Replace 1s in one hot columns with values from another column
up vote
10
down vote
favorite
1
I have a data frame that looks like this:
df = pd.DataFrame({"value": [4, 5, 3], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]})
df
value item1 item2 item3
0 4 0 1 0
1 5 1 0 0
2 3 0 0 1
Basically what I want to do is replace the value of the one hot encoded elements with the value from the "value" column and then delete the "value" column. The resulting data frame should be like this:
df_out = pd.DataFrame({"item1": [0, 5, 0], "item2": [4, 0, 0], "item3": [0, 0, 3]})
item1 item2 item3
0 0 4 0
1 5 0 0
2 0 0 3
python pandas dataframe
edited Dec 5 at 12:55
coldspeed
115k18105182
asked Dec 5 at 12:41
Gorjan Radevski
705
add a comment |
up vote
10
down vote
favorite
1
I have a data frame that looks like this:
df = pd.DataFrame({"value": [4, 5, 3], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]})
df
value item1 item2 item3
0 4 0 1 0
1 5 1 0 0
2 3 0 0 1
Basically what I want to do is replace the value of the one hot encoded elements with the value from the "value" column and then delete the "value" column. The resulting data frame should be like this:
df_out = pd.DataFrame({"item1": [0, 5, 0], "item2": [4, 0, 0], "item3": [0, 0, 3]})
item1 item2 item3
0 0 4 0
1 5 0 0
2 0 0 3
python pandas dataframe
edited Dec 5 at 12:55
coldspeed
115k18105182
asked Dec 5 at 12:41
Gorjan Radevski
705
i think this can be solved if you just use df["columNameToReplace"] = df["value"] and then delete the value from the dataframe ?
– Vaibhav gusain
Dec 5 at 12:47
add a comment |
up vote
10
down vote
favorite
1
up vote
10
down vote
favorite
1
1
I have a data frame that looks like this:
df = pd.DataFrame({"value": [4, 5, 3], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]})
df
value item1 item2 item3
0 4 0 1 0
1 5 1 0 0
2 3 0 0 1
Basically what I want to do is replace the value of the one hot encoded elements with the value from the "value" column and then delete the "value" column. The resulting data frame should be like this:
df_out = pd.DataFrame({"item1": [0, 5, 0], "item2": [4, 0, 0], "item3": [0, 0, 3]})
item1 item2 item3
0 0 4 0
1 5 0 0
2 0 0 3
python pandas dataframe
edited Dec 5 at 12:55
coldspeed
115k18105182
asked Dec 5 at 12:41
Gorjan Radevski
705
I have a data frame that looks like this:
df = pd.DataFrame({"value": [4, 5, 3], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]})
df
value item1 item2 item3
0 4 0 1 0
1 5 1 0 0
2 3 0 0 1
Basically what I want to do is replace the value of the one hot encoded elements with the value from the "value" column and then delete the "value" column. The resulting data frame should be like this:
df_out = pd.DataFrame({"item1": [0, 5, 0], "item2": [4, 0, 0], "item3": [0, 0, 3]})
item1 item2 item3
0 0 4 0
1 5 0 0
2 0 0 3
python pandas dataframe
python pandas dataframe
edited Dec 5 at 12:55
coldspeed
115k18105182
asked Dec 5 at 12:41
Gorjan Radevski
705
edited Dec 5 at 12:55
coldspeed
115k18105182
asked Dec 5 at 12:41
Gorjan Radevski
705
edited Dec 5 at 12:55
coldspeed
115k18105182
edited Dec 5 at 12:55
coldspeed
115k18105182
edited Dec 5 at 12:55
coldspeed
115k18105182
115k18105182
asked Dec 5 at 12:41
Gorjan Radevski
705
asked Dec 5 at 12:41
Gorjan Radevski
705
asked Dec 5 at 12:41
Gorjan Radevski
705
705
i think this can be solved if you just use df["columNameToReplace"] = df["value"] and then delete the value from the dataframe ?
– Vaibhav gusain
Dec 5 at 12:47
add a comment |
i think this can be solved if you just use df["columNameToReplace"] = df["value"] and then delete the value from the dataframe ?
– Vaibhav gusain
Dec 5 at 12:47
i think this can be solved if you just use df["columNameToReplace"] = df["value"] and then delete the value from the dataframe ?
– Vaibhav gusain
Dec 5 at 12:47
i think this can be solved if you just use df["columNameToReplace"] = df["value"] and then delete the value from the dataframe ?
– Vaibhav gusain
Dec 5 at 12:47
add a comment |
4 Answers
4
active
oldest
votes
up vote
14
down vote
accepted
Why not just multiply?
df.pop('value').values * df
item1 item2 item3
0 0 5 0
1 4 0 0
2 0 0 3
DataFrame.pop has the nice effect of in-place removing and returning a column, so you can do this in a single step.
if the "item_*" columns have anything besides 1 in them, then you can multiply with bools:
df.pop('value').values * df.astype(bool)
item1 item2 item3
0 0 5 0
1 4 0 0
2 0 0 3
If your DataFrame has other columns, then do this:
df
value name item1 item2 item3
0 4 John 0 1 0
1 5 Mike 1 0 0
2 3 Stan 0 0 1
# cols = df.columns[df.columns.str.startswith('item')]
cols = df.filter(like='item').columns
df[cols] = df.pop('value').values * df[cols]
df
name item1 item2 item3
0 John 0 5 0
1 Mike 4 0 0
2 Stan 0 0 3
answered Dec 5 at 12:51
coldspeed
115k18105182
4
Most elegant answer so far
– horro
Dec 5 at 12:53
I like it but I should have been more specific with my question. Here is how my data frame actually looks like:df_in = pd.DataFrame({"value": [4, 5, 3], "name": ["John", "Mike", "Stan"], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]})And the output df should be:df_out = pd.DataFrame({"name": ["John", "Mike", "Stan"], "item1": [0, 4, 0], "item2": [5, 0, 0], "item3": [0, 0, 3]})
– Gorjan Radevski
Dec 5 at 13:08
@GorjanRadevski Let me know if the edit does it for you.
– coldspeed
Dec 5 at 13:11
1
I feel so stupid after watching this answer :}
– Mohit Motwani
Dec 5 at 13:28
1
That works! I have no idea why on the sample data frame it worked without the addition. Thank you!
– Gorjan Radevski
Dec 5 at 13:46
|
show 4 more comments
up vote
1
down vote
You could do something like:
df = pd.DataFrame([df['value']*df['item1'],df['value']*df['item2'],df['value']*df['item3']])
df.columns = ['item1','item2','item3']
EDIT:
As this answer will not scale well to many columns as @coldspeed comments, it should be done iterating a loop:
cols = ['item1','item2','item3']
for c in cols:
df[c] *= df['value']
df.drop('value',axis=1,inplace=True)
answered Dec 5 at 12:50
horro
4681727
1
This won't scale well to many columns.
– coldspeed
Dec 5 at 12:52
1
Fair point, it should be done iterating a loop
– horro
Dec 5 at 12:53
add a comment |
up vote
0
down vote
You need:
col = ['item1','item2','item3']
for c in col:
df[c] = df[c] * df['value']
df.drop(['value'],1,inplace=True)
answered Dec 5 at 12:51
Sociopath
3,36471635
1
Surely you can think of something better than iteration...
– coldspeed
Dec 5 at 12:52
add a comment |
up vote
0
down vote
You can use np.where:
items = ['item1', 'item2', 'item3']
for item in items:
df[item] = np.where(df[item]==1, df['value'], 0)
df.drop(columns = ['value'], inplace =True)
df
item1 item2 item3
0 0 4 0
1 5 0 0
2 0 0 3
answered Dec 5 at 12:49
Mohit Motwani
8731320
1
Again, can you think of something better than iteration? Don't forget to drop the "value" column once done.
– coldspeed
Dec 5 at 12:53
@coldspeed You're right
– Mohit Motwani
Dec 5 at 12:56
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
accepted
Why not just multiply?
df.pop('value').values * df
item1 item2 item3
0 0 5 0
1 4 0 0
2 0 0 3
DataFrame.pop has the nice effect of in-place removing and returning a column, so you can do this in a single step.
if the "item_*" columns have anything besides 1 in them, then you can multiply with bools:
df.pop('value').values * df.astype(bool)
item1 item2 item3
0 0 5 0
1 4 0 0
2 0 0 3
If your DataFrame has other columns, then do this:
df
value name item1 item2 item3
0 4 John 0 1 0
1 5 Mike 1 0 0
2 3 Stan 0 0 1
# cols = df.columns[df.columns.str.startswith('item')]
cols = df.filter(like='item').columns
df[cols] = df.pop('value').values * df[cols]
df
name item1 item2 item3
0 John 0 5 0
1 Mike 4 0 0
2 Stan 0 0 3
answered Dec 5 at 12:51
coldspeed
115k18105182
4
Most elegant answer so far
– horro
Dec 5 at 12:53
I like it but I should have been more specific with my question. Here is how my data frame actually looks like:df_in = pd.DataFrame({"value": [4, 5, 3], "name": ["John", "Mike", "Stan"], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]})And the output df should be:df_out = pd.DataFrame({"name": ["John", "Mike", "Stan"], "item1": [0, 4, 0], "item2": [5, 0, 0], "item3": [0, 0, 3]})
– Gorjan Radevski
Dec 5 at 13:08
@GorjanRadevski Let me know if the edit does it for you.
– coldspeed
Dec 5 at 13:11
1
I feel so stupid after watching this answer :}
– Mohit Motwani
Dec 5 at 13:28
1
That works! I have no idea why on the sample data frame it worked without the addition. Thank you!
– Gorjan Radevski
Dec 5 at 13:46
|
show 4 more comments
up vote
14
down vote
accepted
Why not just multiply?
df.pop('value').values * df
item1 item2 item3
0 0 5 0
1 4 0 0
2 0 0 3
DataFrame.pop has the nice effect of in-place removing and returning a column, so you can do this in a single step.
if the "item_*" columns have anything besides 1 in them, then you can multiply with bools:
df.pop('value').values * df.astype(bool)
item1 item2 item3
0 0 5 0
1 4 0 0
2 0 0 3
If your DataFrame has other columns, then do this:
df
value name item1 item2 item3
0 4 John 0 1 0
1 5 Mike 1 0 0
2 3 Stan 0 0 1
# cols = df.columns[df.columns.str.startswith('item')]
cols = df.filter(like='item').columns
df[cols] = df.pop('value').values * df[cols]
df
name item1 item2 item3
0 John 0 5 0
1 Mike 4 0 0
2 Stan 0 0 3
answered Dec 5 at 12:51
coldspeed
115k18105182
4
Most elegant answer so far
– horro
Dec 5 at 12:53
I like it but I should have been more specific with my question. Here is how my data frame actually looks like:df_in = pd.DataFrame({"value": [4, 5, 3], "name": ["John", "Mike", "Stan"], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]})And the output df should be:df_out = pd.DataFrame({"name": ["John", "Mike", "Stan"], "item1": [0, 4, 0], "item2": [5, 0, 0], "item3": [0, 0, 3]})
– Gorjan Radevski
Dec 5 at 13:08
@GorjanRadevski Let me know if the edit does it for you.
– coldspeed
Dec 5 at 13:11
1
I feel so stupid after watching this answer :}
– Mohit Motwani
Dec 5 at 13:28
1
That works! I have no idea why on the sample data frame it worked without the addition. Thank you!
– Gorjan Radevski
Dec 5 at 13:46
|
show 4 more comments
up vote
14
down vote
accepted
up vote
14
down vote
accepted
Why not just multiply?
df.pop('value').values * df
item1 item2 item3
0 0 5 0
1 4 0 0
2 0 0 3
DataFrame.pop has the nice effect of in-place removing and returning a column, so you can do this in a single step.
if the "item_*" columns have anything besides 1 in them, then you can multiply with bools:
df.pop('value').values * df.astype(bool)
item1 item2 item3
0 0 5 0
1 4 0 0
2 0 0 3
If your DataFrame has other columns, then do this:
df
value name item1 item2 item3
0 4 John 0 1 0
1 5 Mike 1 0 0
2 3 Stan 0 0 1
# cols = df.columns[df.columns.str.startswith('item')]
cols = df.filter(like='item').columns
df[cols] = df.pop('value').values * df[cols]
df
name item1 item2 item3
0 John 0 5 0
1 Mike 4 0 0
2 Stan 0 0 3
answered Dec 5 at 12:51
coldspeed
115k18105182
Why not just multiply?
df.pop('value').values * df
item1 item2 item3
0 0 5 0
1 4 0 0
2 0 0 3
DataFrame.pop has the nice effect of in-place removing and returning a column, so you can do this in a single step.
if the "item_*" columns have anything besides 1 in them, then you can multiply with bools:
df.pop('value').values * df.astype(bool)
item1 item2 item3
0 0 5 0
1 4 0 0
2 0 0 3
If your DataFrame has other columns, then do this:
df
value name item1 item2 item3
0 4 John 0 1 0
1 5 Mike 1 0 0
2 3 Stan 0 0 1
# cols = df.columns[df.columns.str.startswith('item')]
cols = df.filter(like='item').columns
df[cols] = df.pop('value').values * df[cols]
df
name item1 item2 item3
0 John 0 5 0
1 Mike 4 0 0
2 Stan 0 0 3
answered Dec 5 at 12:51
coldspeed
115k18105182
edited Dec 5 at 13:09
answered Dec 5 at 12:51
coldspeed
115k18105182
answered Dec 5 at 12:51
coldspeed
115k18105182
answered Dec 5 at 12:51
coldspeed
115k18105182
115k18105182
4
Most elegant answer so far
– horro
Dec 5 at 12:53
I like it but I should have been more specific with my question. Here is how my data frame actually looks like:df_in = pd.DataFrame({"value": [4, 5, 3], "name": ["John", "Mike", "Stan"], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]})And the output df should be:df_out = pd.DataFrame({"name": ["John", "Mike", "Stan"], "item1": [0, 4, 0], "item2": [5, 0, 0], "item3": [0, 0, 3]})
– Gorjan Radevski
Dec 5 at 13:08
@GorjanRadevski Let me know if the edit does it for you.
– coldspeed
Dec 5 at 13:11
1
I feel so stupid after watching this answer :}
– Mohit Motwani
Dec 5 at 13:28
1
That works! I have no idea why on the sample data frame it worked without the addition. Thank you!
– Gorjan Radevski
Dec 5 at 13:46
|
show 4 more comments
4
Most elegant answer so far
– horro
Dec 5 at 12:53
I like it but I should have been more specific with my question. Here is how my data frame actually looks like:df_in = pd.DataFrame({"value": [4, 5, 3], "name": ["John", "Mike", "Stan"], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]})And the output df should be:df_out = pd.DataFrame({"name": ["John", "Mike", "Stan"], "item1": [0, 4, 0], "item2": [5, 0, 0], "item3": [0, 0, 3]})
– Gorjan Radevski
Dec 5 at 13:08
@GorjanRadevski Let me know if the edit does it for you.
– coldspeed
Dec 5 at 13:11
1
I feel so stupid after watching this answer :}
– Mohit Motwani
Dec 5 at 13:28
1
That works! I have no idea why on the sample data frame it worked without the addition. Thank you!
– Gorjan Radevski
Dec 5 at 13:46
4
4
Most elegant answer so far
– horro
Dec 5 at 12:53
Most elegant answer so far
– horro
Dec 5 at 12:53
I like it but I should have been more specific with my question. Here is how my data frame actually looks like:
df_in = pd.DataFrame({"value": [4, 5, 3], "name": ["John", "Mike", "Stan"], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]}) And the output df should be: df_out = pd.DataFrame({"name": ["John", "Mike", "Stan"], "item1": [0, 4, 0], "item2": [5, 0, 0], "item3": [0, 0, 3]})– Gorjan Radevski
Dec 5 at 13:08
I like it but I should have been more specific with my question. Here is how my data frame actually looks like:
df_in = pd.DataFrame({"value": [4, 5, 3], "name": ["John", "Mike", "Stan"], "item1": [0, 1, 0], "item2": [1, 0, 0], "item3": [0, 0, 1]}) And the output df should be: df_out = pd.DataFrame({"name": ["John", "Mike", "Stan"], "item1": [0, 4, 0], "item2": [5, 0, 0], "item3": [0, 0, 3]})– Gorjan Radevski
Dec 5 at 13:08
@GorjanRadevski Let me know if the edit does it for you.
– coldspeed
Dec 5 at 13:11
@GorjanRadevski Let me know if the edit does it for you.
– coldspeed
Dec 5 at 13:11
1
1
I feel so stupid after watching this answer :}
– Mohit Motwani
Dec 5 at 13:28
I feel so stupid after watching this answer :}
– Mohit Motwani
Dec 5 at 13:28
1
1
That works! I have no idea why on the sample data frame it worked without the addition. Thank you!
– Gorjan Radevski
Dec 5 at 13:46
That works! I have no idea why on the sample data frame it worked without the addition. Thank you!
– Gorjan Radevski
Dec 5 at 13:46
|
show 4 more comments
up vote
1
down vote
You could do something like:
df = pd.DataFrame([df['value']*df['item1'],df['value']*df['item2'],df['value']*df['item3']])
df.columns = ['item1','item2','item3']
EDIT:
As this answer will not scale well to many columns as @coldspeed comments, it should be done iterating a loop:
cols = ['item1','item2','item3']
for c in cols:
df[c] *= df['value']
df.drop('value',axis=1,inplace=True)
answered Dec 5 at 12:50
horro
4681727
1
This won't scale well to many columns.
– coldspeed
Dec 5 at 12:52
1
Fair point, it should be done iterating a loop
– horro
Dec 5 at 12:53
add a comment |
up vote
1
down vote
You could do something like:
df = pd.DataFrame([df['value']*df['item1'],df['value']*df['item2'],df['value']*df['item3']])
df.columns = ['item1','item2','item3']
EDIT:
As this answer will not scale well to many columns as @coldspeed comments, it should be done iterating a loop:
cols = ['item1','item2','item3']
for c in cols:
df[c] *= df['value']
df.drop('value',axis=1,inplace=True)
answered Dec 5 at 12:50
horro
4681727
1
This won't scale well to many columns.
– coldspeed
Dec 5 at 12:52
1
Fair point, it should be done iterating a loop
– horro
Dec 5 at 12:53
add a comment |
up vote
1
down vote
up vote
1
down vote
You could do something like:
df = pd.DataFrame([df['value']*df['item1'],df['value']*df['item2'],df['value']*df['item3']])
df.columns = ['item1','item2','item3']
EDIT:
As this answer will not scale well to many columns as @coldspeed comments, it should be done iterating a loop:
cols = ['item1','item2','item3']
for c in cols:
df[c] *= df['value']
df.drop('value',axis=1,inplace=True)
answered Dec 5 at 12:50
horro
4681727
You could do something like:
df = pd.DataFrame([df['value']*df['item1'],df['value']*df['item2'],df['value']*df['item3']])
df.columns = ['item1','item2','item3']
EDIT:
As this answer will not scale well to many columns as @coldspeed comments, it should be done iterating a loop:
cols = ['item1','item2','item3']
for c in cols:
df[c] *= df['value']
df.drop('value',axis=1,inplace=True)
answered Dec 5 at 12:50
horro
4681727
edited Dec 5 at 12:57
answered Dec 5 at 12:50
horro
4681727
answered Dec 5 at 12:50
horro
4681727
answered Dec 5 at 12:50
horro
4681727
4681727
1
This won't scale well to many columns.
– coldspeed
Dec 5 at 12:52
1
Fair point, it should be done iterating a loop
– horro
Dec 5 at 12:53
add a comment |
1
This won't scale well to many columns.
– coldspeed
Dec 5 at 12:52
1
Fair point, it should be done iterating a loop
– horro
Dec 5 at 12:53
1
1
This won't scale well to many columns.
– coldspeed
Dec 5 at 12:52
This won't scale well to many columns.
– coldspeed
Dec 5 at 12:52
1
1
Fair point, it should be done iterating a loop
– horro
Dec 5 at 12:53
Fair point, it should be done iterating a loop
– horro
Dec 5 at 12:53
add a comment |
up vote
0
down vote
You need:
col = ['item1','item2','item3']
for c in col:
df[c] = df[c] * df['value']
df.drop(['value'],1,inplace=True)
answered Dec 5 at 12:51
Sociopath
3,36471635
1
Surely you can think of something better than iteration...
– coldspeed
Dec 5 at 12:52
add a comment |
up vote
0
down vote
You need:
col = ['item1','item2','item3']
for c in col:
df[c] = df[c] * df['value']
df.drop(['value'],1,inplace=True)
answered Dec 5 at 12:51
Sociopath
3,36471635
1
Surely you can think of something better than iteration...
– coldspeed
Dec 5 at 12:52
add a comment |
up vote
0
down vote
up vote
0
down vote
You need:
col = ['item1','item2','item3']
for c in col:
df[c] = df[c] * df['value']
df.drop(['value'],1,inplace=True)
answered Dec 5 at 12:51
Sociopath
3,36471635
You need:
col = ['item1','item2','item3']
for c in col:
df[c] = df[c] * df['value']
df.drop(['value'],1,inplace=True)
answered Dec 5 at 12:51
Sociopath
3,36471635
edited Dec 5 at 12:52
answered Dec 5 at 12:51
Sociopath
3,36471635
answered Dec 5 at 12:51
Sociopath
3,36471635
answered Dec 5 at 12:51
Sociopath
3,36471635
3,36471635
1
Surely you can think of something better than iteration...
– coldspeed
Dec 5 at 12:52
add a comment |
1
Surely you can think of something better than iteration...
– coldspeed
Dec 5 at 12:52
1
1
Surely you can think of something better than iteration...
– coldspeed
Dec 5 at 12:52
Surely you can think of something better than iteration...
– coldspeed
Dec 5 at 12:52
add a comment |
up vote
0
down vote
You can use np.where:
items = ['item1', 'item2', 'item3']
for item in items:
df[item] = np.where(df[item]==1, df['value'], 0)
df.drop(columns = ['value'], inplace =True)
df
item1 item2 item3
0 0 4 0
1 5 0 0
2 0 0 3
answered Dec 5 at 12:49
Mohit Motwani
8731320
1
Again, can you think of something better than iteration? Don't forget to drop the "value" column once done.
– coldspeed
Dec 5 at 12:53
@coldspeed You're right
– Mohit Motwani
Dec 5 at 12:56
add a comment |
up vote
0
down vote
You can use np.where:
items = ['item1', 'item2', 'item3']
for item in items:
df[item] = np.where(df[item]==1, df['value'], 0)
df.drop(columns = ['value'], inplace =True)
df
item1 item2 item3
0 0 4 0
1 5 0 0
2 0 0 3
answered Dec 5 at 12:49
Mohit Motwani
8731320
1
Again, can you think of something better than iteration? Don't forget to drop the "value" column once done.
– coldspeed
Dec 5 at 12:53
@coldspeed You're right
– Mohit Motwani
Dec 5 at 12:56
add a comment |
up vote
0
down vote
up vote
0
down vote
You can use np.where:
items = ['item1', 'item2', 'item3']
for item in items:
df[item] = np.where(df[item]==1, df['value'], 0)
df.drop(columns = ['value'], inplace =True)
df
item1 item2 item3
0 0 4 0
1 5 0 0
2 0 0 3
answered Dec 5 at 12:49
Mohit Motwani
8731320
You can use np.where:
items = ['item1', 'item2', 'item3']
for item in items:
df[item] = np.where(df[item]==1, df['value'], 0)
df.drop(columns = ['value'], inplace =True)
df
item1 item2 item3
0 0 4 0
1 5 0 0
2 0 0 3
answered Dec 5 at 12:49
Mohit Motwani
8731320
edited Dec 5 at 12:54
answered Dec 5 at 12:49
Mohit Motwani
8731320
answered Dec 5 at 12:49
Mohit Motwani
8731320
answered Dec 5 at 12:49
Mohit Motwani
8731320
8731320
1
Again, can you think of something better than iteration? Don't forget to drop the "value" column once done.
– coldspeed
Dec 5 at 12:53
@coldspeed You're right
– Mohit Motwani
Dec 5 at 12:56
add a comment |
1
Again, can you think of something better than iteration? Don't forget to drop the "value" column once done.
– coldspeed
Dec 5 at 12:53
@coldspeed You're right
– Mohit Motwani
Dec 5 at 12:56
1
1
Again, can you think of something better than iteration? Don't forget to drop the "value" column once done.
– coldspeed
Dec 5 at 12:53
Again, can you think of something better than iteration? Don't forget to drop the "value" column once done.
– coldspeed
Dec 5 at 12:53
@coldspeed You're right
– Mohit Motwani
Dec 5 at 12:56
@coldspeed You're right
– Mohit Motwani
Dec 5 at 12:56
add a comment |
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i think this can be solved if you just use df["columNameToReplace"] = df["value"] and then delete the value from the dataframe ?
– Vaibhav gusain
Dec 5 at 12:47