is python possible to create a new data frame from the existing data frame?
0
i would like to ask about this because i have created a code like this but it returned only the header.
Here is the sample of my df name df_all
Doc name count year
[A1,A2] John 1 2018
[A1,A3] Mark 0 2018
[A2,A4] John 3 2018
here is the code that i have tried
n_wsp_71 = [i for i in df_all if i.count != 0]
n_wsp_71
and here is my results
['Doc', 'name', 'count', 'year']
but i expect to see this results
newdf
Doc name count year
[A1,A2] John 1 2018
[A2,A4] John 3 2018
python pandas
asked Nov 23 '18 at 5:12
Hook Im
1248
add a comment |
0
i would like to ask about this because i have created a code like this but it returned only the header.
Here is the sample of my df name df_all
Doc name count year
[A1,A2] John 1 2018
[A1,A3] Mark 0 2018
[A2,A4] John 3 2018
here is the code that i have tried
n_wsp_71 = [i for i in df_all if i.count != 0]
n_wsp_71
and here is my results
['Doc', 'name', 'count', 'year']
but i expect to see this results
newdf
Doc name count year
[A1,A2] John 1 2018
[A2,A4] John 3 2018
python pandas
asked Nov 23 '18 at 5:12
Hook Im
1248
2
That's a dataframe not a list...
– U9-Forward
Nov 23 '18 at 5:14
Trylist_all = list_all[list_all['count'] != 0]
– Sociopath
Nov 23 '18 at 5:16
im blur, sorry. i have updated the post
– Hook Im
Nov 23 '18 at 5:17
add a comment |
0
0
0
i would like to ask about this because i have created a code like this but it returned only the header.
Here is the sample of my df name df_all
Doc name count year
[A1,A2] John 1 2018
[A1,A3] Mark 0 2018
[A2,A4] John 3 2018
here is the code that i have tried
n_wsp_71 = [i for i in df_all if i.count != 0]
n_wsp_71
and here is my results
['Doc', 'name', 'count', 'year']
but i expect to see this results
newdf
Doc name count year
[A1,A2] John 1 2018
[A2,A4] John 3 2018
python pandas
asked Nov 23 '18 at 5:12
Hook Im
1248
i would like to ask about this because i have created a code like this but it returned only the header.
Here is the sample of my df name df_all
Doc name count year
[A1,A2] John 1 2018
[A1,A3] Mark 0 2018
[A2,A4] John 3 2018
here is the code that i have tried
n_wsp_71 = [i for i in df_all if i.count != 0]
n_wsp_71
and here is my results
['Doc', 'name', 'count', 'year']
but i expect to see this results
newdf
Doc name count year
[A1,A2] John 1 2018
[A2,A4] John 3 2018
python pandas
python pandas
asked Nov 23 '18 at 5:12
Hook Im
1248
asked Nov 23 '18 at 5:12
Hook Im
1248
edited Nov 23 '18 at 5:15
asked Nov 23 '18 at 5:12
Hook Im
1248
asked Nov 23 '18 at 5:12
Hook Im
1248
asked Nov 23 '18 at 5:12
Hook Im
1248
1248
2
That's a dataframe not a list...
– U9-Forward
Nov 23 '18 at 5:14
Trylist_all = list_all[list_all['count'] != 0]
– Sociopath
Nov 23 '18 at 5:16
im blur, sorry. i have updated the post
– Hook Im
Nov 23 '18 at 5:17
add a comment |
2
That's a dataframe not a list...
– U9-Forward
Nov 23 '18 at 5:14
Trylist_all = list_all[list_all['count'] != 0]
– Sociopath
Nov 23 '18 at 5:16
im blur, sorry. i have updated the post
– Hook Im
Nov 23 '18 at 5:17
2
2
That's a dataframe not a list...
– U9-Forward
Nov 23 '18 at 5:14
That's a dataframe not a list...
– U9-Forward
Nov 23 '18 at 5:14
Try
list_all = list_all[list_all['count'] != 0]– Sociopath
Nov 23 '18 at 5:16
Try
list_all = list_all[list_all['count'] != 0]– Sociopath
Nov 23 '18 at 5:16
im blur, sorry. i have updated the post
– Hook Im
Nov 23 '18 at 5:17
im blur, sorry. i have updated the post
– Hook Im
Nov 23 '18 at 5:17
add a comment |
2 Answers
2
active
oldest
votes
1
You're doing a list comprehension to pandas data-frames, that will never work
So you have to do:
list_all=list_all[list_all['count']!=0]
And now:
print(list_all)
Is:
Doc name count year
[A1,A2] John 1 2018
[A2,A4] John 3 2018
answered Nov 23 '18 at 5:17
U9-Forward
13k21137
1
thank you for your understanding and comment!
– Hook Im
Nov 23 '18 at 5:28
@HookIm Happy to help, :-), 😃😃😃, Note: mine is faster based on the timingsChirag's: 0.04203809888741396 Mine: 0.010112490249476552
– U9-Forward
Nov 23 '18 at 5:34
1
ah.. i see! thank you for reminding me to consider the runtime too. by the way, i i would like to return count == 0. when i tried, it has nothing come out. why it became like that?
– Hook Im
Nov 23 '18 at 5:41
@HookIm YW, happy to help again, but why it did not work forcount==0?, did you try:df[df['count']==0]
– U9-Forward
Nov 23 '18 at 5:43
1
yes, already did that. but, i will fix this by myself :) thank you so much!
– Hook Im
Nov 23 '18 at 5:50
|
show 1 more comment
2
df = pd.DataFrame({"Doc": [["A1","A2"], ["A1","A3"], ["A2","A4"]], "name": ["John", "Mark", "John"], "count": [1,0,3], "year": [2018, 2018, 2018]})
df2 = df.query("count!=0").reset_index(drop=True)
# for count = 0
df2 = df.query("count==0").reset_index(drop=True)
#method 2
df2 = df[~(df["count"].isin(['0']))].reset_index(drop=True)
# for count = 0
df2 = df[(df["count"].isin(['0']))].reset_index(drop=True)
print(df2)
Output:
Doc name count year
0 [A1, A2] John 1 2018
1 [A2, A4] John 3 2018
Doc name count year
0 [A1, A3] Mark 0 2018
answered Nov 23 '18 at 5:26
Srce Cde
1,136511
i have not considered using the query before, new thing to learn! thank you. by the way, when i try to show the count that equal to 0. why it shows nothing. is it a sensitive case?
– Hook Im
Nov 23 '18 at 5:30
@HookIm Updated the alternative method. Also forcount=0
– Srce Cde
Nov 23 '18 at 5:56
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
You're doing a list comprehension to pandas data-frames, that will never work
So you have to do:
list_all=list_all[list_all['count']!=0]
And now:
print(list_all)
Is:
Doc name count year
[A1,A2] John 1 2018
[A2,A4] John 3 2018
answered Nov 23 '18 at 5:17
U9-Forward
13k21137
1
thank you for your understanding and comment!
– Hook Im
Nov 23 '18 at 5:28
@HookIm Happy to help, :-), 😃😃😃, Note: mine is faster based on the timingsChirag's: 0.04203809888741396 Mine: 0.010112490249476552
– U9-Forward
Nov 23 '18 at 5:34
1
ah.. i see! thank you for reminding me to consider the runtime too. by the way, i i would like to return count == 0. when i tried, it has nothing come out. why it became like that?
– Hook Im
Nov 23 '18 at 5:41
@HookIm YW, happy to help again, but why it did not work forcount==0?, did you try:df[df['count']==0]
– U9-Forward
Nov 23 '18 at 5:43
1
yes, already did that. but, i will fix this by myself :) thank you so much!
– Hook Im
Nov 23 '18 at 5:50
|
show 1 more comment
1
You're doing a list comprehension to pandas data-frames, that will never work
So you have to do:
list_all=list_all[list_all['count']!=0]
And now:
print(list_all)
Is:
Doc name count year
[A1,A2] John 1 2018
[A2,A4] John 3 2018
answered Nov 23 '18 at 5:17
U9-Forward
13k21137
1
thank you for your understanding and comment!
– Hook Im
Nov 23 '18 at 5:28
@HookIm Happy to help, :-), 😃😃😃, Note: mine is faster based on the timingsChirag's: 0.04203809888741396 Mine: 0.010112490249476552
– U9-Forward
Nov 23 '18 at 5:34
1
ah.. i see! thank you for reminding me to consider the runtime too. by the way, i i would like to return count == 0. when i tried, it has nothing come out. why it became like that?
– Hook Im
Nov 23 '18 at 5:41
@HookIm YW, happy to help again, but why it did not work forcount==0?, did you try:df[df['count']==0]
– U9-Forward
Nov 23 '18 at 5:43
1
yes, already did that. but, i will fix this by myself :) thank you so much!
– Hook Im
Nov 23 '18 at 5:50
|
show 1 more comment
1
1
1
You're doing a list comprehension to pandas data-frames, that will never work
So you have to do:
list_all=list_all[list_all['count']!=0]
And now:
print(list_all)
Is:
Doc name count year
[A1,A2] John 1 2018
[A2,A4] John 3 2018
answered Nov 23 '18 at 5:17
U9-Forward
13k21137
You're doing a list comprehension to pandas data-frames, that will never work
So you have to do:
list_all=list_all[list_all['count']!=0]
And now:
print(list_all)
Is:
Doc name count year
[A1,A2] John 1 2018
[A2,A4] John 3 2018
answered Nov 23 '18 at 5:17
U9-Forward
13k21137
answered Nov 23 '18 at 5:17
U9-Forward
13k21137
answered Nov 23 '18 at 5:17
U9-Forward
13k21137
answered Nov 23 '18 at 5:17
U9-Forward
13k21137
13k21137
1
thank you for your understanding and comment!
– Hook Im
Nov 23 '18 at 5:28
@HookIm Happy to help, :-), 😃😃😃, Note: mine is faster based on the timingsChirag's: 0.04203809888741396 Mine: 0.010112490249476552
– U9-Forward
Nov 23 '18 at 5:34
1
ah.. i see! thank you for reminding me to consider the runtime too. by the way, i i would like to return count == 0. when i tried, it has nothing come out. why it became like that?
– Hook Im
Nov 23 '18 at 5:41
@HookIm YW, happy to help again, but why it did not work forcount==0?, did you try:df[df['count']==0]
– U9-Forward
Nov 23 '18 at 5:43
1
yes, already did that. but, i will fix this by myself :) thank you so much!
– Hook Im
Nov 23 '18 at 5:50
|
show 1 more comment
1
thank you for your understanding and comment!
– Hook Im
Nov 23 '18 at 5:28
@HookIm Happy to help, :-), 😃😃😃, Note: mine is faster based on the timingsChirag's: 0.04203809888741396 Mine: 0.010112490249476552
– U9-Forward
Nov 23 '18 at 5:34
1
ah.. i see! thank you for reminding me to consider the runtime too. by the way, i i would like to return count == 0. when i tried, it has nothing come out. why it became like that?
– Hook Im
Nov 23 '18 at 5:41
@HookIm YW, happy to help again, but why it did not work forcount==0?, did you try:df[df['count']==0]
– U9-Forward
Nov 23 '18 at 5:43
1
yes, already did that. but, i will fix this by myself :) thank you so much!
– Hook Im
Nov 23 '18 at 5:50
1
1
thank you for your understanding and comment!
– Hook Im
Nov 23 '18 at 5:28
thank you for your understanding and comment!
– Hook Im
Nov 23 '18 at 5:28
@HookIm Happy to help, :-), 😃😃😃, Note: mine is faster based on the timings
Chirag's: 0.04203809888741396 Mine: 0.010112490249476552– U9-Forward
Nov 23 '18 at 5:34
@HookIm Happy to help, :-), 😃😃😃, Note: mine is faster based on the timings
Chirag's: 0.04203809888741396 Mine: 0.010112490249476552– U9-Forward
Nov 23 '18 at 5:34
1
1
ah.. i see! thank you for reminding me to consider the runtime too. by the way, i i would like to return count == 0. when i tried, it has nothing come out. why it became like that?
– Hook Im
Nov 23 '18 at 5:41
ah.. i see! thank you for reminding me to consider the runtime too. by the way, i i would like to return count == 0. when i tried, it has nothing come out. why it became like that?
– Hook Im
Nov 23 '18 at 5:41
@HookIm YW, happy to help again, but why it did not work for
count==0?, did you try: df[df['count']==0]– U9-Forward
Nov 23 '18 at 5:43
@HookIm YW, happy to help again, but why it did not work for
count==0?, did you try: df[df['count']==0]– U9-Forward
Nov 23 '18 at 5:43
1
1
yes, already did that. but, i will fix this by myself :) thank you so much!
– Hook Im
Nov 23 '18 at 5:50
yes, already did that. but, i will fix this by myself :) thank you so much!
– Hook Im
Nov 23 '18 at 5:50
|
show 1 more comment
2
df = pd.DataFrame({"Doc": [["A1","A2"], ["A1","A3"], ["A2","A4"]], "name": ["John", "Mark", "John"], "count": [1,0,3], "year": [2018, 2018, 2018]})
df2 = df.query("count!=0").reset_index(drop=True)
# for count = 0
df2 = df.query("count==0").reset_index(drop=True)
#method 2
df2 = df[~(df["count"].isin(['0']))].reset_index(drop=True)
# for count = 0
df2 = df[(df["count"].isin(['0']))].reset_index(drop=True)
print(df2)
Output:
Doc name count year
0 [A1, A2] John 1 2018
1 [A2, A4] John 3 2018
Doc name count year
0 [A1, A3] Mark 0 2018
answered Nov 23 '18 at 5:26
Srce Cde
1,136511
i have not considered using the query before, new thing to learn! thank you. by the way, when i try to show the count that equal to 0. why it shows nothing. is it a sensitive case?
– Hook Im
Nov 23 '18 at 5:30
@HookIm Updated the alternative method. Also forcount=0
– Srce Cde
Nov 23 '18 at 5:56
add a comment |
2
df = pd.DataFrame({"Doc": [["A1","A2"], ["A1","A3"], ["A2","A4"]], "name": ["John", "Mark", "John"], "count": [1,0,3], "year": [2018, 2018, 2018]})
df2 = df.query("count!=0").reset_index(drop=True)
# for count = 0
df2 = df.query("count==0").reset_index(drop=True)
#method 2
df2 = df[~(df["count"].isin(['0']))].reset_index(drop=True)
# for count = 0
df2 = df[(df["count"].isin(['0']))].reset_index(drop=True)
print(df2)
Output:
Doc name count year
0 [A1, A2] John 1 2018
1 [A2, A4] John 3 2018
Doc name count year
0 [A1, A3] Mark 0 2018
answered Nov 23 '18 at 5:26
Srce Cde
1,136511
i have not considered using the query before, new thing to learn! thank you. by the way, when i try to show the count that equal to 0. why it shows nothing. is it a sensitive case?
– Hook Im
Nov 23 '18 at 5:30
@HookIm Updated the alternative method. Also forcount=0
– Srce Cde
Nov 23 '18 at 5:56
add a comment |
2
2
2
df = pd.DataFrame({"Doc": [["A1","A2"], ["A1","A3"], ["A2","A4"]], "name": ["John", "Mark", "John"], "count": [1,0,3], "year": [2018, 2018, 2018]})
df2 = df.query("count!=0").reset_index(drop=True)
# for count = 0
df2 = df.query("count==0").reset_index(drop=True)
#method 2
df2 = df[~(df["count"].isin(['0']))].reset_index(drop=True)
# for count = 0
df2 = df[(df["count"].isin(['0']))].reset_index(drop=True)
print(df2)
Output:
Doc name count year
0 [A1, A2] John 1 2018
1 [A2, A4] John 3 2018
Doc name count year
0 [A1, A3] Mark 0 2018
answered Nov 23 '18 at 5:26
Srce Cde
1,136511
df = pd.DataFrame({"Doc": [["A1","A2"], ["A1","A3"], ["A2","A4"]], "name": ["John", "Mark", "John"], "count": [1,0,3], "year": [2018, 2018, 2018]})
df2 = df.query("count!=0").reset_index(drop=True)
# for count = 0
df2 = df.query("count==0").reset_index(drop=True)
#method 2
df2 = df[~(df["count"].isin(['0']))].reset_index(drop=True)
# for count = 0
df2 = df[(df["count"].isin(['0']))].reset_index(drop=True)
print(df2)
Output:
Doc name count year
0 [A1, A2] John 1 2018
1 [A2, A4] John 3 2018
Doc name count year
0 [A1, A3] Mark 0 2018
answered Nov 23 '18 at 5:26
Srce Cde
1,136511
edited Nov 23 '18 at 6:02
answered Nov 23 '18 at 5:26
Srce Cde
1,136511
answered Nov 23 '18 at 5:26
Srce Cde
1,136511
answered Nov 23 '18 at 5:26
Srce Cde
1,136511
1,136511
i have not considered using the query before, new thing to learn! thank you. by the way, when i try to show the count that equal to 0. why it shows nothing. is it a sensitive case?
– Hook Im
Nov 23 '18 at 5:30
@HookIm Updated the alternative method. Also forcount=0
– Srce Cde
Nov 23 '18 at 5:56
add a comment |
i have not considered using the query before, new thing to learn! thank you. by the way, when i try to show the count that equal to 0. why it shows nothing. is it a sensitive case?
– Hook Im
Nov 23 '18 at 5:30
@HookIm Updated the alternative method. Also forcount=0
– Srce Cde
Nov 23 '18 at 5:56
i have not considered using the query before, new thing to learn! thank you. by the way, when i try to show the count that equal to 0. why it shows nothing. is it a sensitive case?
– Hook Im
Nov 23 '18 at 5:30
i have not considered using the query before, new thing to learn! thank you. by the way, when i try to show the count that equal to 0. why it shows nothing. is it a sensitive case?
– Hook Im
Nov 23 '18 at 5:30
@HookIm Updated the alternative method. Also for
count=0– Srce Cde
Nov 23 '18 at 5:56
@HookIm Updated the alternative method. Also for
count=0– Srce Cde
Nov 23 '18 at 5:56
add a comment |
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2
That's a dataframe not a list...
– U9-Forward
Nov 23 '18 at 5:14
Try
list_all = list_all[list_all['count'] != 0]– Sociopath
Nov 23 '18 at 5:16
im blur, sorry. i have updated the post
– Hook Im
Nov 23 '18 at 5:17