How to change the first value in the tuple of a list?

This is my matrix:

b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 

0.023)]]

And I want to let it to be a

b = [[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 

0.023)]]

To add n for the first value in the tuple, and I tried:

for n in b:

    for ee,ww in n:

        ee == ee + 2903

It doesn't work.
How should I keep the change to the original matrix b?

edited Nov 19 '18 at 16:29

jdehesa

23.8k43554

asked Nov 19 '18 at 16:27

賴韋安

142

2

Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.

– chepner
Nov 19 '18 at 16:29

1

I don't understand where the 2903 comes from, and b is three levels deep while your for loop is only two. Also you're using the comparison == rather than the assignment =.

– Acccumulation
Nov 19 '18 at 16:34

If I gave you a function update, would it have to update b in place, so that print(b); update(b); print(b) would output two different values, or can it return a copy new_b = update(b). In the second case, do you need to retain the original object? That is, after new_b = update(b), would you require id(new_b) == id(b)? Would you require that id(new_b[0]) == id(b[0]) as well? (Keeping the existing rows, and only replacing the tuples within each row?)

– chepner
Nov 19 '18 at 16:38

ee== .. is not an assignment. And ee=.. changes the variable but does not change n.

– hpaulj
Nov 19 '18 at 16:47

add a comment |

This is my matrix:

b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 

0.023)]]

And I want to let it to be a

b = [[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 

0.023)]]

To add n for the first value in the tuple, and I tried:

for n in b:

    for ee,ww in n:

        ee == ee + 2903

It doesn't work.
How should I keep the change to the original matrix b?

edited Nov 19 '18 at 16:29

jdehesa

23.8k43554

asked Nov 19 '18 at 16:27

賴韋安

142

2

Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.

– chepner
Nov 19 '18 at 16:29

1

I don't understand where the 2903 comes from, and b is three levels deep while your for loop is only two. Also you're using the comparison == rather than the assignment =.

– Acccumulation
Nov 19 '18 at 16:34

If I gave you a function update, would it have to update b in place, so that print(b); update(b); print(b) would output two different values, or can it return a copy new_b = update(b). In the second case, do you need to retain the original object? That is, after new_b = update(b), would you require id(new_b) == id(b)? Would you require that id(new_b[0]) == id(b[0]) as well? (Keeping the existing rows, and only replacing the tuples within each row?)

– chepner
Nov 19 '18 at 16:38

ee== .. is not an assignment. And ee=.. changes the variable but does not change n.

– hpaulj
Nov 19 '18 at 16:47

add a comment |

This is my matrix:

b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 

0.023)]]

And I want to let it to be a

b = [[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 

0.023)]]

To add n for the first value in the tuple, and I tried:

for n in b:

    for ee,ww in n:

        ee == ee + 2903

It doesn't work.
How should I keep the change to the original matrix b?

edited Nov 19 '18 at 16:29

jdehesa

23.8k43554

asked Nov 19 '18 at 16:27

賴韋安

142

This is my matrix:

b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 

0.023)]]

And I want to let it to be a

b = [[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 

0.023)]]

To add n for the first value in the tuple, and I tried:

for n in b:

    for ee,ww in n:

        ee == ee + 2903

It doesn't work.
How should I keep the change to the original matrix b?

python numpy replace tuples

edited Nov 19 '18 at 16:29

jdehesa

23.8k43554

asked Nov 19 '18 at 16:27

賴韋安

142

edited Nov 19 '18 at 16:29

jdehesa

23.8k43554

asked Nov 19 '18 at 16:27

賴韋安

142

edited Nov 19 '18 at 16:29

jdehesa

23.8k43554

edited Nov 19 '18 at 16:29

jdehesa

23.8k43554

edited Nov 19 '18 at 16:29

jdehesa

23.8k43554

asked Nov 19 '18 at 16:27

賴韋安

142

asked Nov 19 '18 at 16:27

賴韋安

142

asked Nov 19 '18 at 16:27

賴韋安

142

2

Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.

– chepner
Nov 19 '18 at 16:29

1

I don't understand where the 2903 comes from, and b is three levels deep while your for loop is only two. Also you're using the comparison == rather than the assignment =.

– Acccumulation
Nov 19 '18 at 16:34

If I gave you a function update, would it have to update b in place, so that print(b); update(b); print(b) would output two different values, or can it return a copy new_b = update(b). In the second case, do you need to retain the original object? That is, after new_b = update(b), would you require id(new_b) == id(b)? Would you require that id(new_b[0]) == id(b[0]) as well? (Keeping the existing rows, and only replacing the tuples within each row?)

– chepner
Nov 19 '18 at 16:38

ee== .. is not an assignment. And ee=.. changes the variable but does not change n.

– hpaulj
Nov 19 '18 at 16:47

add a comment |

2

Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.

– chepner
Nov 19 '18 at 16:29

1

I don't understand where the 2903 comes from, and b is three levels deep while your for loop is only two. Also you're using the comparison == rather than the assignment =.

– Acccumulation
Nov 19 '18 at 16:34

If I gave you a function update, would it have to update b in place, so that print(b); update(b); print(b) would output two different values, or can it return a copy new_b = update(b). In the second case, do you need to retain the original object? That is, after new_b = update(b), would you require id(new_b) == id(b)? Would you require that id(new_b[0]) == id(b[0]) as well? (Keeping the existing rows, and only replacing the tuples within each row?)

– chepner
Nov 19 '18 at 16:38

ee== .. is not an assignment. And ee=.. changes the variable but does not change n.

– hpaulj
Nov 19 '18 at 16:47

Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.

– chepner
Nov 19 '18 at 16:29

I don't understand where the 2903 comes from, and b is three levels deep while your for loop is only two. Also you're using the comparison == rather than the assignment =.

– Acccumulation
Nov 19 '18 at 16:34

If I gave you a function update, would it have to update b in place, so that print(b); update(b); print(b) would output two different values, or can it return a copy new_b = update(b). In the second case, do you need to retain the original object? That is, after new_b = update(b), would you require id(new_b) == id(b)? Would you require that id(new_b[0]) == id(b[0]) as well? (Keeping the existing rows, and only replacing the tuples within each row?)

– chepner
Nov 19 '18 at 16:38

ee== .. is not an assignment. And ee=.. changes the variable but does not change n.

– hpaulj
Nov 19 '18 at 16:47

add a comment |

2 Answers
2

active

oldest

votes

Tuples are immutable. You can use a list comprehension instead:

res = [[(i+5, j) for i, j in tup] for tup in b]



[[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 0.023)]]

answered Nov 19 '18 at 16:29

jpp

100k2161111

3

This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.

– chepner
Nov 19 '18 at 16:32

@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.

– jpp
Nov 19 '18 at 16:33

1

I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takes b as an argument and updates it in-place.

– chepner
Nov 19 '18 at 16:35

add a comment |

You cannot modify tuples, they are immutable in Python. You can however replace the existing tuples with other tuples.

b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 0.023)]]

for n in b:

    for i, (ee, ww) in enumerate(n):

        n[i] = (ee + 2903, ww)

print(b)

Output:

[[(2904, 0.044), (2905, 0.042)], [(2907, 0.18), (2909, 0.023)], [(2907, 0.03), (2908, 0.023)]]

answered Nov 19 '18 at 16:31

jdehesa

23.8k43554

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Tuples are immutable. You can use a list comprehension instead:

res = [[(i+5, j) for i, j in tup] for tup in b]



[[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 0.023)]]

answered Nov 19 '18 at 16:29

jpp

100k2161111

3

This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.

– chepner
Nov 19 '18 at 16:32

@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.

– jpp
Nov 19 '18 at 16:33

1

I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takes b as an argument and updates it in-place.

– chepner
Nov 19 '18 at 16:35

add a comment |

Tuples are immutable. You can use a list comprehension instead:

res = [[(i+5, j) for i, j in tup] for tup in b]



[[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 0.023)]]

answered Nov 19 '18 at 16:29

jpp

100k2161111

3

This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.

– chepner
Nov 19 '18 at 16:32

@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.

– jpp
Nov 19 '18 at 16:33

1

I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takes b as an argument and updates it in-place.

– chepner
Nov 19 '18 at 16:35

add a comment |

Tuples are immutable. You can use a list comprehension instead:

res = [[(i+5, j) for i, j in tup] for tup in b]



[[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 0.023)]]

answered Nov 19 '18 at 16:29

jpp

100k2161111

Tuples are immutable. You can use a list comprehension instead:

res = [[(i+5, j) for i, j in tup] for tup in b]



[[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 0.023)]]

answered Nov 19 '18 at 16:29

jpp

100k2161111

answered Nov 19 '18 at 16:29

jpp

100k2161111

answered Nov 19 '18 at 16:29

jpp

100k2161111

answered Nov 19 '18 at 16:29

jpp

100k2161111

3

This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.

– chepner
Nov 19 '18 at 16:32

@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.

– jpp
Nov 19 '18 at 16:33

1

I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takes b as an argument and updates it in-place.

– chepner
Nov 19 '18 at 16:35

add a comment |

3

This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.

– chepner
Nov 19 '18 at 16:32

@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.

– jpp
Nov 19 '18 at 16:33

1

I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takes b as an argument and updates it in-place.

– chepner
Nov 19 '18 at 16:35

This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.

– chepner
Nov 19 '18 at 16:32

@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.

– jpp
Nov 19 '18 at 16:33

I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takes b as an argument and updates it in-place.

– chepner
Nov 19 '18 at 16:35

add a comment |

You cannot modify tuples, they are immutable in Python. You can however replace the existing tuples with other tuples.

b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 0.023)]]

for n in b:

    for i, (ee, ww) in enumerate(n):

        n[i] = (ee + 2903, ww)

print(b)

Output:

[[(2904, 0.044), (2905, 0.042)], [(2907, 0.18), (2909, 0.023)], [(2907, 0.03), (2908, 0.023)]]

answered Nov 19 '18 at 16:31

jdehesa

23.8k43554

add a comment |

You cannot modify tuples, they are immutable in Python. You can however replace the existing tuples with other tuples.

b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 0.023)]]

for n in b:

    for i, (ee, ww) in enumerate(n):

        n[i] = (ee + 2903, ww)

print(b)

Output:

[[(2904, 0.044), (2905, 0.042)], [(2907, 0.18), (2909, 0.023)], [(2907, 0.03), (2908, 0.023)]]

answered Nov 19 '18 at 16:31

jdehesa

23.8k43554

add a comment |

You cannot modify tuples, they are immutable in Python. You can however replace the existing tuples with other tuples.

b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 0.023)]]

for n in b:

    for i, (ee, ww) in enumerate(n):

        n[i] = (ee + 2903, ww)

print(b)

Output:

[[(2904, 0.044), (2905, 0.042)], [(2907, 0.18), (2909, 0.023)], [(2907, 0.03), (2908, 0.023)]]

answered Nov 19 '18 at 16:31

jdehesa

23.8k43554

You cannot modify tuples, they are immutable in Python. You can however replace the existing tuples with other tuples.

b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 0.023)]]

for n in b:

    for i, (ee, ww) in enumerate(n):

        n[i] = (ee + 2903, ww)

print(b)

Output:

[[(2904, 0.044), (2905, 0.042)], [(2907, 0.18), (2909, 0.023)], [(2907, 0.03), (2908, 0.023)]]

answered Nov 19 '18 at 16:31

jdehesa

23.8k43554

answered Nov 19 '18 at 16:31

jdehesa

23.8k43554

answered Nov 19 '18 at 16:31

jdehesa

23.8k43554

answered Nov 19 '18 at 16:31

jdehesa

23.8k43554

add a comment |

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