

87.5k72653

2

This is a funny way to go about it :). I once looked into that awful replace function... so needlessly complex the mind boggles. The docs have this "gem": You are encouraged to experiment and play with this method to gain intuition about how it works.
â€“Â jpp
1 hour ago

@jpp yep ,I am still waiting for them to fix the bug /;-(
â€“Â Wen
1 hour ago

add a commentÂ |Â

up vote
4
down vote

Using replace : why we need a C here , check this

df[['val1','val2']].replace('A':'C','B':'A','C':'B')
Out[263]: 
 val1 val2
0 B A
1 A A
2 B B
3 B B

edited 1 hour ago

answered 1 hour ago

Wen

87.5k72653

2

This is a funny way to go about it :). I once looked into that awful replace function... so needlessly complex the mind boggles. The docs have this "gem": You are encouraged to experiment and play with this method to gain intuition about how it works.
â€“Â jpp
1 hour ago

@jpp yep ,I am still waiting for them to fix the bug /;-(
â€“Â Wen
1 hour ago

add a commentÂ |Â

up vote
4
down vote

Using replace : why we need a C here , check this

df[['val1','val2']].replace('A':'C','B':'A','C':'B')
Out[263]: 
 val1 val2
0 B A
1 A A
2 B B
3 B B

edited 1 hour ago

answered 1 hour ago

Wen

87.5k72653

Using replace : why we need a C here , check this

df[['val1','val2']].replace('A':'C','B':'A','C':'B')
Out[263]: 
 val1 val2
0 B A
1 A A
2 B B
3 B B

edited 1 hour ago

answered 1 hour ago

Wen

87.5k72653

edited 1 hour ago

answered 1 hour ago

Wen

87.5k72653

answered 1 hour ago

Wen

87.5k72653

answered 1 hour ago

Wen

87.5k72653

2

This is a funny way to go about it :). I once looked into that awful replace function... so needlessly complex the mind boggles. The docs have this "gem": You are encouraged to experiment and play with this method to gain intuition about how it works.
â€“Â jpp
1 hour ago

@jpp yep ,I am still waiting for them to fix the bug /;-(
â€“Â Wen
1 hour ago

add a commentÂ |Â

2

This is a funny way to go about it :). I once looked into that awful replace function... so needlessly complex the mind boggles. The docs have this "gem": You are encouraged to experiment and play with this method to gain intuition about how it works.
â€“Â jpp
1 hour ago

@jpp yep ,I am still waiting for them to fix the bug /;-(
â€“Â Wen
1 hour ago

This is a funny way to go about it :). I once looked into that awful replace function... so needlessly complex the mind boggles. The docs have this "gem": You are encouraged to experiment and play with this method to gain intuition about how it works.
â€“Â jpp
1 hour ago

@jpp yep ,I am still waiting for them to fix the bug /;-(
â€“Â Wen
1 hour ago

add a commentÂ |Â

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