我来自Python背景，使用Python的Decimal模块进行任意精度计算的重要一点是知道何时执行，而不必指定Decimal类型的数字，例如，概率计算。
在Julia计算中，您经常需要为数字指定任意精度类型吗？
我通过实例学习得最好，我想要解决的例子如下。
计算随机抽样时至少一次成功的概率，而无需更换一桶26 ABC至Z块26！，3（26）！和5（26）！倍。
鉴于：
一次试验成功=以正确的顺序随机抽出块：A，B，C，到Z.
随机试验的数量是n = 26！或3n或5n。
一次随机试验成功的概率，p = 1 / n
一次随机试验失败的概率是f = 1-p。
计算（使用类型规范尽可能少）：
使用CL = 1-f ^ n的n次试验中至少一次成功的置信水平CL
使用CL = 1 - f ^（3n）进行3n次试验的CL
使用CL = 1 - f ^（5n）进行5n次试验的CL
Julia在通过计算流动“大”数字类型规范方面表现很好。例如，在下面的代码中，我只将一个数字指定为大“26”，其他一切都是自动的。
我刚刚开始使用Julia，但多年来一直在以各种方式进行任意精度计算。朱莉娅用这种我曾经拥有过的东西提供最亲切的体验。
# Set precision to 150 bits which should be adequate precision.
setprecision(150)
# Note that we only have to specify a "big" number here.
n = factorial(big"26")
println("n = factorial(big\"26\") = ", n)
println("Note that we never have to use \"big\" again in the following code.")
println("typeof(n) = ", typeof(n), "\n")
# p is the probability of success on 1 trial.
p = 1/n
println("p = 1/n = ", p)
# Note we did not have to specify the type of p.
println("typeof(p) = ", typeof(p), "\n")
# f is the probability of failure on 1 trial.
f = 1 - p
println("f = 1 - p = ", f)
println("typeof(f) = ", typeof(f), "\n")
# CL is the probability of at least 1 success in n trials.
# CL stands for confidence level.
CL = 1 - f^n
println("The 63% CL for n trials = 1 - f^n = ", CL)
println("typeof(CL) = ", typeof(CL), "\n")
# Here is the 95% conf. level using 3n random trials.
CL95 = 1 - f^(3n)
println("The 95% CL for 3n trials = ", CL95)
println("typeof(CL95) = ", typeof(CL95), "\n")
# Here is the 99% conf. level using 5n random trials.
CL99 = 1 - f^(5n)
println("The 99% CL for 5n trials = ", CL99)
println("typeof(CL99) = ", typeof(CL99), "\n")
""" ============================= Output ==============================
n = factorial(big"26") = 403291461126605635584000000
Note that we never have to use "big" again in the following code.
typeof(n) = BigInt
p = 1/n = 2.4795962632247974600749435458479566174226555415e-27
typeof(p) = BigFloat
f = 1 - p = 9.9999999999999999999999999752040373677520254001e-01
typeof(f) = BigFloat
The 63% CL for n trials = 1 - f^n = 6.3212055882855767839219205578358958187929158048e-01
typeof(CL) = BigFloat
The 95% CL for 3n trials = 9.5021293163213605701567013782477488392169554992e-01
typeof(CL95) = BigFloat
The 99% CL for 5n trials = 9.9326205300091453290223898909666750856240017783e-01
typeof(CL99) = BigFloat
"""
如果你想要像Python中的Decimal一样的精确十进制浮点数，我自己写了一个。您可以将小数精度设置为任何正整数值。如果您需要该模块的副本，请与我联系。
julia> factorial(BigInt(26))
403291461126605635584000000
julia> println("How many decimal digits do we need?")
How many decimal digits do we need?
julia> length(string( factorial(BigInt(26)) ))
27
julia> using PDFPs
julia> PDFP_setDefaultPrecision(40)
40
julia> n = PDFP( factorial(BigInt(26)) )
PDFP(0, 26, [4, 0, 3, 2, 9, 1, 4, 6, 1, 1, 2, 6, 6, 0, 5, 6, 3, 5, 5, 8, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])
julia> disp(a::PDFP) = println(PDFP_toShortCommonString(a))
disp (generic function with 1 method)
julia> p = 1/n
PDFP(0, -27, [2, 4, 7, 9, 5, 9, 6, 2, 6, 3, 2, 2, 4, 7, 9, 7, 4, 6, 0, 0, 7, 4, 9, 4, 3, 5, 4, 5, 8, 4, 7, 9, 5, 6, 6, 1, 7, 4, 2, 2])
julia> disp(p)
2.47960E-27
julia> println("Let's call failure q as per convention in probability questions")
Let's call failure q as per normal in probability questions
julia> q = 1 - p
PDFP(0, -1, [9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 7, 5, 2, 0, 4, 0, 3, 7, 3, 6, 7, 7, 5, 2])
julia> disp(q)
1.00000
julia> PDFP_toFortranString(q)
"9.999999999999999999999999975204037367752E-1"
julia> at_least_one_success_in_n_trials = 1 - q^n
PDFP(0, -1, [6, 3, 2, 1, 2, 0, 5, 5, 8, 8, 2, 8, 5, 5, 8, 0, 5, 5, 6, 0, 2, 0, 5, 7, 9, 5, 3, 4, 5, 6, 9, 0, 2, 1, 9, 8, 7, 6, 4, 9])
julia> disp(at_least_one_success_in_n_trials)
0.632121
julia> at_least_one_success_in_3n_trials = 1 - q^(3*n)
PDFP(0, -1, [9, 5, 0, 2, 1, 2, 9, 3, 1, 6, 3, 2, 1, 3, 6, 2, 1, 0, 1, 6, 5, 3, 1, 1, 5, 3, 8, 4, 6, 6, 4, 3, 9, 0, 4, 8, 2, 9, 1, 0])
julia> disp(at_least_one_success_in_3n_trials)
0.950213
julia> at_least_one_success_in_5n_trials = 1 - q^(5*n)
PDFP(0, -1, [9, 9, 3, 2, 6, 2, 0, 5, 3, 0, 0, 0, 9, 1, 4, 5, 6, 7, 4, 4, 6, 4, 3, 7, 4, 3, 4, 3, 4, 4, 7, 4, 8, 6, 8, 9, 6, 2, 8, 1])
julia> disp(at_least_one_success_in_5n_trials)
0.993262