Python numpy exponential notation

Question

np.log(someMatrix) generates values of the type 1.220536239336080619e+01. I'd like actually see all the digits intead of +e01. Is there a way to tell numpy to do so please?

I tried around() but doesn't work for me.

Why do you say you're not seeing "all the digits"? If you've got huge numbers, don't forget that floating-point numbers have limited precision, so the ones place on an e+20 float is essentially noise. — user2357112
– user2357112, Commented Mar 5, 2016 at 8:53
A float double has a max 52 bits of precision, you can check how many you have with something like len(bin(<int>)) if you print it via. one of the answers suggested below. For example the number in your question would take 63 bits to represent exactly. — user3467349
– user3467349, Commented Mar 5, 2016 at 9:19

manu190466 · Accepted Answer · 2016-03-05 09:15:08Z

1

You can see as many digits of a float as you want with a formatted print. But I agree with user2357112, the last digits may be out of precision limits and are not significant.

r=1.220536239336080619e+01
print "%-.24f"%(r)

12.205362393360806194664292

answered Mar 5, 2016 at 9:15

manu190466

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Daniel · Accepted Answer · 2016-03-05 09:16:32Z

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when you write your data to file, you can choose between different formattings, %f is the one, that don't use exponential notation:

np.savetxt('matrix.txt', np.log(someMatrix), '%f')

answered Mar 5, 2016 at 9:16

Daniel

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MSeifert · Accepted Answer · 2016-03-05 12:23:49Z

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You can use np.array2str with a formatter that displays it as number without decimals:

import numpy as np
a = np.array([1434e24])
np.array2string(a, formatter={'float_kind':lambda x: "%.0f" % x})
# prints '[1433999999999999897558319104]'

notice however that the number is dominated by the floating point precision and therefore will not be exactly what you typed in.

answered Mar 5, 2016 at 12:23

MSeifert

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