make lower half of a n*n list zero without using any functions in python - python-3.x
I tried to solve it by using 2 for loops and an if statement . But i was unable to get the desired output.
INPUT-
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
thislist=[1]*10
thislist=[thislist]*10
print(thislist)
for i in range(10):
for j in range(10):
print(thislist[i][j], end = " ")
print()
print()
for i in range(10):
for j in range(10):
if i>j:
thislist[i][j]=0
for i in range(10):
for j in range(10):
print(thislist[i][j], end = " ")
print()
This was the output i got:
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 1
but when i made a list using the below method i got the desired output.
thislist=[[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1],
[1,1,1,1,1,1,1,1,1,1]]
print(thislist)
for i in range(10):
for j in range(10):
if i>j:
thislist[i][j]=0
for i in range(10):
for j in range(10):
print(thislist[i][j], end = " ")
print()
note-This is the desired OUTPUT-
1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1
0 0 1 1 1 1 1 1 1 1
0 0 0 1 1 1 1 1 1 1
0 0 0 0 1 1 1 1 1 1
0 0 0 0 0 1 1 1 1 1
0 0 0 0 0 0 1 1 1 1
0 0 0 0 0 0 0 1 1 1
0 0 0 0 0 0 0 0 1 1
0 0 0 0 0 0 0 0 0 1
Can someone explain whats the difference between the above 2 codes?
As you pointed out, the problem comes from the manner you created your list of list. In your first example, you do something like this:
list1 = [1]*10
list_of_list1=[list1]*10
list_of_list1 is actually a list of shallow copies of the original list1. Then if you modify a value in list_of_list1, the modification will occurs in all the rows of list_of_list1.
The opposit of a shallow copy is a deep copy. You might want to search more info on the Internet about this topic
In the mean time, you can simply try this.
thislist = []
for row in range(10):
list1 = [1]*10
thislist.append(list1)
But I usually go with numpy when it is available.
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Check with idxmcx and for loop with shift
s=df.set_index('empl_ID')
idx=s.columns.get_indexer(s.idxmax(1))
l=[(s.iloc[t, x :y].eq(s.iloc[t, x :y].shift())&s.iloc[t, x :y].eq(0)).any() for t , x ,y in zip(df.index,idx,idx+5)]
df['Label']=l
df
empl_ID day_1 day_2 day_3 day_4 ... day_7 day_8 day_9 day_10 Label
0 1 1 1 1 1 ... 0 1 1 1 False
1 2 0 0 1 1 ... 1 1 1 0 False
2 3 0 1 0 0 ... 1 1 1 1 True
3 4 1 0 1 0 ... 1 0 1 0 False
4 5 1 0 0 1 ... 1 1 1 1 True
5 6 0 0 0 0 ... 1 1 1 1 False
[6 rows x 12 columns]
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Running the four commands results in below output, from a dataframe called cancer.
$ print("\n target")
$ print(cancer.target)
$ print("\n target_names")
$ print(cancer.target_names)
target
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1 0 1 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 1 1 1 1 0 0 0 1 1
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0 1 0 0 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 0 1 1 0 0 1 1 1 1 1 1 0 1 1 1 1 1 1
1 0 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 0 1 0 1 1 1 1 1 0 1 1
0 1 0 1 1 0 1 0 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1
1 1 1 1 1 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 0 1 0 1 1 1 1 1 0 1 1 0 1 0 1 0 0
1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
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d={0:'malignant',1:'benign'}
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print(target)
['malignant', 'malignant', 'malignant', 'malignant', 'malignant', 'benign', 'benign', 'benign', 'malignant', 'benign', 'malignant', 'malignant', 'benign']
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This builds off a solution provided by #Jezrael : link
df['Output_col'] = df.dot(df.columns.str.cat(['_']*len(df.columns),sep='')).str.strip('_')
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2 0 0 1 0 C
3 1 0 1 0 A_C
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How to find which combinations have the most frequency?
I have an SPSS data set with 500+ respondents and 18 symptoms that they could have. Each symptom has its own variable Symptom01 = 1 means they have the symptom 1 Symptom02 = 0 means they dont have the symptom 2 etc etc What I want to know is what combination of 3 symptoms is more frequent in my data set. For example how many people have symptom 1, 5 and 6; how many people have symptom 1, 2 and 3, etc. I doesn't mean that they only have those symptoms. Theey could have others. I just want to know which group of 3 symptoms is more frequent in my dataset. It's a lot of combinations so how would you do this? Can someone help me?
Please note the macro below uses the variable names Symptom1, Symptom2 etc' instead of "Symptom01", "Symptom02"...
First creating some sample data to work on:
data list list/Symptom1 to Symptom18.
begin data
1 0 0 1 1 0 0 1 1 0 0 0 0 0 1 1 1 1
1 1 1 1 0 1 0 0 1 0 1 1 0 0 1 1 0 0
0 1 1 0 1 1 1 1 1 1 1 0 1 0 0 1 0 0
1 0 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0
1 0 1 0 0 1 0 1 0 0 0 0 0 1 0 1 0 0
0 0 0 1 1 1 0 0 0 0 1 0 0 0 1 0 0 1
1 0 1 1 1 1 1 0 1 1 0 0 0 1 1 1 0 1
1 0 0 0 1 0 1 1 0 0 0 1 0 1 0 0 1 0
0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 1 1 1
1 0 0 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0
end data.
Now defining a macro with three loops:
EDIT - this version accounts for repeating combinations of symptoms
define AllCombsOf3 ()
!do !vr1=1 !to 18
!do !vr2=!vr1 !to 18
!do !vr3=!vr2 !to 18
!if (!vr2<>!vr1 !and !vr2<>!vr3) !then
compute !concat("C_",!vr1,"_",!vr2,"_",!vr3)= !concat("Symptom",!vr1)=1 & !concat("Symptom",!vr2)=1 & !concat("Symptom",!vr3)=1 .
!ifend
!doend
!doend
!doend
!enddefine.
Running the macro and displaying wanted results:
AllCombsOf3.
means C_1_2_3 to C_16_17_18.
EDIT 2 - new macro for a four symptom version
define AllCombsOf4 ()
!do !vr1=1 !to 18
!do !vr2=!vr1 !to 18
!do !vr3=!vr2 !to 18
!do !vr4=!vr3 !to 18
!if (!vr2<>!vr1 !and !vr2<>!vr3 !and !vr3<>!vr4) !then
compute !concat("C_",!vr1,"_",!vr2,"_",!vr3,"_",!vr4)=
!concat("Symptom",!vr1)=1 & !concat("Symptom",!vr2)=1 &
!concat("Symptom",!vr3)=1 & !concat("Symptom",!vr4)=1 .
!ifend
!doend !doend !doend !doend
!enddefine.
AllCombsOf4.
means C_1_2_3_4 to C_15_16_17_18.